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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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24-game | Short python solution | short-python-solution-by-lhengi-88st | Saw somebody is creating a whole num class and using operator overload to avoid double comparison error, but I think doing espilon comparison is good enough..\n | lhengi | NORMAL | 2019-02-11T19:27:49.915209+00:00 | 2019-02-11T19:27:49.915255+00:00 | 189 | false | Saw somebody is creating a whole num class and using operator overload to avoid double comparison error, but I think doing espilon comparison is good enough..\n"""\nclass Solution:\n def judgePoint24(self, nums: \'List[int]\') -> \'bool\':\n return self.helper(nums)\n \n def helper(self,lst):\n i... | 2 | 0 | [] | 0 |
24-game | Backtracking Simple Python Solution - Without Estimation | backtracking-simple-python-solution-with-6g9v | The logic behind this solution is that every number is stored as (numerator, denominator). \n From the array of length n, 2 numbers are picked\n\t All 4 operati | dewank | NORMAL | 2018-11-25T19:48:43.848131+00:00 | 2018-11-25T19:48:43.848214+00:00 | 410 | false | The logic behind this solution is that every number is stored as `(numerator, denominator)`. \n* From the array of length `n`, 2 numbers are picked\n\t* All 4 operations are done on the picked numbers\n\t* Answer is stored back in the array: array now becomes of length `n-1`\n* Repeated till array length is 1 -> Then t... | 2 | 0 | [] | 1 |
24-game | [Java][DFS] | javadfs-by-neruppu_da-r25x | Given: (a, b, c, d) - (A tuple of 4)\nGenerate: \nEdited: Thanks to @bellevue \n ((a+b),c,d) ((a-b),c,d) ((b-a),c,d) ((ab),c,d) ((a/b),c,d) ((b/a),c, | neruppu_da | NORMAL | 2017-09-17T06:03:32.798000+00:00 | 2017-09-17T06:03:32.798000+00:00 | 970 | false | **Given**: (a, b, c, d) - (A tuple of 4)\n**Generate**: \nEdited: Thanks to @bellevue \n* ((a+b),c,d) ((a-b),c,d) ((b-a),c,d) ((a*b),c,d) ((a/b),c,d) ((b/a),c,d)\n* ((a+c),b,d) ................................................................. ((c/a),b,d)\n* ((a+d),b,c) ....................................... | 2 | 1 | [] | 1 |
24-game | Javascript recursive solution | javascript-recursive-solution-by-lccpc10-zmrm | \nvar judgePoint24 = function(cards) {\n if(cards.length===1) {\n var roundValue = Math.round(cards[0]*100)/100;\n return roundValue === 24;\n }\n\n fo | lccpc10 | NORMAL | 2017-11-25T16:31:30.257000+00:00 | 2017-11-25T16:31:30.257000+00:00 | 618 | false | ```\nvar judgePoint24 = function(cards) {\n if(cards.length===1) {\n var roundValue = Math.round(cards[0]*100)/100;\n return roundValue === 24;\n }\n\n for(var i=0; i<cards.length; i++) {\n for(var j=0; j<cards.length; j++) {\n if(i===j) continue;\n var small = Math.min(i,j);\n var large = Ma... | 2 | 0 | [] | 1 |
24-game | Simple & easy solution🔥🔥 | simple-easy-solution-by-aayu_t-8uxe | \n\n# Code\npython3 []\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n n = len(cards)\n if n == 1:\n if abs(2 | aayu_t | NORMAL | 2024-09-05T12:27:53.458754+00:00 | 2024-09-05T12:27:53.458788+00:00 | 58 | false | \n\n# Code\n```python3 []\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n n = len(cards)\n if n == 1:\n if abs(24 - cards[0]) < 0.001:\n return True\n return False\n \n for i in range(n):\n for j in range(i + 1, n):\n ... | 1 | 0 | ['Array', 'Python3'] | 0 |
24-game | Backtracking | backtracking-by-rpsingh21-1xtx | Complexity\n- Time complexity: O(n!)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(n^2)\n Add your space complexity here, e.g. O(n) \n\n# | rpsingh21 | NORMAL | 2024-05-20T18:46:12.161310+00:00 | 2024-05-20T18:46:12.161333+00:00 | 87 | false | # Complexity\n- Time complexity: $$O(n!)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n^2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n \n def generat_results(a, b):\... | 1 | 0 | ['Backtracking', 'Python3'] | 0 |
24-game | Permutation + Backtracking | permutation-backtracking-by-coderpriest-npuf | Intuition\nTo solve this problem, we need to check if it\'s possible to combine the given numbers in such a way that they result in 24. We can achieve this by t | coderpriest | NORMAL | 2024-03-20T07:56:27.130481+00:00 | 2024-03-20T07:56:27.130509+00:00 | 390 | false | # Intuition\nTo solve this problem, we need to check if it\'s possible to combine the given numbers in such a way that they result in 24. We can achieve this by trying out all possible permutations of the numbers and applying mathematical operations to them. If any permutation yields 24, we return true; otherwise, we r... | 1 | 0 | ['Array', 'Math', 'Backtracking', 'C++'] | 1 |
24-game | Simple go solution | simple-go-solution-by-singhsaubhik-m17u | Code\n\nfunc abs(n float64) float64 { if n < 0 { return -n }; return n }\n\nfunc operationsResult(a, b float64) []float64 {\n res := []float64{a-b, a+b, b-a, | singhsaubhik | NORMAL | 2024-03-09T17:12:02.623227+00:00 | 2024-03-09T17:12:02.623254+00:00 | 13 | false | # Code\n```\nfunc abs(n float64) float64 { if n < 0 { return -n }; return n }\n\nfunc operationsResult(a, b float64) []float64 {\n res := []float64{a-b, a+b, b-a, a*b}\n if a > 0 {\n\t\tres = append(res, b / a)\n\t}\n\n\tif b > 0 {\n\t\tres = append(res, a / b)\n\t}\n\n return res\n}\n\nfunc helper(cards []flo... | 1 | 0 | ['Go'] | 0 |
24-game | DFS: traversing all permutations and all operations | dfs-traversing-all-permutations-and-all-io4i2 | \n# Code\n\nclass Solution {\npublic:\n vector<float> ops(float a, float b)\n {\n return {a+b,a-b,b-a,a*b,a/b,b/a};\n }\n bool dfs(vector<int | flyljx | NORMAL | 2023-06-05T04:45:46.099282+00:00 | 2023-06-05T04:45:46.099322+00:00 | 159 | false | \n# Code\n```\nclass Solution {\npublic:\n vector<float> ops(float a, float b)\n {\n return {a+b,a-b,b-a,a*b,a/b,b/a};\n }\n bool dfs(vector<int> nums, int k, float res)\n {\n if (k==3)\n {\n for (float f:ops(res,nums[3]))\n if (fabs(f-24.0)<0.001) return tr... | 1 | 0 | ['C++'] | 0 |
24-game | Python || Solver can optionally print *all* solutions | python-solver-can-optionally-print-all-s-ip5s | Intuition\nPick any two numbers apply any operation, and replace the two numbers with their result. We now have a smaller problem.\n\n# Code\n\nclass Solution:\ | stomachion | NORMAL | 2023-06-02T17:52:33.841927+00:00 | 2023-08-23T05:32:50.005108+00:00 | 928 | false | # Intuition\nPick any two numbers apply any operation, and replace the two numbers with their result. We now have a smaller problem.\n\n# Code\n```\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n def compute(a,b,op):\n if op==0:\n return a+b\n if op=... | 1 | 0 | ['Python3'] | 1 |
24-game | A lot of solutions,A lot of people are copying | a-lot-of-solutionsa-lot-of-people-are-co-ug6c | I didn\'t have an idea at first\n Describe your first thoughts on how to solve this problem. \n\n# Use a backtracking method\n Describe your approach to solving | appleamzon | NORMAL | 2022-11-22T07:06:51.300290+00:00 | 2022-11-22T07:06:51.300339+00:00 | 200 | false | # I didn\'t have an idea at first\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Use a backtracking method\n<!-- Describe your approach to solving the problem. -->\n\n# Time:O(1) Space:O(n3)\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-... | 1 | 0 | ['Math', 'Java'] | 0 |
24-game | Go. Backtracking | go-backtracking-by-sobik-q90b | \nvar epsilon = math.Pow10(-6)\n\nfunc judgePoint24(cards []int) bool {\n\n\tgenerateAllPossibleResults := func(a, b float64) []float64 {\n\t\tres := [] float64 | sobik | NORMAL | 2022-08-14T11:02:21.048267+00:00 | 2022-08-14T11:02:21.048308+00:00 | 365 | false | ```\nvar epsilon = math.Pow10(-6)\n\nfunc judgePoint24(cards []int) bool {\n\n\tgenerateAllPossibleResults := func(a, b float64) []float64 {\n\t\tres := [] float64 {a+b, a*b, a-b, b-a}\n\t if b != 0 {res = append(res, a/b) }\n\t if a != 0 {res = append(res, b/a) }\n\t\treturn res\n\t}\n\n\tisValid := func(a []flo... | 1 | 0 | ['Backtracking', 'Go'] | 0 |
24-game | C++ + INTUTION + EASY | c-intution-easy-by-mistake1-bn61 | we take any two number and do operations( consider all pairs ) \n 1.a ->oprarion are a+b, a-b, b-a, a*b , a/b (if b!=0 ) and b/a if (b!=0).\n2. take | mistake1 | NORMAL | 2022-07-25T04:24:31.390904+00:00 | 2022-07-25T04:24:31.390937+00:00 | 200 | false | 1. we take any two number and do operations( consider all pairs ) \n 1.a ->oprarion are a+b, a-b, b-a, a*b , a/b (if b!=0 ) and b/a if (b!=0).\n2. take any operation and rest of the array element and do cheak for possible\n3. at the end size of the array will be 1 and if elenment is 24 return true else ret... | 1 | 0 | [] | 0 |
24-game | C++ Backtracking solution | c-backtracking-solution-by-uttu_dce-rva7 | \tif we have the array\n [a, b, c, d]\n\n and if we chose two numbers lets say we chose a and b, then all the possible operations are\n a + b (1)\n | uttu_dce | NORMAL | 2022-06-29T06:24:55.416754+00:00 | 2022-06-29T06:25:04.652933+00:00 | 216 | false | \tif we have the array\n [a, b, c, d]\n\n and if we chose two numbers lets say we chose a and b, then all the possible operations are\n a + b (1)\n a - b (2)\n b + a (3)\n b - a (4)\n a*b (5)\n b*a (6)\n a/b (7)\n b/a (8)\n\n but 1 and 3 are same and [5, 6] are same\n so we count the... | 1 | 0 | [] | 0 |
24-game | Easy Understanding Python DFS | easy-understanding-python-dfs-by-rtlc202-n052 | \n\n def dfs(cards):\n if len(cards) == 1:\n return math.isclose(cards[0], 24) # do NOT use \'==\'\n \n # | rtlc2020 | NORMAL | 2022-05-10T01:49:32.468282+00:00 | 2022-05-10T01:49:32.468311+00:00 | 277 | false | \n```\n def dfs(cards):\n if len(cards) == 1:\n return math.isclose(cards[0], 24) # do NOT use \'==\'\n \n # allow to start with any card\n for k in range(len(cards)):\n cards[k], cards[0] = cards[0], cards[k]\n \n ... | 1 | 0 | ['Depth-First Search'] | 1 |
24-game | well commented easy to understand solution | well-commented-easy-to-understand-soluti-967o | \nclass Solution {\npublic:\n \n bool isPossible(vector<float>& cards) {\n int n = cards.size();\n // If we are left with only 1 cards then | vishalrathi02 | NORMAL | 2022-03-21T02:02:41.269666+00:00 | 2022-03-21T02:02:41.269694+00:00 | 225 | false | ```\nclass Solution {\npublic:\n \n bool isPossible(vector<float>& cards) {\n int n = cards.size();\n // If we are left with only 1 cards then it should be 24 else return false\n if (n == 1) {\n // max number can be 9*9*9*9 i.e 10 power 4 around, so we can float till 4 digits\n ... | 1 | 0 | [] | 0 |
24-game | Python one line solution | python-one-line-solution-by-vanya_shkvir-jdko | Thanks to @karrenbel for his three loop solution\nhttps://leetcode.com/problems/24-game/discuss/1399166/Python3-clean-code\n\n\nfrom fractions import Fraction\n | vanya_shkvir | NORMAL | 2022-02-02T21:40:47.362657+00:00 | 2022-02-02T21:40:47.362685+00:00 | 730 | false | Thanks to @karrenbel for his three loop solution\nhttps://leetcode.com/problems/24-game/discuss/1399166/Python3-clean-code\n\n```\nfrom fractions import Fraction\n\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool: \n return any([any([True if f1(d, f2(c, f3(a, b))) == 24 or f1(f2(a, b), f... | 1 | 0 | ['Python'] | 1 |
24-game | recursion and permutation Python code | recursion-and-permutation-python-code-by-8gzw | ```\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n def helper(nums, left, right):\n if left == right:\n | WaydeZ | NORMAL | 2022-01-03T15:57:40.879764+00:00 | 2022-01-03T15:59:09.769777+00:00 | 218 | false | ```\nclass Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n def helper(nums, left, right):\n if left == right:\n return set([nums[left]])\n res = set()\n for i in range(left, right):\n res1 = helper(nums, left, i)\n r... | 1 | 0 | [] | 1 |
24-game | Python Backtracking with Explaination | python-backtracking-with-explaination-by-a1wm | python\nfrom itertools import permutations\n\nclass Solution:\n def __init__(self):\n self.eps = 0.001\n \n def judgePoint24(self, cards: List[i | taranthemonk | NORMAL | 2021-12-21T17:21:36.725475+00:00 | 2021-12-21T17:21:36.725518+00:00 | 267 | false | ```python\nfrom itertools import permutations\n\nclass Solution:\n def __init__(self):\n self.eps = 0.001\n \n def judgePoint24(self, cards: List[int]) -> bool:\n \n # solve the nums to get 24\n def solve(nums: int) -> bool:\n # no operation\n if len(nums) == 1... | 1 | 0 | [] | 0 |
24-game | (C++) 679. 24 Game | c-679-24-game-by-qeetcode-6cr1 | \n\nclass Solution {\npublic:\n bool judgePoint24(vector<int>& cards) {\n \n function<bool(vector<double>&)> fn = [&] (vector<double>& nums) {\ | qeetcode | NORMAL | 2021-12-02T23:08:16.887620+00:00 | 2021-12-02T23:08:16.887648+00:00 | 201 | false | \n```\nclass Solution {\npublic:\n bool judgePoint24(vector<int>& cards) {\n \n function<bool(vector<double>&)> fn = [&] (vector<double>& nums) {\n if (nums.size() == 1) return abs(nums[0] - 24) < 1e-6; \n for (int i = 0; i < nums.size(); ++i) {\n for (int j = i+1; ... | 1 | 0 | ['C'] | 0 |
24-game | Cannot get this case to work [3,3,8,8] what combination would this be? | cannot-get-this-case-to-work-3388-what-c-g3t9 | class Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n \n #basically create a stack of operations and indices used and based \n | ariboi27 | NORMAL | 2021-11-04T03:48:50.286852+00:00 | 2021-11-04T03:50:27.819423+00:00 | 178 | false | class Solution:\n def judgePoint24(self, cards: List[int]) -> bool:\n \n #basically create a stack of operations and indices used and based \n self.s = []\n \n #initial condition\n for i,card in enumerate(cards):\n self.s.append(([i],card))\n \n ... | 1 | 0 | [] | 4 |
24-game | Shortest and fastest solution in C++ | shortest-and-fastest-solution-in-c-by-ru-196g | Runtime 0ms, beats 100% of C++ submissions.\n\n\nclass Solution {\n bool eq24(double i) {\n return (i > 23.9999 && i < 24.0001);\n }\n double op | rufeng2000 | NORMAL | 2021-10-18T07:31:20.650737+00:00 | 2021-10-18T07:33:54.526646+00:00 | 420 | false | Runtime 0ms, beats 100% of C++ submissions.\n\n```\nclass Solution {\n bool eq24(double i) {\n return (i > 23.9999 && i < 24.0001);\n }\n double op(double a, double b, int op) {\n if (op == 0) return a + b;\n if (op == 1) return a - b;\n if (op == 2) return a * b;\n return a ... | 1 | 0 | ['C'] | 3 |
24-game | [Python] 2 clean recursive solutions O(N^N) | python-2-clean-recursive-solutions-onn-b-71ww | Solution 1: Use itertools to get all the combinations. Then we recursively execute each possible operator in the list until we have one element leftover.\n\n\nd | quantum73 | NORMAL | 2021-10-09T17:15:56.747467+00:00 | 2021-10-09T17:19:03.373840+00:00 | 407 | false | __Solution 1__: Use itertools to get all the combinations. Then we recursively execute each possible operator in the list until we have one element leftover.\n\n```\ndef judgePoint24(self, cards: List[int]) -> bool:\n def dfs(pos):\n if len(pos) == 1 and round(pos[0], 1) == 24:\n return True\n ... | 1 | 0 | ['Backtracking', 'Recursion', 'Combinatorics', 'Python'] | 0 |
24-game | C++ | Brute Force | c-brute-force-by-sonusharmahbllhb-l798 | \nclass Solution {\npublic:\n bool ans=0;\n void gun(vector<float> v,vector<char> ch){\n if(ans) return;\n if(v.size()==1){\n if( | sonusharmahbllhb | NORMAL | 2021-08-30T14:58:06.218624+00:00 | 2021-08-30T15:02:16.961599+00:00 | 545 | false | ```\nclass Solution {\npublic:\n bool ans=0;\n void gun(vector<float> v,vector<char> ch){\n if(ans) return;\n if(v.size()==1){\n if(abs(24-v[0])<0.1) ans=1;\n return;\n }\n for(int i=0;i<v.size()-1;i++){\n if(ans) return;\n vector<float> tmpv... | 1 | 1 | ['C', 'C++'] | 0 |
24-game | Simple DFS java solution without rounding issue | simple-dfs-java-solution-without-roundin-co6e | \n\nclass Solution {\n public boolean judgePoint24(int[] cards) {\n List<Number> nums = List.of(\n new Number(cards[0]),\n new N | fredking | NORMAL | 2021-08-18T05:49:55.513533+00:00 | 2021-08-18T05:55:35.973085+00:00 | 590 | false | \n```\nclass Solution {\n public boolean judgePoint24(int[] cards) {\n List<Number> nums = List.of(\n new Number(cards[0]),\n new Number(cards[1]),\n new Number(cards[2]),\n new Number(cards[3]));\n return dfs(nums);\n }\n \n private boolean dfs(List... | 1 | 0 | [] | 0 |
24-game | Python Easy 95% Runtime Memoization with Comments | python-easy-95-runtime-memoization-with-pbjll | \nimport functools\n\nclass Solution:\n # Helper function to get the values for all possible cominations of num1 and num2 using different mathematical operat | mahaksarin21 | NORMAL | 2021-07-31T11:46:49.898328+00:00 | 2021-07-31T11:53:07.028181+00:00 | 335 | false | ```\nimport functools\n\nclass Solution:\n # Helper function to get the values for all possible cominations of num1 and num2 using different mathematical operators\n def getRes(self, num1, num2):\n res = [num1+num2, num1-num2, num2-num1, num1*num2]\n if num1 != 0:\n res.append(num2/num1)\... | 1 | 0 | [] | 1 |
24-game | C++ iteration based | c-iteration-based-by-zehziur-3lv9 | Iterate every permutation and expression tree forms.\n\n\nclass Solution {\npublic:\n double eval(double a, double b, int op) {\n switch (op) {\n | zehziur | NORMAL | 2021-07-15T11:14:44.240383+00:00 | 2021-07-15T11:14:44.240431+00:00 | 219 | false | Iterate every permutation and expression tree forms.\n\n```\nclass Solution {\npublic:\n double eval(double a, double b, int op) {\n switch (op) {\n case 0: return a + b;\n case 1: return a - b;\n case 2: return a * b;\n case 3: return a / b;\n }\n ret... | 1 | 0 | [] | 0 |
24-game | Python solution easy-understanding | python-solution-easy-understanding-by-fl-5r1l | class Solution:\n\n # calculte the different results of two numbers\n def two_number(a, b):\n s = set()\n s.add(a+b)\n s.add(a-b)\n | flyingspa | NORMAL | 2021-06-16T10:57:24.857486+00:00 | 2021-06-16T10:57:24.857529+00:00 | 409 | false | class Solution:\n\n # calculte the different results of two numbers\n def two_number(a, b):\n s = set()\n s.add(a+b)\n s.add(a-b)\n s.add(b-a)\n s.add(a*b)\n if b!=0:\n ans = a/b\n if abs(ans-round(ans))<10**(-3):\n s.add(round(ans))\n... | 1 | 1 | ['Python'] | 0 |
24-game | Python - Backtracking on card permutations | python-backtracking-on-card-permutations-cgai | if we have N operands, we have N - 1 points where that would be the "last" operator\nSo for start st and end en, we iterate through all mid points and recursive | mahsaelyasi | NORMAL | 2021-06-02T01:59:09.398972+00:00 | 2021-06-02T02:00:23.405199+00:00 | 226 | false | if we have N operands, we have N - 1 points where that would be the "last" operator\nSo for start `st` and end `en`, we iterate through all `mid` points and recursively evaluate on `[st, mid]` and `[mid + 1, en]`. We apply all four operators. We add sub solutions to sets and return and expand the solution sets. 24 shou... | 1 | 0 | [] | 1 |
count-sorted-vowel-strings | [Java/C++/Python] DP, O(1) Time & Space | javacpython-dp-o1-time-space-by-lee215-k5w0 | Explanation\ndp[n][k] means the number of strings constructed by at most k different characters.\n\nIf k = 1, use only u\nIf k = 2, use only o,u\nIf k = 3, use | lee215 | NORMAL | 2020-11-01T04:08:11.249862+00:00 | 2020-11-02T03:05:28.093641+00:00 | 41,521 | false | # **Explanation**\n`dp[n][k]` means the number of strings constructed by at most `k` different characters.\n\nIf `k = 1`, use only `u`\nIf `k = 2`, use only `o,u`\nIf `k = 3`, use only `i,o,u`\nIf `k = 4`, use only `e,i,o,u`\nIf `k = 5`, use only `a,e,i,o,u`\n<br>\n\n\n# Solution 1: Top Down\nTime `O(nk)`\nSpace `O(nk)... | 419 | 15 | [] | 41 |
count-sorted-vowel-strings | Java DP O(n) time Easy to understand | java-dp-on-time-easy-to-understand-by-fr-ente | time: O(n)\nspace: O(1)\njust think about the mathematical induction\nif we know all the string of len k, so the string of len k + 1 will be\n1 add a before all | FrankLu321 | NORMAL | 2020-11-01T04:09:17.365625+00:00 | 2020-11-01T04:22:58.465348+00:00 | 11,328 | false | time: O(n)\nspace: O(1)\njust think about the mathematical induction\nif we know all the string of len k, so the string of len k + 1 will be\n1 add a before all string of len k\n2 add e before the string of len k, which is starts with or after e\n3 add i before the string of len k, which starts with or after i\n4 add... | 353 | 10 | [] | 41 |
count-sorted-vowel-strings | ✅🔥DP & 5 Variables Approach - C++/Java/Python | dp-5-variables-approach-cjavapython-by-d-tkx1 | Intuition\n\n- Here we can observe a pattern to this problem.\n\n\n\t\t a e i o u\n n=1 1 1 1 1 1 /a, e, i, o, u\n n=2 5 4 3 2 1 /a-> a | dhruba-datta | NORMAL | 2022-05-11T02:59:40.146515+00:00 | 2023-12-03T22:24:44.675893+00:00 | 16,181 | false | # Intuition\n\n- Here we can observe a pattern to this problem.\n\n```\n\t\t a e i o u\n n=1 1 1 1 1 1 /a, e, i, o, u\n n=2 5 4 3 2 1 /a-> aa,ae,ai,ao,au | e-> ee,ei,eo,eu | i-> ii,io,iu | o-> oo,ou | u-> uu\n n=3 15 10 6 3 1\n```\n\n- If we observe from last there will be only 1 element... | 217 | 2 | ['Dynamic Programming', 'Python', 'C++', 'Java', 'Python3'] | 14 |
count-sorted-vowel-strings | One-line solution | Math | No DP, no Big Integers | O(1) time, space | one-line-solution-math-no-dp-no-big-inte-plgi | Let me start with the answer itself, you just need to return the following:\n\n\n\nFor example, in Ruby:\n\ndef count_vowel_strings(n)\n (n + 1) * (n + 2) * (n | ykurmangaliyev | NORMAL | 2021-01-17T13:02:29.867847+00:00 | 2021-01-17T13:09:04.134852+00:00 | 8,954 | false | Let me start with the answer itself, you just need to return the following:\n\n\n\nFor example, in Ruby:\n```\ndef count_vowel_strings(n)\n (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24\nend\n```\n\nSee the explai... | 198 | 4 | ['Math', 'Combinatorics'] | 16 |
count-sorted-vowel-strings | Vowels count - simple math - faster | vowels-count-simple-math-faster-by-suren-0nis | Hi,\n\nWhen we do some paper work for this we could see that vowel \'a\' will have all the values including \'a\', where as vowel \'e\' will have all the letter | Surendaar | NORMAL | 2022-05-11T05:28:54.739238+00:00 | 2022-05-11T06:47:28.822434+00:00 | 5,812 | false | Hi,\n\nWhen we do some paper work for this we could see that vowel \'a\' will have all the values including \'a\', where as vowel \'e\' will have all the letter except \'a\'.. as it is already covered in \'a\'.\n\na -> aa, ae, ai, ao, au\ne -> ee, ei, eo, eu\ni -> ii, io, iu\n0 -> oo, ou\nu -> uu\n\nBecause if we add e... | 117 | 1 | ['Java'] | 13 |
count-sorted-vowel-strings | Dynamic Programming - Python [100%] [Explanation + code] | dynamic-programming-python-100-explanati-izpp | In this problem, the pattern I observed was prepending \'a\' to all the strings of length n-1 does not affect any order. Similarly, \'e\' can be prepended to st | mihirrane | NORMAL | 2020-11-01T06:08:56.406936+00:00 | 2021-01-17T22:20:12.898143+00:00 | 7,589 | false | In this problem, the pattern I observed was prepending \'a\' to all the strings of length n-1 does not affect any order. Similarly, \'e\' can be prepended to strings starting with \'e\' and greater vowels and so on.\n\nSo we have our subproblem.\n\n**How do we fill the DP Table?**\nLets, take an example of n = 3\n\n![i... | 106 | 2 | ['Dynamic Programming', 'Python3'] | 12 |
count-sorted-vowel-strings | Very Easy Solution | No DP | No Math Formula | Solving for n = 3 reveals the pattern | very-easy-solution-no-dp-no-math-formula-yjmf | \nThe problem seems like it will have a typical recursive solution,\nbut in reality it has a very trivial solution.\n\nImportant part of this problem is to rego | interviewrecipes | NORMAL | 2020-11-01T04:01:20.630165+00:00 | 2020-11-01T04:29:52.331516+00:00 | 8,191 | false | \nThe problem seems like it will have a typical recursive solution,\nbut in reality it has a very trivial solution.\n\nImportant part of this problem is to regonize the pattern.\n\nFor `n = 3`, let\'s see how many strings can be formed.\n\nIf start of the string is "aa", the third letter can be any of\n`{a, e, i, o, u}... | 76 | 4 | [] | 14 |
count-sorted-vowel-strings | Intuitive C++ Solution (With explaination) | intuitive-c-solution-with-explaination-b-mlp2 | The solution uses DP but in a way that I found much more intuitive. We start the solution from the right i.e. generate suffixes and keep prepending it to either | yashjain | NORMAL | 2020-11-01T04:39:12.456477+00:00 | 2021-08-31T10:08:33.548537+00:00 | 3,544 | false | The solution uses DP but in a way that I found much more intuitive. We start the solution from the right i.e. generate suffixes and keep prepending it to either a, e, i, o, or u.\na = number of strings that start with a\ne = number of strings that start with e\ni = number of strings that start with i\no = number of str... | 73 | 0 | ['Dynamic Programming', 'C', 'C++'] | 7 |
count-sorted-vowel-strings | Count Sorted Vowel Strings | O(1) 1-Line Mathematical Solution w/ Explanation | count-sorted-vowel-strings-o1-1-line-mat-2sqq | Idea:\n\nThe mathematical solution is derived from Pascal\'s Triangle.\n\nThe formula for a triangular number is:\n\n x * (x + 1)\n\t -------------\n\t | sgallivan | NORMAL | 2021-01-17T11:47:17.505221+00:00 | 2021-01-17T12:21:34.532006+00:00 | 2,231 | false | ***Idea:***\n\nThe mathematical solution is derived from Pascal\'s Triangle.\n\nThe formula for a triangular number is:\n```\n x * (x + 1)\n\t -------------\n\t 1 * 2\n```\n...and the formula for a tetrahedral number is:\n```\n x * (x + 1) * (x + 2)\n\t -----------------------\n\t 1 * 2 * 3\n... | 58 | 3 | [] | 8 |
count-sorted-vowel-strings | [Python] simple DP solution - O(N) / O(1) | python-simple-dp-solution-on-o1-by-alanl-brnm | Python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n \n a = e = i = o = u = 1\n for _ in range(n - 1):\n a | alanlzl | NORMAL | 2020-11-01T04:01:34.889411+00:00 | 2020-11-01T04:01:34.889461+00:00 | 1,670 | false | ```Python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n \n a = e = i = o = u = 1\n for _ in range(n - 1):\n a, e, i, o, u = a+e+i+o+u, e+i+o+u, i+o+u, o+u, u\n \n return a+e+i+o+u\n``` | 55 | 11 | [] | 6 |
count-sorted-vowel-strings | ✅ Python 4 approaches (DP, Maths) | python-4-approaches-dp-maths-by-constant-n35m | DP Tabulation\n\nclass Solution:\n def countVowelStrings(self, n: int) -> int: \n dp = [[0] * 6 for _ in range(n+1)]\n for i in range(1, | constantine786 | NORMAL | 2022-05-11T00:47:21.808058+00:00 | 2022-05-12T00:58:43.761781+00:00 | 5,492 | false | 1. ##### DP Tabulation\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int: \n dp = [[0] * 6 for _ in range(n+1)]\n for i in range(1, 6):\n dp[1][i] = i\n \n for i in range(2, n+1):\n dp[i][1]=1\n for j in range(2, 6):\n ... | 52 | 1 | ['Python', 'Python3'] | 6 |
count-sorted-vowel-strings | Beautiful C++ solution (100% faster) | beautiful-c-solution-100-faster-by-pig14-s6a2 | (Updated)\n\nThe variable a, e, i, o, or u means the number of strings ended with an a, e, I, o, or u with a given length.\n\nIn each iteration, we expend the l | pig14726 | NORMAL | 2020-11-25T16:25:36.995497+00:00 | 2020-12-15T01:49:17.678458+00:00 | 3,288 | false | (Updated)\n\nThe variable a, e, i, o, or u means the number of strings ended with an a, e, I, o, or u with a given length.\n\nIn each iteration, we expend the length by 1. \u2018a\u2019 can only come after string ended with \u2018a\u2019, so the number of strings ended with an \'a\' remains unchanged. \u2018e\u2019 can... | 48 | 1 | ['C'] | 6 |
count-sorted-vowel-strings | DP | C++ | Recursion->Memoization->Tabulation | Well commented | dp-c-recursion-memoization-tabulation-we-jdb1 | Recursive Solution [ Time-Complexity = O(n^2), Space-Complexity = O(n) ]\n\t\n \n\tint count(int i, int n, string &vowels){\n\n\t\t\tif(n==0) return 1; // if a | _walterwhite | NORMAL | 2022-05-11T04:23:34.888945+00:00 | 2022-05-11T04:32:02.711319+00:00 | 3,285 | false | **Recursive Solution** **[ Time-Complexity = O(n^2), Space-Complexity = O(n) ]**\n\t\n ```\n\tint count(int i, int n, string &vowels){\n\n\t\t\tif(n==0) return 1; // if a string of length n is formed\n\n\t\t\tif(i>=5) return 0; // if i exceeds the length of vowels\n\n\t\t\tint pick, notPick;\n\t\t\t// pick i.e. pick t... | 45 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 3 |
count-sorted-vowel-strings | [Python] Start and bars, explained | python-start-and-bars-explained-by-dbabi-ndqj | Actually, this is very classical combinatorical problem, which can be solved, using so-called stars and bars method: https://en.wikipedia.org/wiki/Stars_and_bar | dbabichev | NORMAL | 2021-01-17T09:02:19.924795+00:00 | 2022-05-11T07:18:04.765601+00:00 | 7,342 | false | Actually, this is very classical combinatorical problem, which can be solved, using so-called **stars and bars** method: https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics). Imagine, that we have string `aa..aee..eii..ioo..ouu..u`, where we have some numbers of `a, e, i, o` and `u`. Then we can look at this pr... | 44 | 3 | [] | 4 |
count-sorted-vowel-strings | [Java/Python] Top Down DP - Clean & Concise | javapython-top-down-dp-clean-concise-by-758uk | Python 3\npython\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n @lru_cache(None)\n def dp(n, i):\n if n == 0: retu | hiepit | NORMAL | 2020-11-01T04:04:17.517363+00:00 | 2020-11-01T04:23:28.760786+00:00 | 3,555 | false | **Python 3**\n```python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n @lru_cache(None)\n def dp(n, i):\n if n == 0: return 1 # Found a valid answer\n if i == 5: return 0 # Reach to length of vowels [a, e, i, o, u]\n ans = dp(n, i + 1) # Skip vowel... | 43 | 2 | [] | 3 |
count-sorted-vowel-strings | Python. One liner solution. O(1) time & space. | python-one-liner-solution-o1-time-space-4jhmt | Some combinatorics:\nIt is known that the characters should appear in ascending lexicographic order.\nIt is known that there are a total of 5 characters.\nConcl | m-d-f | NORMAL | 2021-01-17T10:14:15.381379+00:00 | 2021-01-17T15:42:31.873856+00:00 | 2,394 | false | Some combinatorics:\nIt is known that the characters should appear in ascending lexicographic order.\nIt is known that there are a total of 5 characters.\nConclusion: There are only 4 places where the character is replaced by the one followed by the sorted order.\nWe have n + 4 locations in total.\nHence, the answer is... | 36 | 7 | ['Python', 'Python3'] | 8 |
count-sorted-vowel-strings | 3 different solutions (backtracking, dynamic programming, math) | 3-different-solutions-backtracking-dynam-noou | \n\nBacktracking solution:\nTime complexity: O(n^5)\nWe try all the possibilities where adding the new vowel doesn\'t break the lexicographic order. For example | syph | NORMAL | 2021-05-06T03:29:20.045364+00:00 | 2021-05-06T20:15:17.408950+00:00 | 2,360 | false | \n\n**Backtracking solution:**\n**Time complexity:** `O(n^5)`\nWe try all the possibilities where adding the new vowel doesn\'t break the lexicographic order. For example, if the last character of our actual possibility is \'e\', we can\'t add an \'a\' after it.\n```\ndef count(n, last=\'\'):\n if n == 0:\n r... | 31 | 0 | ['Math', 'Dynamic Programming', 'Backtracking', 'Python', 'Python3'] | 4 |
count-sorted-vowel-strings | C++ Backtracking vs Dynamic Programming beats 100.0% | c-backtracking-vs-dynamic-programming-be-vq9o | First I tried backtracking and find a solution but runtime is ~1200 ms and beats 5.0% only:\n\n\nclass Solution {\npublic:\n vector<char> vowels = {\'a\', \' | yilmaz | NORMAL | 2020-11-10T19:32:35.057390+00:00 | 2020-11-10T19:51:08.036312+00:00 | 2,895 | false | First I tried backtracking and find a solution but runtime is ~1200 ms and beats 5.0% only:\n\n```\nclass Solution {\npublic:\n vector<char> vowels = {\'a\', \'e\', \'i\', \'o\', \'u\'};\n int countVowelStrings(int n) {\n int count = 0;\n countVowelStringsHelper(n, \' \', count);\n return cou... | 29 | 3 | ['Dynamic Programming', 'Backtracking', 'C', 'C++'] | 6 |
count-sorted-vowel-strings | [C++] 4 lines DP solution | c-4-lines-dp-solution-by-rudy-zj6c | When n = 2, the only valid string starting with u is uu, valid strings starting with o are [oo, ou], and so on. The numbers (we use a vector cnt to store them) | rudy__ | NORMAL | 2020-11-01T04:41:21.902740+00:00 | 2020-11-04T06:19:30.905687+00:00 | 2,652 | false | When `n = 2`, the only valid string starting with `u` is `uu`, valid strings starting with `o` are `[oo, ou]`, and so on. The numbers (we use a vector `cnt` to store them) of valid strings starting with `a, e, i, o, u` are respectively `[5,4,3,2,1]`.
When `n = 3`, the only valid string starting with `u` is `uuu`, val... | 28 | 1 | [] | 0 |
count-sorted-vowel-strings | Java : 0 ms Faster than 100%, Easy to understand with explanation | java-0-ms-faster-than-100-easy-to-unders-nlxs | Time Complexity - O(n)\nSpace Complexity - O(1)\n\nOn carefully observing the sequence getting formed you will notice a pattern here:\n\nFor n = 1 : a , e , i, | anand004 | NORMAL | 2020-11-05T13:11:21.135839+00:00 | 2020-11-05T13:14:16.051475+00:00 | 1,321 | false | Time Complexity - O(n)\nSpace Complexity - O(1)\n\nOn carefully observing the sequence getting formed you will notice a pattern here:\n\nFor n = 1 : a , e , i, o, u \nString starting with a = 1, e = 1, i = 1, o = 1, u = 1 : Total = 5\n\nFor n = 2 : aa, ae, ai, ao, au, ee, ei, eo, eu, ii, io, iu, oo, ou, uu \nString s... | 26 | 0 | ['Java'] | 4 |
count-sorted-vowel-strings | C++ | Faster than 100% | DP and MATH Approach | c-faster-than-100-dp-and-math-approach-b-hc14 | The problem can be solved with dynamic programming approach or from a direct math formula. Here I have solved using DP approach.\n dp[vowels[index] - 97] at eve | persistentBeast | NORMAL | 2022-05-11T05:05:38.916267+00:00 | 2022-05-16T07:42:24.682769+00:00 | 3,084 | false | * The problem can be solved with **dynamic programming approach** or from a **direct math formula**. Here I have solved using DP approach.\n* `dp[vowels[index] - 97]` at every value of `i` stores the number of lexicographically sorted strings which start from `vowels[index]` and is of length `i`.\n* `vowels[index]` ca... | 24 | 1 | ['C'] | 4 |
count-sorted-vowel-strings | C++ Top Down DP and Formula | c-top-down-dp-and-formula-by-votrubac-v0xp | Top Down DP\nTop down DP was the easiest for me to get started. \n\ncpp\nint dp[51][5] = {};\nint countVowelStrings(int n, int prev = 0) {\n if (n == 1)\n | votrubac | NORMAL | 2020-11-01T22:08:31.698677+00:00 | 2020-11-01T23:07:07.702734+00:00 | 1,612 | false | #### Top Down DP\nTop down DP was the easiest for me to get started. \n\n```cpp\nint dp[51][5] = {};\nint countVowelStrings(int n, int prev = 0) {\n if (n == 1)\n return 5 - prev;\n if (dp[n][prev] == 0)\n for (int i = prev; i < 5; ++i)\n dp[n][prev] += countVowelStrings(n - 1, i);\n r... | 24 | 1 | [] | 0 |
count-sorted-vowel-strings | Java O(n) | Visual Explanation | java-on-visual-explanation-by-raunaq21-s7we | The solution is pretty straightforward once a pattern is visible.\n\nFor n = 2\na => (a, e,i,o,u)\ne => (e,i,o,u)\ni => (i, o, u)\no => (o, u)\nu => (u) \n\n\nT | raunaq21 | NORMAL | 2022-05-11T11:56:04.853510+00:00 | 2022-07-28T17:59:37.290965+00:00 | 1,551 | false | The solution is pretty straightforward once a pattern is visible.\n```\nFor n = 2\na => (a, e,i,o,u)\ne => (e,i,o,u)\ni => (i, o, u)\no => (o, u)\nu => (u) \n```\n```\nThe resultant strings are:\naa, ae, ai, ao, au\nee, ei, eo, eu\t\t\t\t\nii, io, iu\t\t\t\t\t\t\noo, ou\t\t\t\t\t\nuu\n```\n```\nFor n = 3\naa => (a,e,i,... | 19 | 0 | ['Java'] | 3 |
count-sorted-vowel-strings | Image Explanation: Distributing N Balls into 5 Boxes, Some Boxes May Be Empty | image-explanation-distributing-n-balls-i-4hrh | Leetcode 1641. Count Sorted Vowel Strings.\n\n# Math \n\nEvery vowel string is start by $a, e, i, o, u$, and there are $N$ strings. \n\nHow to distribute $N$ st | longluo | NORMAL | 2022-05-11T01:32:44.396196+00:00 | 2022-11-16T07:38:52.237591+00:00 | 1,620 | false | [Leetcode](https://leetcode.com/) [1641. Count Sorted Vowel Strings](https://leetcode.com/problems/count-sorted-vowel-strings/).\n\n# Math \n\nEvery vowel string is start by $a, e, i, o, u$, and there are $N$ strings. \n\nHow to distribute $N$ strings into the $5$ boxes, some boxes may be empty?\n\nImaging that there a... | 18 | 0 | ['Math', 'Dynamic Programming', 'Combinatorics', 'C++', 'Java'] | 1 |
count-sorted-vowel-strings | C++ code (Explained using permutation concept, faster than 100%) | c-code-explained-using-permutation-conce-tzbd | Code using permutation \nWe have total n places to fill with 5 vowels to fill with.In short we have n identical thigs to be distributed among 5 people for which | sakg | NORMAL | 2021-01-17T11:42:28.113454+00:00 | 2021-01-17T17:12:05.193417+00:00 | 948 | false | Code using permutation \nWe have total n places to fill with 5 vowels to fill with.In short we have n identical thigs to be distributed among 5 people for which the formulae is \n(n+5-1)C(5-1)= (n+4)C4=(n+1)*(n+2)*(n+3)*(n+4)/24\n```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n /// We can ass... | 18 | 1 | [] | 1 |
count-sorted-vowel-strings | C++ two solutions DP and Math | c-two-solutions-dp-and-math-by-midnights-ob1z | Solved LIVE ON stream, link in profile. Math explained by a mathematician (not me) in my video\n\nHere is the DP solution:\n\nclass Solution {\n \n int d | midnightsimon | NORMAL | 2022-05-11T01:43:57.239134+00:00 | 2022-05-11T01:46:08.479019+00:00 | 1,543 | false | **Solved LIVE ON stream, link in profile. Math explained by a mathematician (not me) in my video**\n\nHere is the DP solution:\n```\nclass Solution {\n \n int dfs(int n, int i, vector<vector<int>>& memo) {\n if(n == 0) return 1;\n if(memo[n][i] != -1) return memo[n][i];\n int ret = 0;\n ... | 15 | 0 | [] | 2 |
count-sorted-vowel-strings | C++ / 1 line / mathematical solution / with brief explanations | c-1-line-mathematical-solution-with-brie-40v0 | We have counts of a + e + i + o + u = n\nTherefore, we can use combination with repetition formula to solve it as:\nanswer = H(5, n) = C(5 + n - 1, n) = C(4 + n | liao119 | NORMAL | 2020-11-01T04:04:27.120269+00:00 | 2020-11-01T04:59:00.487879+00:00 | 497 | false | We have counts of a + e + i + o + u = n\nTherefore, we can use combination with repetition formula to solve it as:\nanswer = H(5, n) = C(5 + n - 1, n) = C(4 + n, n) = C(n + 4, 4) = (n+4) * (n+3) * (n+2) * (n+1) / 4!\n\nMore information of combination with repetition formula can be found here:\nhttps://en.wikipedia.org/... | 15 | 12 | [] | 5 |
count-sorted-vowel-strings | Python with Explanation (Dynamic Programming) | python-with-explanation-dynamic-programm-ud0s | Explanation:\nn = 1 -> 2 -> 3-> 4 ...\n____\na =\t 1 -> 5 -> 15 -> 35 ...\ne =\t 1 -> 4 -> 10 -> 20 ...\ni =\t 1 -> 3 -> 6 -> 10 ...\no =\t | prithuls | NORMAL | 2022-05-11T01:34:30.195247+00:00 | 2022-07-30T17:21:40.148986+00:00 | 1,129 | false | Explanation:\nn = 1 -> 2 -> 3-> 4 ...\n____________________________\na =\t 1 -> 5 -> 15 -> 35 ...\ne =\t 1 -> 4 -> 10 -> 20 ...\ni =\t 1 -> 3 -> 6 -> 10 ...\no =\t 1 -> 2 -> 3 -> 4 ...\nu =\t 1 -> 1 -> 1 -> 1 ...\n____________________________\nsum = 5 -> 15 -> 35 -> 70 ... (that\'s the an... | 14 | 0 | ['Dynamic Programming', 'Python'] | 1 |
count-sorted-vowel-strings | Beginner friendly [Java/JavaScript/Python] Solution | beginner-friendly-javajavascriptpython-s-dr2z | Time Complexity : O(n)\nJava\n\nclass Solution {\n public int countVowelStrings(int n) {\n int a=1, e=1, i=1, o=1, u=1;\n while(n-- > 1){\n | HimanshuBhoir | NORMAL | 2022-02-22T04:54:57.690902+00:00 | 2022-08-13T13:09:51.346691+00:00 | 903 | false | **Time Complexity : O(n)**\n**Java**\n```\nclass Solution {\n public int countVowelStrings(int n) {\n int a=1, e=1, i=1, o=1, u=1;\n while(n-- > 1){\n a = a + e + i + o + u;\n e = e + i + o + u;\n i = i + o + u;\n o = o + u;\n u = u;\n }\n ... | 14 | 0 | ['Python', 'Java', 'JavaScript'] | 1 |
count-sorted-vowel-strings | Most Simple solution with O(1) time complexity and 4 line of code in C++ | most-simple-solution-with-o1-time-comple-1z7g | In my solution I used a basic concept of Permuation and combination that we studied in 11th class for JEE exam. It is about counting the number of ways we can d | charan_is_king | NORMAL | 2021-02-15T05:27:38.618872+00:00 | 2021-02-15T05:27:38.618922+00:00 | 338 | false | In my solution I used a basic concept of Permuation and combination that we studied in 11th class for JEE exam. It is about counting the number of ways we can distribute \'n\' identical balls in \'k\' identical boxes. \n\nx1 + x2 + x3 + .......+ xk = n\n\nand the formula for this comes out to be (n+k-1)C(k-1) \n\nSo im... | 13 | 0 | [] | 0 |
count-sorted-vowel-strings | One line, O(1) space/time, no DP/strings | one-line-o1-spacetime-no-dpstrings-by-an-ppx3 | Solution\n\ncsharp\npublic int CountVowelStrings(int n) =>\n (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24;\n\n\nExplanation\n\nThis is a pure math exercise wit | anikit | NORMAL | 2022-05-11T12:55:23.159408+00:00 | 2023-01-03T21:50:52.444645+00:00 | 396 | false | **Solution**\n\n```csharp\npublic int CountVowelStrings(int n) =>\n (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24;\n```\n\n**Explanation**\n\nThis is a pure math exercise with a single variable, so using DP or actually constructing strings is an overkill. Here is the formula to count [combinations with repetitions](htt... | 12 | 0 | ['Math', 'Combinatorics', 'C#'] | 0 |
count-sorted-vowel-strings | C++ || 3 Approaches ||Recursion || DP || Combinations | c-3-approaches-recursion-dp-combinations-vpda | Approach 1 : Recursion\n\n\nclass Solution {\npublic:\n void dfs(int idx, int length, int n, int &count){\n ++length;\n \n if (length == | luvthakral247 | NORMAL | 2021-04-10T07:12:04.765738+00:00 | 2021-04-10T07:12:04.765768+00:00 | 1,211 | false | **Approach 1 : Recursion**\n\n```\nclass Solution {\npublic:\n void dfs(int idx, int length, int n, int &count){\n ++length;\n \n if (length == n){\n ++count;\n return;\n }\n \n for (int i = idx ; i < 5 ; i++){\n dfs(i,length,n,count);\n ... | 12 | 0 | ['Dynamic Programming', 'Recursion', 'C', 'Combinatorics'] | 1 |
count-sorted-vowel-strings | Python 3 Dynamic Programming Explained | python-3-dynamic-programming-explained-b-odne | \nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n """\n Given a parameter n and assuming 5 vowels, this program\n uses d | hsweiner54 | NORMAL | 2021-01-17T17:57:40.863755+00:00 | 2021-01-18T19:56:35.115218+00:00 | 590 | false | ```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n """\n Given a parameter n and assuming 5 vowels, this program\n uses dynamic programming to determine the number of\n possible sorted vowel strings of length n.\n \n The theory behind the dynamic programmin... | 12 | 1 | ['Dynamic Programming', 'Python3'] | 0 |
count-sorted-vowel-strings | [C++] Super Easy Soln, with explaination. O(n) | O(1) | c-super-easy-soln-with-explaination-on-o-irkq | After working on this problem for quite some time, I figured out the following:\n1) The vowel \'a\' always has 5 children\n2) The vowel \'e\' always has 4 child | xxdil | NORMAL | 2020-12-03T18:24:21.730664+00:00 | 2021-01-18T08:14:40.633196+00:00 | 509 | false | After working on this problem for quite some time, I figured out the following:\n1) The vowel \'a\' always has 5 children\n2) The vowel \'e\' always has 4 children\n3) The vowel \'i\' always has 3 children\n4) and so on ...\n\n\n------\nSo if i was to draw a tree it would look like this:\n\n -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i | prajapati21 | NORMAL | 2023-11-19T05:30:43.172754+00:00 | 2023-11-19T05:30:43.172787+00:00 | 592 | false | \n# Code\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i)//j\n j+=1\n return mul\n \n``` | 11 | 0 | ['Python3'] | 1 |
count-sorted-vowel-strings | ✅C++ || EASY DP Solution 🚀 | c-easy-dp-solution-by-chiikuu-mzxu | Code\n\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<vector<int>>dp(n+1,vector<int>(5+1,0));\n for(int i=1;i<=5;i++)dp[1 | CHIIKUU | NORMAL | 2023-03-22T07:35:51.687183+00:00 | 2023-03-22T07:37:31.216748+00:00 | 1,075 | false | # Code\n```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<vector<int>>dp(n+1,vector<int>(5+1,0));\n for(int i=1;i<=5;i++)dp[1][i]=i;\n for(int i=2;i<=n;i++){\n for(int j=1;j<=5;j++){\n dp[i][j]= dp[i-1][j] + dp[i][j-1];\n }\n }\n... | 11 | 0 | ['Dynamic Programming', 'C++'] | 1 |
count-sorted-vowel-strings | ✅ 5 C++ Solutions || 1 Backtracking,1 Recursion(TLE), 1 DP and 2 other Approach || Easy | 5-c-solutions-1-backtracking1-recursiont-h7px | As you scroll down, there is a optimized solution given than the previous one.\n\n ##### Approach-1: Recursion (Shows TLE for n>40 and works for n<=40)\n\nHere | shivam-214 | NORMAL | 2022-05-11T09:28:15.391521+00:00 | 2022-05-11T09:28:15.391558+00:00 | 900 | false | ### **As you scroll down, there is a optimized solution given than the previous one.**\n\n* ##### **Approach-1**: **Recursion** *(Shows **TLE for n>40 and works for n<=40**)*\n\nHere, every next call for a decreased value of n gives us the required result and then we will add vowels at the start of result we get to ge... | 11 | 0 | ['Math', 'Dynamic Programming', 'Backtracking', 'Recursion', 'C', 'Combinatorics'] | 3 |
count-sorted-vowel-strings | DP Solution [ Tabulation ] ✔ | dp-solution-tabulation-by-shivamishra21-kv6u | The solution has dynamic programming approach.\nWe have 5 vowels : [a,e,i,o,u]\nIf n = 1:\n\t\tsolution will have {a,e,i,o,u} as possible valid string\nIf n == | Shivamishra21 | NORMAL | 2022-02-19T17:28:40.994057+00:00 | 2022-02-19T17:28:40.994102+00:00 | 323 | false | The solution has **dynamic programming approach**.\nWe have 5 vowels : [a,e,i,o,u]\n*If n = 1:*\n\t\tsolution will have {a,e,i,o,u} as possible valid string\n*If n == 2*:\nsolution string will have ["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"]\nthis can be explained as:\nfor a we can add ... | 11 | 0 | ['Dynamic Programming'] | 0 |
count-sorted-vowel-strings | C++| DP-Beats100% - 0ms Solution | PrefixSum | c-dp-beats100-0ms-solution-prefixsum-by-ex3jq | IDEA:- \nInitialise dp of size 5(u,o,i,e,a) with value 1.\n\nLogic:\nFor n = 1\nStrings possible "a","e","i","o","u".\n\nNow in case of n = 2 (i.e string length | garvitsharma05 | NORMAL | 2021-05-16T22:37:32.692364+00:00 | 2021-05-16T22:37:32.692402+00:00 | 509 | false | IDEA:- \nInitialise dp of size 5(u,o,i,e,a) with value 1.\n```\nLogic:\nFor n = 1\nStrings possible "a","e","i","o","u".\n\nNow in case of n = 2 (i.e string length == 2);\nFor "a" we have 5 choices to make it in sorted order i.e "a","e","i","o","u".\nFor "e", similarly we have 4 choices "e","i","o","u".\nSimilarly for ... | 11 | 0 | ['Dynamic Programming', 'C', 'Prefix Sum'] | 4 |
count-sorted-vowel-strings | "Python" easy explanation blackboard | python-easy-explanation-blackboard-by-ha-g32y | The code is well explained in the image below\n Simple and easy python3 solution*\n\n\n\n\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n | harish_sahu | NORMAL | 2021-01-17T14:13:22.247633+00:00 | 2021-01-17T14:14:54.144808+00:00 | 1,114 | false | * **The code is well explained in the image below**\n* **Simple and easy python3 solution**\n\n\n\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n dp = [[0 for j in range(5)] for i... | 11 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 3 |
count-sorted-vowel-strings | [JavaScript] Backtracking | javascript-backtracking-by-sma8282-uihg | This is basically the same as the other "generate permutations" problems with an extra constraint. (By the way, I quite literally learned 99% of my backtracking | sma8282 | NORMAL | 2020-11-01T11:23:39.068474+00:00 | 2020-12-04T14:21:19.535088+00:00 | 913 | false | This is basically the same as the other "generate permutations" problems with an extra constraint. (By the way, I quite literally learned 99% of my backtracking and "generate permutations" techniques from watching this video series: https://www.youtube.com/watch?v=RkXl5iYoQn4&ab_channel=happygirlzt)\n\nSo anyways, what... | 11 | 0 | ['JavaScript'] | 1 |
count-sorted-vowel-strings | ✅Multiple C++ Solutions 🚀 | multiple-c-solutions-by-chiikuu-b942 | Code\n\n//Solution #1:\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n int a=1, e=1, i=1, o=1, u=1;\n while(--n){\n o | CHIIKUU | NORMAL | 2023-03-22T07:45:37.957567+00:00 | 2023-03-22T07:45:37.957594+00:00 | 795 | false | # Code\n```\n//Solution #1:\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n int a=1, e=1, i=1, o=1, u=1;\n while(--n){\n o += u;\n i += o;\n e += i;\n a += e;\n }\n return a+e+i+o+u;\n }\n};\n\n//Solution #2\nclass Solution {\np... | 9 | 0 | ['C++'] | 0 |
count-sorted-vowel-strings | Java solution explanation with photos beats %100 of submissions | java-solution-explanation-with-photos-be-09ki | \nclass Solution {\n public int countVowelStrings(int n) {\n int[] arr = new int[5];\n for (int i = 0; i<5; i++)\n arr[i] = 1;\n | CS_OFM | NORMAL | 2021-01-17T16:12:50.576442+00:00 | 2021-01-17T16:12:50.576469+00:00 | 357 | false | ```\nclass Solution {\n public int countVowelStrings(int n) {\n int[] arr = new int[5];\n for (int i = 0; i<5; i++)\n arr[i] = 1;\n for (int k = 0; k<n; k++)\n for (int j = 1; j<5;j++)\n arr[j] = arr[j-1] + arr[j];\n return arr[4]; \n }\n}\n... | 9 | 0 | ['Java'] | 1 |
count-sorted-vowel-strings | C++ with comments. O(N) time O(1) space. Beats 100%. | c-with-comments-on-time-o1-space-beats-1-tbhn | The idea is to count the strings ending on a particular vowel. Here are how these counts go:\n\n\n a e i o u\n\t1 1 1 | 28gevu | NORMAL | 2020-12-17T02:06:00.053402+00:00 | 2020-12-17T02:20:29.857686+00:00 | 378 | false | The idea is to count the strings ending on a particular vowel. Here are how these counts go:\n\n```\n a e i o u\n\t1 1 1 1 1\n\t1 2 3 4 5\n\t1 3 6 10 15\n```\n\nYou get the idea: at each iteration there is a single string e... | 9 | 1 | [] | 1 |
count-sorted-vowel-strings | 0ms||100%||C++ Solution||O(1) Time Complexity | 0ms100c-solutiono1-time-complexity-by-ay-tv7m | \nclass Solution {\npublic:\n int countVowelStrings(int n) {\n return (n+1)*(n+2)*(n+3)*(n+4)/24;\n }\n};\n\n\n\nplease upvote!!! | Ayush_Modi43 | NORMAL | 2022-05-11T12:06:33.688551+00:00 | 2022-05-11T12:29:41.040289+00:00 | 390 | false | ```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n return (n+1)*(n+2)*(n+3)*(n+4)/24;\n }\n};\n```\n\n\nplease upvote!!! | 8 | 1 | [] | 1 |
count-sorted-vowel-strings | My Java Solutions 1) DP 2) Math formula | my-java-solutions-1-dp-2-math-formula-by-laqd | \n\nclass Solution {\n public int countVowelStrings(int n) {\n // create a dp array of size 6\n int [] dp = new int [6];\n // dp[0] must | vrohith | NORMAL | 2020-11-02T09:23:52.673008+00:00 | 2020-11-02T09:23:52.673053+00:00 | 751 | false | ```\n\nclass Solution {\n public int countVowelStrings(int n) {\n // create a dp array of size 6\n int [] dp = new int [6];\n // dp[0] must be 0 and the rest 1\n for (int i=1; i<dp.length; i++)\n dp[i] = 1;\n for (int i=1; i<=n; i++) {\n for (int k=1; k<=5; k+... | 8 | 0 | ['Math', 'Dynamic Programming', 'Java'] | 1 |
count-sorted-vowel-strings | C++| Intuitive Solution | DP and Memoization | c-intuitive-solution-dp-and-memoization-ib0r4 | Logic :\nThis problem is based on simple pick and non-pick concept. At every index we have a choice either to pick a vowel or skip it. So, here we get the idea | gojosaturo512 | NORMAL | 2022-07-13T19:23:54.072839+00:00 | 2022-07-14T14:52:35.567395+00:00 | 403 | false | **Logic :**\nThis problem is based on simple pick and non-pick concept. At every index we have a choice either to pick a vowel or skip it. So, here we get the idea of recursion, but while doing so we encounter overlapping problems and hence we go for the memoization part.\n\n```\nclass Solution {\npublic:\n int solv... | 7 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C', 'C++'] | 0 |
count-sorted-vowel-strings | DP Solution with Explanation | dp-solution-with-explanation-by-sai_pras-jw60 | It is an awesome problem. Lets deep dive in it.\n\nSo, they are asking the number of ways in which we can generate lexographically sorted strings of length n, | sai_prashant | NORMAL | 2022-05-11T15:26:18.804472+00:00 | 2022-05-11T15:32:48.025037+00:00 | 530 | false | It is an awesome problem. Lets deep dive in it.\n\nSo, they are asking the number of ways in which we can generate lexographically sorted strings of length n, which only contains vowels.\n\nNow lets think about it.\nSuppose if `n=1` then the possible strings would be ```"a","e","i","o","u"```.\n\n\nNow lets make an a... | 7 | 0 | ['Java'] | 1 |
count-sorted-vowel-strings | [C++] O(1), 1 Line Code | SIMPLEST SOLUTION | c-o1-1-line-code-simplest-solution-by-an-myji | \nclass Solution {\npublic:\n int countVowelStrings(int n) {\n \n return ((n+4)*(n+3)*(n+2)*(n+1))/24;\n \n }\n};\n\n\nExplanation :\ | anveshreddy18 | NORMAL | 2022-05-11T06:04:36.623569+00:00 | 2022-05-11T06:04:36.623593+00:00 | 145 | false | ```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n \n return ((n+4)*(n+3)*(n+2)*(n+1))/24;\n \n }\n};\n```\n\n**Explanation :**\n\nThink of the problem in this way. We have 5 objects and n slots to fill with these 5 objects (a,e,i,o,u). Objects can be repeated. Since we want on... | 7 | 0 | ['Math', 'Probability and Statistics'] | 2 |
count-sorted-vowel-strings | [C++] Recursive Vs. Math Solutions Compared and Explained, 100% Time, 100% Space | c-recursive-vs-math-solutions-compared-a-ymbl | The core intuition here, came to me by just looking at the second example with n == 2.\n\ncpp\n"aa", "ae", "ai", "ao", "au", // all the combinations of {\'a\', | ajna | NORMAL | 2021-01-17T12:56:54.695810+00:00 | 2021-01-17T12:56:54.695839+00:00 | 521 | false | The core intuition here, came to me by just looking at the second example with `n == 2`.\n\n```cpp\n"aa", "ae", "ai", "ao", "au", // all the combinations of {\'a\', \'e\', \'i\', \'o\', \'u\'} from the level before with \'a\' prepended\n"ee", "ei", "eo", "eu", // all the combinations of {\'e\', \'i\', \'o\', \'u\... | 7 | 2 | ['C', 'Combinatorics', 'Probability and Statistics', 'C++'] | 0 |
count-sorted-vowel-strings | [c++] commented | c-commented-by-vishwasgajawada-00s0 | \nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<int>v(5,1); //[1,1,1,1,1] -> (a,e,i,o,u)\n //for the n=2 ,consider this, | vishwasgajawada | NORMAL | 2021-01-17T11:32:44.677011+00:00 | 2021-01-17T11:32:44.677041+00:00 | 228 | false | ```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<int>v(5,1); //[1,1,1,1,1] -> (a,e,i,o,u)\n //for the n=2 ,consider this, \n // if starting is a, then the suffix can be any of the string in vector v from a\n // i.e aa,ae,ai,ao,au\n //for e, ee,ei,eo,eu and e... | 7 | 0 | [] | 0 |
count-sorted-vowel-strings | one line [C++]beats 100% in time | less memory than 98.67% | one-line-cbeats-100-in-time-less-memory-4huj1 | This probelm has a pattern you need to find,and when you did it would look like something like this:\n\nint countVowelStrings(int n) {\n return ((n+1)*(n | mkhasib1 | NORMAL | 2021-01-17T09:40:36.945542+00:00 | 2021-01-17T09:40:36.945582+00:00 | 331 | false | This probelm has a `pattern` you need to find,and when you did it would look like something like this:\n```\nint countVowelStrings(int n) {\n return ((n+1)*(n+2)*(n+3)*(n+4))/24;\n }\n```\nIf you have any questions drop them below \uD83D\uDC47 and if liked it ,kindly **Upvote** it \u2B06 | 7 | 1 | [] | 4 |
count-sorted-vowel-strings | JAVA 1-D DP beat 100% with explainations | java-1-d-dp-beat-100-with-explainations-5yjc0 | \n public int countVowelStrings(int n) {\n // dp[i][j]: for first i vowels of length j, what is the unique lexicon pattern count\n // divid int | 66dante | NORMAL | 2021-01-17T07:44:18.978564+00:00 | 2021-01-17T07:44:18.978595+00:00 | 562 | false | ```\n public int countVowelStrings(int n) {\n // dp[i][j]: for first i vowels of length j, what is the unique lexicon pattern count\n // divid into two groups, one with only i - vowels dp[i-1][j], the other that must contain at least 1 i-th vowel\n // the second group\'s last letter must be i-th... | 7 | 1 | [] | 0 |
count-sorted-vowel-strings | [C++] Clean O(n) code & Detailed Explanation | c-clean-on-code-detailed-explanation-by-n50m1 | When n = 1\na e i o u\n1 1 1 1 1\n\nWhen n = 2\naa \nae ee\nai ei ii\nao eo io oo\nau eu iu ou uu\n5 4 3 2 1\n\nWhen n = 3\nYou might realise, the # o | bluebirdy | NORMAL | 2020-11-01T05:19:36.303820+00:00 | 2020-11-01T05:27:06.898032+00:00 | 285 | false | **When n = 1**\n`a e i o u`\n`1 1 1 1 1 `\n\n**When n = 2**\n`aa` \n`ae ee`\n`ai ei ii`\n`ao eo io oo`\n`au eu iu ou uu`\n`5 4 3 2 1`\n\n**When n = 3**\nYou might realise, the # of three letter sequences begin as `aaa` and ends as `auu`, is exactly the same as the **sum** of all possible two letter sequences.... | 7 | 0 | [] | 1 |
count-sorted-vowel-strings | JAVA Back Track | java-back-track-by-aoali77-xmf7 | \nclass Solution {\n //String vowels = "aeiou";\n int count = 0;\n public int countVowelStrings(int n) {\n backTrack(n, 0);\n return coun | aoali77 | NORMAL | 2020-11-01T04:06:31.631262+00:00 | 2021-02-09T01:07:17.486536+00:00 | 651 | false | ```\nclass Solution {\n //String vowels = "aeiou";\n int count = 0;\n public int countVowelStrings(int n) {\n backTrack(n, 0);\n return count;\n }\n private void backTrack(int n, int indexToStart){\n if(n == 0){\n count++;\n return;\n }\n for(int i... | 7 | 0 | ['Backtracking', 'Java'] | 2 |
count-sorted-vowel-strings | Python one line (math.comb) - with explanation updated | python-one-line-mathcomb-with-explanatio-vnsb | Idea: \n\nWe can simplify the question to how many "vowel"-ending strings at each n. \n"-e" at n = 2: "ae", "ee"\n\nLet\'s see how many "vowel"-ending strings p | mp0530 | NORMAL | 2020-11-01T04:02:39.524210+00:00 | 2020-11-01T05:24:54.401974+00:00 | 530 | false | **Idea**: \n\nWe can simplify the question to how many "vowel"-ending strings at each n. \n"-e" at n = 2: "ae", "ee"\n\nLet\'s see how many "vowel"-ending strings possible at n:\n\nEx:\n```\nn = 1 2 3 4 5 \n------------------------\n"-a": 1 1 1 1 1 \n"-e": 1 2 3 4 5 \n"-i": 1 3 6 10 15 \n"-o": 1... | 7 | 0 | [] | 1 |
count-sorted-vowel-strings | [C++]-1 LINER-EASY TO UNDERSTAND-BEGINNER-SHORT CONCISE-INTERVIEW-WITH EXPLANATION | c-1-liner-easy-to-understand-beginner-sh-vmwd | \n/*\n\nEXPLANATION: [C++]-1 LINER-EASY TO UNDERSTAND-BEGINNER-SHORT CONCISE-INTERVIEW-WITH EXPLANATION\n\nEXPLANATION:\n\nBasically n==1 - 5 combinations. Each | justcodingandcars | NORMAL | 2020-11-01T04:00:38.460237+00:00 | 2020-11-01T04:05:22.257437+00:00 | 794 | false | ```\n/*\n\nEXPLANATION: [C++]-1 LINER-EASY TO UNDERSTAND-BEGINNER-SHORT CONCISE-INTERVIEW-WITH EXPLANATION\n\nEXPLANATION:\n\nBasically n==1 - 5 combinations. Each ending with a vowel\n\nending with a-5,e-4,i-3,o-2,u-1\nn==2 = 5+4+3+2+1\n\nsimilarly n==3 == 35... 15+ (aa-5 , ae-4, ai-3,ao-2, au-1 and so on), which will... | 7 | 1 | [] | 0 |
count-sorted-vowel-strings | ✅C++ | Backtracking- recursion | Easy to Understand | c-backtracking-recursion-easy-to-underst-y36z | \nclass Solution {\npublic:\n \n void helper(int n, vector<char> &temp,int &count, int idx, vector<char> &vowels)\n {\n if(temp.size()==n)\n | Yash2arma | NORMAL | 2022-05-11T19:21:25.173952+00:00 | 2022-05-11T19:21:25.173998+00:00 | 287 | false | ```\nclass Solution {\npublic:\n \n void helper(int n, vector<char> &temp,int &count, int idx, vector<char> &vowels)\n {\n if(temp.size()==n)\n {\n count++;\n return;\n }\n \n for(int i=idx; i<vowels.size(); i++)\n {\n temp.push_back(vo... | 6 | 0 | ['Backtracking', 'Recursion', 'C', 'C++'] | 0 |
count-sorted-vowel-strings | Easy Approach || Simple to understand | easy-approach-simple-to-understand-by-ka-ep0z | Hope you liked it :)\n\n```\nclass Solution {\npublic:\n void solve(vector &v, int n, int currindex, int &count, int l)\n {\n if(currindex >= n)\n | kajalgupta_0701 | NORMAL | 2022-05-11T05:02:09.975813+00:00 | 2022-05-11T05:03:17.122041+00:00 | 451 | false | **Hope you liked it :)**\n\n```\nclass Solution {\npublic:\n void solve(vector<char> &v, int n, int currindex, int &count, int l)\n {\n if(currindex >= n)\n {\n count++;\n return;\n }\n for(int i=l; i<v.size(); i++)\n {\n solve(v, n, currindex+1,... | 6 | 0 | ['Backtracking', 'C', 'C++'] | 0 |
count-sorted-vowel-strings | Count Sorted Vowel Strings | count-sorted-vowel-strings-by-chandakji2-y187 | i) proceed to build string recursively.\nii) insert next element in the string only if it is greater than the last element present and then decerease the n by 1 | chandakji2204 | NORMAL | 2022-05-11T00:30:29.846932+00:00 | 2022-05-11T00:30:29.846960+00:00 | 651 | false | i) proceed to build string recursively.\nii) insert next element in the string only if it is greater than the last element present and then decerease the n by 1.\niii) finally that will be our required string if n==0 i.e string of size n is form.\n\nremember : in recursion we have to write a base condition , recursion ... | 6 | 1 | ['Backtracking', 'Recursion', 'C'] | 1 |
count-sorted-vowel-strings | C++ | Dynamic Programming | 100% Faster | | c-dynamic-programming-100-faster-by-dhru-fr35 | \nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<int> v(n+1); v[0] = 1;\n for(int j = 0; j < 5; ++j){\n for | dhruv_73 | NORMAL | 2021-09-26T16:32:22.339919+00:00 | 2021-09-26T16:32:22.339957+00:00 | 355 | false | ```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n vector<int> v(n+1); v[0] = 1;\n for(int j = 0; j < 5; ++j){\n for(int i = 1; i <= n; ++i){\n v[i] += v[i-1];\n }\n }\n return v[n];\n }\n};\n```\n\n**Result-**\n```\nRuntime: 0 ms... | 6 | 0 | ['Dynamic Programming', 'C'] | 1 |
count-sorted-vowel-strings | Python DP O(N) time O(1) space w/ comment | python-dp-on-time-o1-space-w-comment-by-jyk32 | py\n\'\'\'\nw: dp\nh: when n = 1, we only have 5 choice,\n wheb n = 2, we have 15 options, how they related\n consider: [a, e, i, o, u]\n [aa, a | leovam | NORMAL | 2021-01-18T05:08:47.219347+00:00 | 2021-01-18T05:08:47.219399+00:00 | 290 | false | ```py\n\'\'\'\nw: dp\nh: when n = 1, we only have 5 choice,\n wheb n = 2, we have 15 options, how they related\n consider: [a, e, i, o, u]\n [aa, ae, ai, ao, au -- 5 strings\n ee, ei, eo, eu -- 4 strings\n ii, io, iu -- 3 strings\n oo, ou... | 6 | 0 | ['Dynamic Programming', 'Python'] | 0 |
count-sorted-vowel-strings | Count Sorted Vowel Strings: Python, 1-D DP | count-sorted-vowel-strings-python-1-d-dp-8v5o | python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n a = [1] * 5\n for _ in range(1, n):\n for i in range(1, 5):\ | sjoin | NORMAL | 2021-01-17T08:52:41.166666+00:00 | 2021-01-17T08:59:20.306092+00:00 | 422 | false | ```python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n a = [1] * 5\n for _ in range(1, n):\n for i in range(1, 5):\n a[i] += a[i-1]\n return sum(a)\n``` | 6 | 1 | ['Dynamic Programming', 'Python'] | 1 |
count-sorted-vowel-strings | java backtrack solution very simple | java-backtrack-solution-very-simple-by-o-12aj | \n int ans;\n public int countVowelStrings(int n) {\n // char[] ch ={\'a\',\'e\',\'i\',\'o\',\'u\'};\n ans=0;\n count(0,n);\n retu | osama010 | NORMAL | 2020-11-10T04:33:22.799369+00:00 | 2020-11-10T04:33:22.799471+00:00 | 646 | false | ```\n int ans;\n public int countVowelStrings(int n) {\n // char[] ch ={\'a\',\'e\',\'i\',\'o\',\'u\'};\n ans=0;\n count(0,n);\n return ans;\n }\n void count(int src,int n){\n if(n==0){ans++;return;}\n for(int i=src;i<5;i++){\n count(i,n-1);\n }\n }... | 6 | 0 | ['Backtracking', 'Java'] | 3 |
count-sorted-vowel-strings | Python DP `O(1)` space, super EASY. | python-dp-o1-space-super-easy-by-herokil-h6dn | python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n a, e, i, o, u = 1, 1, 1, 1, 1\n for _ in range(n):\n a, e, i | herokillerever | NORMAL | 2020-11-09T13:10:04.566535+00:00 | 2020-11-09T13:10:04.566569+00:00 | 228 | false | ```python\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n a, e, i, o, u = 1, 1, 1, 1, 1\n for _ in range(n):\n a, e, i, o, u = a+e+i+o+u, e+i+o+u, i+o+u, o+u, u\n return a\n``` | 6 | 1 | [] | 1 |
count-sorted-vowel-strings | ✅C++ || DP Beats 100% 🚀 || (Memoization and Recursion) | c-dp-beats-100-memoization-and-recursion-0an7 | Code\n\nclass Solution {\npublic:\n int t(vector<vector<int>>&dp,int n,int i){\n if(i==0){\n dp[n][i]=0;\n return dp[n][i];\n | CHIIKUU | NORMAL | 2023-03-22T07:28:01.336628+00:00 | 2023-03-22T07:37:13.689948+00:00 | 1,606 | false | # Code\n```\nclass Solution {\npublic:\n int t(vector<vector<int>>&dp,int n,int i){\n if(i==0){\n dp[n][i]=0;\n return dp[n][i];\n }\n if(n==1){\n dp[n][i]=i;\n return dp[n][i];\n }\n if(dp[n][i]!=-1)return dp[n][i];\n dp[n][i] = t... | 5 | 0 | ['Dynamic Programming', 'Combinatorics', 'C++'] | 0 |
count-sorted-vowel-strings | ✅C++ || Easy Recursion || Simple Explanation || 🗓️ Daily LeetCoding Challenge May, Day 11 | c-easy-recursion-simple-explanation-dail-2gjr | Please Upvote If It Helps\n\n\nclass Solution {\npublic:\n void helper(vector<char> &vowels, vector<char> &tmp, int &ans, int n, int start)\n {\n / | mayanksamadhiya12345 | NORMAL | 2022-05-11T16:56:00.197038+00:00 | 2022-05-11T19:32:50.601358+00:00 | 145 | false | **Please Upvote If It Helps**\n\n```\nclass Solution {\npublic:\n void helper(vector<char> &vowels, vector<char> &tmp, int &ans, int n, int start)\n {\n // if our tmp.size(), in which we are storing our combination reaches to the n then push that combination into our ans\n if(n==tmp.size())\n ... | 5 | 0 | [] | 0 |
count-sorted-vowel-strings | Java || From recursion to memoization to tabulation || easy and explained | java-from-recursion-to-memoization-to-ta-h0tc | Recursion -->\n[ Time-Complexity = exponential, Space-Complexity = O(n) ]\n\nclass Solution {\n int count =0;\n public int countVowelStrings(int n) {\n | gourav02 | NORMAL | 2022-05-11T10:45:21.703225+00:00 | 2022-05-12T07:25:25.261567+00:00 | 351 | false | Recursion -->\n[ Time-Complexity = exponential, Space-Complexity = O(n) ]\n```\nclass Solution {\n int count =0;\n public int countVowelStrings(int n) {\n String[] s= new String[]{"a","e","i","o","u"};\n \n int ans = solve(0,n,s);\n return ans;\n }\n public int solve(int ind , int ... | 5 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 1 |
count-sorted-vowel-strings | python easy and a hard way | backtracking | DP | | python-easy-and-a-hard-way-backtracking-2v0xx | Using backtracking\n\n def countVowelStrings(self, n):\n vowels = [\'a\', \'e\', \'i\', \'o\', \'u\']\n res = []\n idx = 0\n\n de | b44hrt | NORMAL | 2022-05-11T10:27:45.605475+00:00 | 2022-05-11T10:30:08.535878+00:00 | 415 | false | **Using backtracking**\n```\n def countVowelStrings(self, n):\n vowels = [\'a\', \'e\', \'i\', \'o\', \'u\']\n res = []\n idx = 0\n\n def rec(s, idx):\n if len(s) == n:\n res.append(s)\n return\n for i in vowels[idx:]:\n r... | 5 | 0 | ['Dynamic Programming', 'Backtracking', 'Recursion', 'Python'] | 0 |
count-sorted-vowel-strings | C++ logic based 2 line solution | c-logic-based-2-line-solution-by-nimesh-8h2aa | Logic based self explanatory code :\n\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n \n n += 4;\n \n int temp = n | Nimesh-Srivastava | NORMAL | 2022-05-11T05:37:29.594538+00:00 | 2022-05-11T05:43:23.688633+00:00 | 159 | false | Logic based self explanatory code :\n```\nclass Solution {\npublic:\n int countVowelStrings(int n) {\n \n n += 4;\n \n int temp = n * (n - 1) * (n - 2) * (n - 3);\n \n return temp / 24;\n }\n};\n```\n\nOR\n\n```\nclass Solution {\npublic:\n int countVowelStrings(int n)... | 5 | 0 | ['C', 'C++'] | 0 |
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