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minimum-operations-to-make-the-array-k-increasing
✅✅Faster || Easy To Understand || C++ Code
faster-easy-to-understand-c-code-by-__kr-jf9g
Using Binary Search\n\n Time Complexity :- O(NlogN * K)\n\n Space Complexity :- O(N * K)\n\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& nums,
__KR_SHANU_IITG
NORMAL
2022-07-04T06:00:11.612637+00:00
2022-07-04T06:00:36.532620+00:00
141
false
* ***Using Binary Search***\n\n* ***Time Complexity :- O(NlogN * K)***\n\n* ***Space Complexity :- O(N * K)***\n\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& nums, int k) {\n \n int n = nums.size();\n \n // all the k subsets are independent of each other, so we can perfo...
0
0
['Binary Search', 'C', 'Binary Tree']
0
minimum-operations-to-make-the-array-k-increasing
python LIS - TC(nlogn)
python-lis-tcnlogn-by-vincent_great-7es9
\ndef kIncreasing(self, arr: List[int], k: int) -> int: \n\tdef lis(nums):\n\t\tans = [nums[0]]\n\t\tfor n in nums[1:]:\n\t\t\tif ans[-1]<=n:\n\t\t\t\tans.appen
vincent_great
NORMAL
2022-06-30T16:25:57.341528+00:00
2022-06-30T16:25:57.341578+00:00
58
false
```\ndef kIncreasing(self, arr: List[int], k: int) -> int: \n\tdef lis(nums):\n\t\tans = [nums[0]]\n\t\tfor n in nums[1:]:\n\t\t\tif ans[-1]<=n:\n\t\t\t\tans.append(n)\n\t\t\telse:\n\t\t\t\tidx = bisect.bisect(ans, n)\n\t\t\t\tans[idx] = n\n\t\treturn len(ans)\n\n\tres = 0\n\tfor i in range(k):\n\t\tcans = []\n\t\twhil...
0
0
[]
0
minimum-operations-to-make-the-array-k-increasing
Golang 100/80
golang-10080-by-adam-hoelscher-a86y
Logic in comments. O(n * log n/k)time; O(n/k) space\n\nfunc kIncreasing(arr []int, k int) int {\n\n // the name of this problem is not optimal. We\'re not ac
adam-hoelscher
NORMAL
2022-06-28T00:46:42.433946+00:00
2022-06-28T00:46:42.433978+00:00
26
false
Logic in comments. `O(n * log n/k)`time; `O(n/k)` space\n```\nfunc kIncreasing(arr []int, k int) int {\n\n // the name of this problem is not optimal. We\'re not actually trying to make the suqsequences increasing\n // we\'re trying to make them non-decreasing\n\n // the general approach is the same idea as ht...
0
0
['Binary Search', 'Dynamic Programming']
0
minimum-operations-to-make-the-array-k-increasing
K grps LIS
k-grps-lis-by-ishubhamrana-6bjp
\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n vector<vector<int> > lis(k);\n int N=arr.size();\n \n
iShubhamRana
NORMAL
2022-06-24T05:30:58.906827+00:00
2022-06-24T05:30:58.906869+00:00
19
false
```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n vector<vector<int> > lis(k);\n int N=arr.size();\n \n for(int i=0;i<N;i++){\n int grp = i%k;\n grp_size[grp]++;\n \n auto &lis_grp = lis[grp];\n \n ...
0
0
[]
0
minimum-operations-to-make-the-array-k-increasing
[Python 10 lines] Binary Search with thought process when being asked during interviews
python-10-lines-binary-search-with-thoug-82fe
Binary Search\nLet\'s simply the problem first. What if k is large enough such that there is only one pair for each i to construct the sequence? We just need to
jandk
NORMAL
2022-06-12T10:57:05.684654+00:00
2022-06-12T10:57:05.684695+00:00
96
false
### Binary Search\nLet\'s simply the problem first. What if `k` is large enough such that there is only one pair for each `i` to construct the sequence? We just need to check each pair to see if it\'s valid or not, and increase the final result by 1 if the pair is invalid.\n\nThen let\'s decrement `k` , the number of i...
0
0
['Binary Tree']
0
minimum-operations-to-make-the-array-k-increasing
[JAVA] Easy to understand O(NlogN)
java-easy-to-understand-onlogn-by-chenya-cxie
very very similar to longest increasing subsequence\n\n\nclass Solution {\n public int kIncreasing(int[] A, int k) {\n // Longest increasing subsequen
chenyang_johan
NORMAL
2022-05-11T07:41:34.499707+00:00
2022-05-11T07:41:34.499747+00:00
144
false
very very similar to longest increasing subsequence\n\n```\nclass Solution {\n public int kIncreasing(int[] A, int k) {\n // Longest increasing subsequence for\n // A[i]..A[i + k].. A[i + 2k]\n int n = A.length;\n // int[] memo = new int[n];\n int res = 0;\n for (int s = 0; ...
0
0
['Binary Tree']
0
minimum-operations-to-make-the-array-k-increasing
[C++] LIS with binary search
c-lis-with-binary-search-by-kaminyou-b6bi
\nclass Solution {\npublic:\n int upperBound(vector<int>& nums, int x) {\n int left = 0;\n int right = nums.size();\n while (left < righ
kaminyou
NORMAL
2022-04-29T16:57:21.787866+00:00
2022-04-29T16:57:21.787901+00:00
117
false
```\nclass Solution {\npublic:\n int upperBound(vector<int>& nums, int x) {\n int left = 0;\n int right = nums.size();\n while (left < right) {\n int mid = left + (right - left) / 2;\n if (nums[mid] > x) right = mid;\n else left = mid + 1;\n }\n ret...
0
0
['C', 'Binary Tree']
0
minimum-operations-to-make-the-array-k-increasing
[C++] Beats 97% | LIS | Just ignore K for a while :)
c-beats-97-lis-just-ignore-k-for-a-while-zw3u
We can skip K for a while, which makes this question as Minimum operations to Make the Array Increasing, which can be solved by finding the LIS (Longest Increas
meKabhi
NORMAL
2022-04-23T12:30:17.820753+00:00
2022-04-23T12:30:47.916800+00:00
157
false
We can skip `K` for a while, which makes this question as `Minimum operations to Make the Array Increasing`, which can be solved by finding the LIS (Longest Increasing Subsequence), using binary search, and subtracting the size of LIS from size of the array. Performing this step `K` times will give the desired result.\...
0
0
['Dynamic Programming', 'C', 'Binary Tree', 'C++']
0
minimum-operations-to-make-the-array-k-increasing
C++ 11 line
c-11-line-by-sanzenin_aria-mye8
```\n int kIncreasing(vector& arr, int k) {\n int numGood = 0;\n for(int i = 0; i<k; i++){\n vector v;\n for(int j=i;j<ar
sanzenin_aria
NORMAL
2022-04-14T00:54:08.938944+00:00
2022-04-14T00:54:08.938985+00:00
82
false
```\n int kIncreasing(vector<int>& arr, int k) {\n int numGood = 0;\n for(int i = 0; i<k; i++){\n vector<int> v;\n for(int j=i;j<arr.size();j+=k){\n auto it = upper_bound(v.begin(), v.end(), arr[j]);\n if(it == v.end()) v.push_back(arr[j]);\n ...
0
0
[]
0
minimum-operations-to-make-the-array-k-increasing
C++ lis
c-lis-by-wufengxuan1230-n67q
\nclass Solution {\npublic:\n int kIncreasing(vector<int>& nums, int k) {\n int res = 0;\n vector<int> lis(nums.size() / k + 2, INT_MAX);\n
wufengxuan1230
NORMAL
2022-03-29T04:44:19.541007+00:00
2022-03-29T04:47:56.219112+00:00
66
false
```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& nums, int k) {\n int res = 0;\n vector<int> lis(nums.size() / k + 2, INT_MAX);\n for (int i = 0; i < k; ++i) {\n std::fill(lis.begin(), lis.end(), INT_MAX);\n int maxLen = 1;\n lis[maxLen] = nums[i];\n...
0
0
[]
0
minimum-operations-to-make-the-array-k-increasing
Golang LIS Binary Search O(nlog(n)) time complexity
golang-lis-binary-search-onlogn-time-com-r2a7
\nfunc kIncreasing(arr []int, k int) int {\n buckets := partition(arr, k)\n \n // T -> O(n)\n // S -> O(n)\n var numberOfOperations int\n for
alyxjperkins
NORMAL
2022-03-27T19:24:51.636592+00:00
2022-03-27T19:24:51.636635+00:00
50
false
```\nfunc kIncreasing(arr []int, k int) int {\n buckets := partition(arr, k)\n \n // T -> O(n)\n // S -> O(n)\n var numberOfOperations int\n for _, bucket := range buckets {\n numberOfOperations += longestIncreasingSequence(bucket)\n }\n \n return numberOfOperations\n}\n\n// T -> O(n)\...
0
0
[]
0
minimum-operations-to-make-the-array-k-increasing
[Java] Clean LC-300 LIS version with Collections.binarySearch
java-clean-lc-300-lis-version-with-colle-wywn
\nclass Solution {\n private int lis(List<Integer> nums) {\n List<Integer> sub = new ArrayList<>();\n sub.add(nums.get(0));\n \n
srinathcoder
NORMAL
2022-03-22T05:41:59.786653+00:00
2022-03-28T02:05:11.321757+00:00
186
false
```\nclass Solution {\n private int lis(List<Integer> nums) {\n List<Integer> sub = new ArrayList<>();\n sub.add(nums.get(0));\n \n for (int i = 1; i < nums.size(); i++) {\n int num = nums.get(i);\n if (num >= sub.get(sub.size() - 1)) {\n sub.add(num);...
0
0
['Binary Tree', 'Java']
0
minimum-operations-to-make-the-array-k-increasing
C++ | Binary Search | LIS
c-binary-search-lis-by-offamikumar-nh0t
\nclass Solution {\n int getPos(int val, vector<int>&v) {\n int high = v.size()-1, low = 0; \n int idx{}; \n while (low < high) {\n
offamikumar
NORMAL
2022-03-15T08:23:48.219284+00:00
2022-03-15T08:23:48.219317+00:00
80
false
```\nclass Solution {\n int getPos(int val, vector<int>&v) {\n int high = v.size()-1, low = 0; \n int idx{}; \n while (low < high) {\n int mid = (low + high) / 2; \n if (val >= v[mid]) {\n low = mid + 1; \n } else {\n high = mid; \n ...
0
0
['Binary Tree']
0
binary-prefix-divisible-by-5
Detailed Explanation using Modular Arithmetic O(n)
detailed-explanation-using-modular-arith-omhg
Intuition\n First, let us see how to append an extra bit at the end of a binary number\n Any binary number akak-1...a0 has the form a020 + a121 + ........ a02k
just__a__visitor
NORMAL
2019-03-31T05:02:07.125180+00:00
2019-03-31T05:02:07.125242+00:00
9,796
false
**Intuition**\n* First, let us see how to append an extra bit at the end of a binary number\n* Any binary number **a<sub>k</sub>a<sub>k-1</sub>...a<sub>0</sub>** has the form a<sub>0</sub>*2<sup>0</sup> + a<sub>1</sub>*2<sup>1</sup> + ........ a<sub>0</sub>*2<sup>k</sup>\n\n\n* To free up an extra slot at the end, we ...
147
10
[]
12
binary-prefix-divisible-by-5
[Java/Python 3] 7/1 liners - left shift, bitwise or, and mod.
javapython-3-71-liners-left-shift-bitwis-tgbi
Since A includes only 0s and 1s, we can imitate binary operations.\n\njava\n public List<Boolean> prefixesDivBy5(int[] A) {\n int k = 0;\n List
rock
NORMAL
2019-03-31T04:30:38.062831+00:00
2022-02-28T18:58:20.309831+00:00
5,261
false
Since A includes only `0s` and `1s`, we can imitate binary operations.\n\n```java\n public List<Boolean> prefixesDivBy5(int[] A) {\n int k = 0;\n List<Boolean> ans = new ArrayList<>();\n for (int a : A) {\n k = (k << 1 | a) % 5; // left shift k by 1 bit and plus current element a is t...
66
3
[]
10
binary-prefix-divisible-by-5
Java solution with best explanation(don't know why some explanation not concise)
java-solution-with-best-explanationdont-xp36i
Prerequisite\n\nYou need to know how to contruct a number with bits in some numeric representation. \n\nFor example: \n1. Contruct the number from a string in d
416486188
NORMAL
2019-05-21T03:20:34.471163+00:00
2020-01-08T21:37:34.805515+00:00
2,539
false
**Prerequisite**\n\nYou need to know how to contruct a number with bits in some numeric representation. \n\nFor example: \n1. Contruct the number from a string in decimal like `"12345678"`, key: `num = num * 10 + (c - \'0\')` where `c` is the current character.\n2. Contruct the number from a string in binary like `"010...
54
1
[]
5
binary-prefix-divisible-by-5
[Python] Calculate Prefix Mod
python-calculate-prefix-mod-by-lee215-1sgk
Solution 1\npy\n def prefixesDivBy5(self, A):\n for i in xrange(1, len(A)):\n A[i] += A[i - 1] * 2 % 5\n return [a % 5 == 0 for a in
lee215
NORMAL
2019-03-31T04:03:41.623466+00:00
2021-03-29T04:32:25.103402+00:00
4,997
false
# Solution 1\n```py\n def prefixesDivBy5(self, A):\n for i in xrange(1, len(A)):\n A[i] += A[i - 1] * 2 % 5\n return [a % 5 == 0 for a in A]\n```\n`O(1)` space\n```\n def prefixesDivBy5(self, A):\n sums = 0\n for i, a in enumerate(A):\n sums = sums * 2 % 5 + a\n ...
38
5
[]
11
binary-prefix-divisible-by-5
C++ 4 lines
c-4-lines-by-votrubac-owmy
Just build your number from the binary string. The only trick here is to track modulo 5, not the actual number, to avoid overflow.\n\nvector<bool> prefixesDivBy
votrubac
NORMAL
2019-03-31T04:27:50.842973+00:00
2019-03-31T04:27:50.843024+00:00
3,701
false
Just build your number from the binary string. The only trick here is to track modulo 5, not the actual number, to avoid overflow.\n```\nvector<bool> prefixesDivBy5(vector<int>& A, vector<bool> a = {}, int num = 0) {\n for (auto i : A) {\n num = (num * 2 + i) % 5;\n a.push_back(!num);\n }\n return a;\n}\n```
31
2
[]
6
binary-prefix-divisible-by-5
Simple method(Java)
simple-methodjava-by-poorvank-087n
When a binary number is appended by 0 , the new remainder can be calculated based on current remainder only.\nremainder = (remainder * 2) % 5;\n\nAnd when a bin
poorvank
NORMAL
2019-03-31T04:23:55.131992+00:00
2019-03-31T04:23:55.132055+00:00
1,685
false
When a binary number is appended by 0 , the new remainder can be calculated based on current remainder only.\nremainder = (remainder * 2) % 5;\n\nAnd when a binary number is appended by 1.\n remainder = (remainder * 2 + 1) % 5;\n \n For example:\n If it is appended by 0 it will become 10 (2 in decimal) means twice of t...
29
3
[]
3
binary-prefix-divisible-by-5
C++ and Python simple solution
c-and-python-simple-solution-by-tovam-ltcq
C++ :\n\n\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool> res;\n int num;\n \n
TovAm
NORMAL
2021-10-06T10:25:12.370295+00:00
2021-10-06T10:25:12.370484+00:00
2,688
false
**C++ :**\n\n```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool> res;\n int num;\n \n for(int i = 0; i < nums.size(); ++i)\n {\n num = (num * 2 + nums[i]) % 5;\n res.push_back(num == 0);\n }\n ...
28
0
['C', 'Python', 'C++', 'Python3']
2
binary-prefix-divisible-by-5
Detailed explanation for mod calculation and equation (with super simple Java solution)
detailed-explanation-for-mod-calculation-gm4s
Java solution\n\njava\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> result = new ArrayList<>();\n int r =
phoenixpan
NORMAL
2019-04-02T00:32:22.054228+00:00
2019-04-02T00:32:22.054270+00:00
757
false
**Java solution**\n\n```java\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> result = new ArrayList<>();\n int r = 0;\n for (int bit : A) {\n r = (r * 2 + bit) % 5;\n result.add(r % 5 == 0);\n }\n return result;\n }\n}\n``...
14
0
[]
1
binary-prefix-divisible-by-5
[Java] 100% Runtime easy just 1 iteration
java-100-runtime-easy-just-1-iteration-b-uw5e
\n\nUPVOTE IF YOU FIND IT USEFUL\n\n\n\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> list = new ArrayList<>();\n
pulkitswami7
NORMAL
2020-09-05T07:06:12.250363+00:00
2020-09-05T07:06:12.250424+00:00
451
false
<hr>\n\n***UPVOTE IF YOU FIND IT USEFUL***\n<hr>\n\n```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> list = new ArrayList<>();\n \n int val = 0;\n for(int i=0;i<A.length;i++){\n val = val * 2 + A[i];\n \n if((val %= 5)...
9
0
[]
2
binary-prefix-divisible-by-5
Simple Java Solution With Explanation
simple-java-solution-with-explanation-by-bxgn
Idea\ncurr keeping track of decimal representation so far. When scanning from right, if we multiply curr by 2, it will shift all the bits in curr to left to mak
naveen_kothamasu
NORMAL
2019-03-31T05:00:40.870081+00:00
2019-03-31T05:00:40.870116+00:00
608
false
**Idea**\n`curr` keeping track of decimal representation so far. When scanning from right, if we multiply `curr` by 2, it will shift all the bits in `curr` to left to make place for the incoming digit `i`. That way, we are reusing the computations. We are not interested in the actual number, so to avoid overflow, we ca...
8
2
[]
3
binary-prefix-divisible-by-5
[Java] 2ms, 100% + math theory
java-2ms-100-math-theory-by-stefanelstan-0oyk
\nclass Solution {\n /** Algorithm/Theory\n 1. In order to determine if a binary is divisible by 5, we have to build the 10 base & check\n 2. A
StefanelStan
NORMAL
2022-05-27T15:11:31.969295+00:00
2022-05-27T15:11:31.969334+00:00
997
false
```\nclass Solution {\n /** Algorithm/Theory\n 1. In order to determine if a binary is divisible by 5, we have to build the 10 base & check\n 2. As the problem states, there will be a stream of bits, 1 and 0 that will build n numbers\n In order to build the next number, start with a value of ...
7
0
['Java']
1
binary-prefix-divisible-by-5
Python3 using yield
python3-using-yield-by-wingsoflight2003-w0fy
\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n x = 0\n for i in range(len(A)):\n x = x*2+A[i]\n
wingsoflight2003
NORMAL
2020-02-10T15:23:15.525592+00:00
2020-02-10T15:23:15.525784+00:00
526
false
```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n x = 0\n for i in range(len(A)):\n x = x*2+A[i]\n yield x%5==0\n```
7
0
['Python3']
2
binary-prefix-divisible-by-5
[C++] Simple Solution
c-simple-solution-by-pankajgupta20-ty24
\tclass Solution {\n\tpublic:\n\t\tvector prefixesDivBy5(vector& nums) {\n\t\t\tvector res;\n\t\t\tint t = 0;\n\t\t\tfor(int i = 0; i < nums.size(); i++){\n\t\t
pankajgupta20
NORMAL
2021-05-31T11:58:04.882830+00:00
2021-05-31T11:58:04.882875+00:00
700
false
\tclass Solution {\n\tpublic:\n\t\tvector<bool> prefixesDivBy5(vector<int>& nums) {\n\t\t\tvector<bool> res;\n\t\t\tint t = 0;\n\t\t\tfor(int i = 0; i < nums.size(); i++){\n\t\t\t\tt = (t * 2 + nums[i]) % 5;\n\t\t\t\tres.push_back(t == 0);\n\t\t\t}\n\t\t\treturn res;\n\t\t}\n\t};
5
0
['C', 'C++']
2
binary-prefix-divisible-by-5
1018 | JavaScript 1 Line Solution
1018-javascript-1-line-solution-by-spork-1vwz
Runtime: 72 ms, faster than 59.00% of JavaScript online submissions\n> Memory Usage: 38.2 MB, less than 92.31% of JavaScript online submissions\n\nNot the faste
sporkyy
NORMAL
2019-08-12T15:06:54.475362+00:00
2021-08-31T13:44:28.390736+00:00
727
false
> Runtime: **72 ms**, faster than *59.00%* of JavaScript online submissions\n> Memory Usage: **38.2 MB**, less than *92.31%* of JavaScript online submissions\n\nNot the fastest.\n\n```\n/**\n * @param {number[]} A\n * @param {number} acc\n * @return {boolean[]}\n */\nconst prefixesDivBy5 = (A, acc = 0) => A.map(d => !(...
5
0
['JavaScript']
1
binary-prefix-divisible-by-5
A couple of different solutions in Python
a-couple-of-different-solutions-in-pytho-13tk
Here is the first solution:\n\npython\nclass Solution:\n def prefixesDivBy5(self, A):\n result = []\n number = 0\n for bit in A:\n
lxnn
NORMAL
2019-07-14T11:52:29.299869+00:00
2019-07-14T14:23:55.380058+00:00
417
false
Here is the first solution:\n\n```python\nclass Solution:\n def prefixesDivBy5(self, A):\n result = []\n number = 0\n for bit in A:\n number = 2*number + bit\n result.append(number % 5 == 0)\n return result\n```\n\nThe idea is simple: we accumulate the bits into a bi...
5
0
[]
2
binary-prefix-divisible-by-5
Easy Solution
easy-solution-by-algoartisan-q7j6
\nvector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool>v;\n \n int sum=0;\n \n for(auto x : nums){\n
AlgoArtisan
NORMAL
2022-09-10T21:24:43.254888+00:00
2022-09-10T21:24:43.254926+00:00
921
false
```\nvector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool>v;\n \n int sum=0;\n \n for(auto x : nums){\n \n sum=(sum*2+x)%5;\n \n v.push_back(sum==0);\n }\n \n return v;\n }\n```
4
0
['C', 'C++']
0
binary-prefix-divisible-by-5
Java Solution with Explanation, faster than 100.00%
java-solution-with-explanation-faster-th-fngb
\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int sum = 0;\n int length = nums.length;\n List<Boolean> result
kingston880420
NORMAL
2021-05-29T04:04:49.546906+00:00
2021-05-29T04:04:49.546937+00:00
500
false
```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int sum = 0;\n int length = nums.length;\n List<Boolean> result = new ArrayList<>();\n //Go through the nums notice that shifting a binary number to the left is equals to multiplying a decimal number by 2\n ...
4
0
['Java']
2
binary-prefix-divisible-by-5
Python | Fast > 99%
python-fast-99-by-ddhnnng-feri
\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n answer = [None] * len(A)\n S = 0\n for idx, i in enumerate(A
ddhnnng
NORMAL
2021-03-12T21:32:16.498000+00:00
2021-03-12T21:33:45.078709+00:00
290
false
```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n answer = [None] * len(A)\n S = 0\n for idx, i in enumerate(A):\n S <<= 1 # bit shift left\n S += i # add new least-sig bit\n S %= 5...
4
0
['Python']
0
binary-prefix-divisible-by-5
Python O(n) use <<= to solve
python-on-use-to-solve-by-abandonblue101-6g1g
\ndef prefixesDivBy5(self, A: List[int]) -> List[bool]:\n \n cum = 0\n ans = []\n \n for i in range(len(A)):\n cum
abandonblue1015
NORMAL
2020-07-23T12:31:40.496490+00:00
2020-07-23T12:31:40.496519+00:00
143
false
```\ndef prefixesDivBy5(self, A: List[int]) -> List[bool]:\n \n cum = 0\n ans = []\n \n for i in range(len(A)):\n cum <<= 1 # based on binary\n cum += (A[i])\n if cum % 5 == 0:\n ans.append(True)\n else:\n ans...
4
0
['Python3']
1
binary-prefix-divisible-by-5
Solution in Python 3 (beats ~98%) (three lines) ( O(1) space )
solution-in-python-3-beats-98-three-line-lo0c
```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n \tn = 0\n \tfor i in range(len(A)): A[i], n = (2n + A[i]) % 5 == 0, (2n
junaidmansuri
NORMAL
2019-08-11T14:45:13.659533+00:00
2019-08-11T14:46:09.491418+00:00
474
false
```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n \tn = 0\n \tfor i in range(len(A)): A[i], n = (2*n + A[i]) % 5 == 0, (2*n + A[i]) % 5\n \treturn A\n\t\t\n\t\t\n- Junaid Mansuri\n(LeetCode ID)@hotmail.com
4
1
['Python', 'Python3']
1
binary-prefix-divisible-by-5
🔥VERY SIMPLE || JAVA || BEATS 99.11% 🔥
very-simple-java-beats-9911-by-tamilselv-ilf1
Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n) \n# Code\n\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n L
Tamilselvan_Srini
NORMAL
2024-01-05T10:14:16.904827+00:00
2024-01-05T10:20:29.227185+00:00
680
false
# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$ \n# Code\n```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n List<Boolean>list=new ArrayList<>();\n int sum=0;\n for(int i=0;i<nums.length;i++)\n {\n sum = (sum * 2 + nums[i]) ...
3
0
['Array', 'Java']
1
binary-prefix-divisible-by-5
|| Easy Solution || DFA ||
easy-solution-dfa-by-leetcode_ashutosh-yscp
Intuition\nUsing Deterministic Finite Automata \n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nConstant\n\n# Code\n\nclass Solution {\n pu
leetcode_Ashutosh
NORMAL
2023-07-19T18:10:48.648386+00:00
2023-10-16T15:03:46.150974+00:00
173
false
# Intuition\nUsing Deterministic Finite Automata \n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nConstant\n\n# Code\n```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n List<Boolean> res = new ArrayList<>();\n int daigram[][] ={{0,1},{2,3},{4,0},{1,2},{3,4}};\...
3
0
['Math', 'Brainteaser', 'C++', 'Java', 'Python3']
1
binary-prefix-divisible-by-5
Java, C++, Python | 🚀 A pinch of math in 7 lines explained
java-c-python-a-pinch-of-math-in-7-lines-v8ec
\u26A0\uFE0F Disclaimer: The original solution was crafted in Java.\n\nDont forget to upvote if you like the content below. \uD83D\uDE43\n\n# Intuition\nThe tas
wallandteen
NORMAL
2023-05-04T21:16:18.902431+00:00
2023-05-09T04:46:05.358313+00:00
824
false
> \u26A0\uFE0F **Disclaimer**: The original solution was crafted in Java.\n\nDont forget to upvote if you like the content below. \uD83D\uDE43\n\n# Intuition\nThe task requires us to check whether the binary prefix of each number in an array is divisible by 5. We need to return a boolean array where each index correspo...
3
0
['C++', 'Java', 'Python3']
0
binary-prefix-divisible-by-5
With Explanation Comments: Time: 3 ms (100.00%), Space: 14 MB (46.11%)
with-explanation-comments-time-3-ms-1000-f2i3
Like it? ->Upvote please! \u30C4\n\n\'\'\'\nclass Solution {\npublic:\n vector prefixesDivBy5(vector& nums) {\n \n //initialize a new vector wi
deleted_user
NORMAL
2022-09-05T19:56:24.177570+00:00
2022-09-08T18:53:35.197734+00:00
343
false
**Like it? ->Upvote please!** \u30C4\n\n\'\'\'\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n //initialize a new vector with boolean values\n vector<bool> res;\n int num=0;\n \n //loop over the whole array numbers\n for(int i=0;i<nu...
3
0
['Array', 'C', 'C++']
0
binary-prefix-divisible-by-5
C++|| EASY TO UNDERSTAND || fast and efficient
c-easy-to-understand-fast-and-efficient-pdtdg
\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n int n=nums.size();\n vector<bool> vec;\n int num=0; \n f
aarindey
NORMAL
2021-10-19T22:09:13.610677+00:00
2021-10-19T22:12:29.091601+00:00
184
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n int n=nums.size();\n vector<bool> vec;\n int num=0; \n for(int i=0;i<n;i++)\n {\n num=(num*2+nums[i])%10;\n vec.push_back(num%5==0);\n }\n return vec;\n }\n};\n```\n**Please upvote ...
3
0
[]
0
binary-prefix-divisible-by-5
C++ || Fast and Easy Method
c-fast-and-easy-method-by-suniti0804-hzdi
\n vector prefixesDivBy5(vector& A) \n {\n vector res;\n int n = A.size();\n \n int num = 0;\n \n for(int i = 0;
suniti0804
NORMAL
2021-04-13T14:49:18.587362+00:00
2021-04-14T17:37:57.121093+00:00
271
false
\n vector<bool> prefixesDivBy5(vector<int>& A) \n {\n vector<bool> res;\n int n = A.size();\n \n int num = 0;\n \n for(int i = 0; i < n; i++)\n {\n num = num * 2 + A[i];\n if(num % 5 == 0)\n res.push_back(true);\n els...
3
0
['C', 'C++']
0
binary-prefix-divisible-by-5
Java 100% fast, simple
java-100-fast-simple-by-umedjan-0f75
\nclass Solution {\n\tpublic List<Boolean> prefixesDivBy5(int[] A) {\n\t\tList<Boolean> result = new ArrayList<>(A.length);\n\t\tint s = 0;\n\t\tfor (int i : A)
umedjan
NORMAL
2021-02-10T09:53:27.520991+00:00
2021-02-10T09:53:27.521022+00:00
383
false
```\nclass Solution {\n\tpublic List<Boolean> prefixesDivBy5(int[] A) {\n\t\tList<Boolean> result = new ArrayList<>(A.length);\n\t\tint s = 0;\n\t\tfor (int i : A)\n\t\t\tresult.add((s = (s * 2 + i) % 5) == 0);\n\n\t\treturn result;\n\t}\n}\n```
3
0
['Java']
3
binary-prefix-divisible-by-5
C++ 12ms 100% Fast
c-12ms-100-fast-by-kumaranuj1303-u9fr
\nvector<bool> prefixesDivBy5(vector<int>& A) \n {\n int size = A.size();\n vector<bool> B(size);\n for (auto i = 1; i < size; i++)\n
kumaranuj1303
NORMAL
2020-10-20T13:10:42.541165+00:00
2020-10-20T13:10:42.541205+00:00
219
false
```\nvector<bool> prefixesDivBy5(vector<int>& A) \n {\n int size = A.size();\n vector<bool> B(size);\n for (auto i = 1; i < size; i++)\n {\n A[i] = (A[i] + A[i-1] * 2) % 10;\n if (A[i] % 5)\n B[i] = 0;\n else\n B[i] = 1;\n ...
3
0
[]
0
binary-prefix-divisible-by-5
[Java] Simple bit shifting | O(n) time, O(1) space
java-simple-bit-shifting-on-time-o1-spac-ijo7
```\nclass Solution {\n public List prefixesDivBy5(int[] A) {\n \n List result = new ArrayList<>(A.length);\n int num = 0;\n for(i
mdshahnawaz459
NORMAL
2020-09-08T12:20:28.273713+00:00
2020-09-08T12:23:28.173775+00:00
278
false
```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n \n List<Boolean> result = new ArrayList<>(A.length);\n int num = 0;\n for(int i = 0; i < A.length; i++) {\n num <<= 1;\n num |= A[i];\n result.add((num = (num % 5)) == 0 ? true: false)...
3
0
['Bit Manipulation', 'Java']
0
binary-prefix-divisible-by-5
Javascript simple solution with explanation
javascript-simple-solution-with-explanat-3yjk
\nvar prefixesDivBy5 = function (A) {\n let stv = 0;\n return A.map((bit) => {\n stv = stv * 2 + bit;\n\t//adding 1 binary bit means whether the number get
leomacode
NORMAL
2020-08-25T18:37:01.540257+00:00
2020-08-25T18:40:52.686430+00:00
453
false
```\nvar prefixesDivBy5 = function (A) {\n let stv = 0;\n return A.map((bit) => {\n stv = stv * 2 + bit;\n\t//adding 1 binary bit means whether the number gets doubled (if add 0) or gets doubled and plus 1 (if add 1)\n\t//Example: decimal 2 is 10 in binary. If add 0 in binary: 10 will be 100 which is 4 in decimal ...
3
0
['JavaScript']
0
binary-prefix-divisible-by-5
Python Simple solution
python-simple-solution-by-lokeshsk1-7og0
\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n s=\'\';l=[]\n for i in A:\n s+=str(i)\n l.app
lokeshsk1
NORMAL
2020-08-14T08:52:44.943453+00:00
2020-08-14T10:25:28.688260+00:00
412
false
```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n s=\'\';l=[]\n for i in A:\n s+=str(i)\n l.append(int(s,2)%5==0)\n return l\n \n```
3
0
['Python', 'Python3']
0
binary-prefix-divisible-by-5
C++ solution
c-solution-by-oleksam-u7wm
\nvector<bool> prefixesDivBy5(vector<int>& A) {\n\tint iCurr = 0;\n\tvector<bool> vRes;\n\tfor (auto n : A) {\n\t\tiCurr = (iCurr << 1) | n;\n\t\tvRes.push_back
oleksam
NORMAL
2020-06-30T12:49:06.506140+00:00
2020-06-30T12:49:06.506173+00:00
414
false
```\nvector<bool> prefixesDivBy5(vector<int>& A) {\n\tint iCurr = 0;\n\tvector<bool> vRes;\n\tfor (auto n : A) {\n\t\tiCurr = (iCurr << 1) | n;\n\t\tvRes.push_back(iCurr % 5 == 0);\n\t\tiCurr %= 10;\n\t}\n\treturn vRes;\n}\n```\n
3
0
['C', 'C++']
0
binary-prefix-divisible-by-5
JAVA Solution with a Trick to prevent Overflow
java-solution-with-a-trick-to-prevent-ov-7m3z
\npublic List<Boolean> prefixesDivBy5(int[] A) {\n\tint currentNum = 0;\n\tList<Boolean> sol = new ArrayList<>();\n\tfor(int a : A) {\n\t\t// (a * b + c) % d =
anubhavjindal
NORMAL
2019-11-16T06:21:50.433788+00:00
2019-11-16T06:21:50.433845+00:00
227
false
```\npublic List<Boolean> prefixesDivBy5(int[] A) {\n\tint currentNum = 0;\n\tList<Boolean> sol = new ArrayList<>();\n\tfor(int a : A) {\n\t\t// (a * b + c) % d = ((a % d) * (b % d) + c % d) % d\n\t\tcurrentNum = (currentNum << 1 | a)%5;\n\t\tsol.add(currentNum==0);\n\t}\n\treturn sol;\n}\n```
3
0
[]
0
binary-prefix-divisible-by-5
Simple, C++ Solution [beats 98%, 100%]
simple-c-solution-beats-98-100-by-pooja0-y5hc
Runtime: 8 ms, faster than 98.57% of C++ online submissions for Binary Prefix Divisible By 5.\nMemory Usage: 10.6 MB, less than 100.00% of C++ online submission
pooja0406
NORMAL
2019-09-01T08:18:45.228178+00:00
2019-09-01T08:18:45.228210+00:00
477
false
Runtime: 8 ms, faster than 98.57% of C++ online submissions for Binary Prefix Divisible By 5.\nMemory Usage: 10.6 MB, less than 100.00% of C++ online submissions for Binary Prefix Divisible By 5.\n\n```\n vector<bool> prefixesDivBy5(vector<int>& A) {\n \n vector<bool> res;\n int n = A.size();\n ...
3
0
['C++']
1
binary-prefix-divisible-by-5
C solution with DFA; general purpose
c-solution-with-dfa-general-purpose-by-j-kwo7
A general purpose solution written in C with DFA:\n\nc\nbool* prefixesDivBy5(int* A, int ASize, int* returnSize){\n *returnSize = ASize;\n bool *res = (bo
jules3in
NORMAL
2019-08-07T05:18:03.682804+00:00
2019-08-07T05:18:03.682840+00:00
234
false
A general purpose solution written in C with DFA:\n\n```c\nbool* prefixesDivBy5(int* A, int ASize, int* returnSize){\n *returnSize = ASize;\n bool *res = (bool *)malloc(ASize * sizeof(bool));\n\n int q0 = 0, q1 = 1, q2 = 2, q3 = 3, q4 = 4;\n int DFA[5][2] = { {q0, q1}, {q2, q3}, {q4, q0}, {q1, q2}, {q3, q4}...
3
0
['C']
0
binary-prefix-divisible-by-5
Python Solution - Bit shift with mod check
python-solution-bit-shift-with-mod-check-vyky
Each time you look you add a bit, you are multiplying the previous sum by 2 and adding the value of the current bit. This can be achieved by a simple left bit s
raosrika89
NORMAL
2019-05-26T18:47:06.274556+00:00
2019-05-26T19:44:47.826249+00:00
320
false
Each time you look you add a bit, you are multiplying the previous sum by 2 and adding the value of the current bit. This can be achieved by a simple left bit shift by 1. \n\nIt is then just a matter of checking if the total is divisible by 5 or not. \n\n```\nclass Solution:\n \n def prefixesDivBy5(self, A: List...
3
0
[]
0
binary-prefix-divisible-by-5
JavaScript Solution, Straight forward
javascript-solution-straight-forward-by-kmlzs
javascript\\\nvar prefixesDivBy5 = function(A) {\n let cur = 0;\n const r = []\n for (const a of A) {\n cur = ((cur << 1) | a) % 5;\n r.p
ruinan
NORMAL
2019-04-21T05:19:31.000013+00:00
2019-04-21T05:19:31.000181+00:00
282
false
```javascript\\\nvar prefixesDivBy5 = function(A) {\n let cur = 0;\n const r = []\n for (const a of A) {\n cur = ((cur << 1) | a) % 5;\n r.push(cur === 0);\n }\n return r;\n};\n```
3
0
[]
1
binary-prefix-divisible-by-5
个人觉得很好理解的java解法
ge-ren-jue-de-hen-hao-li-jie-de-javajie-5313z
\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> result = new ArrayList<>();\n int current = 0;\n for
zhf123
NORMAL
2019-04-17T02:50:20.007613+00:00
2019-04-17T02:50:20.007656+00:00
248
false
```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> result = new ArrayList<>();\n int current = 0;\n for (int i:A){\n current = (current*2+i)%10;\n result.add(current%5==0);\n }\n return result;\n}\n}``\n```
3
0
[]
0
binary-prefix-divisible-by-5
Java, simple 3 lines, no modulo
java-simple-3-lines-no-modulo-by-jpv-kvmb
I see lots of solutions using %, which is an expensive operation. If you have something between 0 and 4, then double it and add 0 or 1, your result is between
jpv
NORMAL
2019-03-31T13:49:57.440502+00:00
2019-03-31T13:49:57.440544+00:00
355
false
I see lots of solutions using %, which is an expensive operation. If you have something between 0 and 4, then double it and add 0 or 1, your result is between 0 and 9. Just subtracting 5 if needed is cheaper/faster than using modulo.\n```\n public List<Boolean> prefixesDivBy5(int[] A) {\n List<Boolean> ans ...
3
2
[]
1
binary-prefix-divisible-by-5
Simple and Easy Java Solution || Beats 99% || With intuition and approach
simple-and-easy-java-solution-beats-99-w-myjn
🔍 Intuition: The given binary array represents a number whose digits are read sequentially. We need to check if the number formed by the prefix (from index0toi)
Yashvendra
NORMAL
2025-02-18T17:45:39.448830+00:00
2025-02-18T17:45:39.448830+00:00
157
false
## 🔍 Intuition: - The given binary array represents a number whose digits are read sequentially. - We need to check if the number formed by the prefix (from index `0` to `i`) is divisible by `5`. - Instead of converting the entire prefix into an integer (which may cause overflow for large inputs), we can maintain the ...
2
0
['Array', 'Bit Manipulation', 'Java']
0
binary-prefix-divisible-by-5
python solution
python-solution-by-bhavanabharatisingh-ybrv
Intuitionconvert it into numeric value and check the visibilityComplexity Time complexity: O(n) Space complexity: O(n)Code
bhavanabharatisingh
NORMAL
2024-12-22T12:20:04.859431+00:00
2024-12-22T12:20:04.859431+00:00
182
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> convert it into numeric value and check the visibility # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> O(n) - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> O(n) # Code ```p...
2
0
['Python3']
0
binary-prefix-divisible-by-5
Two Solutions | Bit Manipulation & DFA | Both O(n)
two-solutions-bit-manipulation-dfa-both-fy95r
Solution 1 | Bit Manipulation:\n## Intuition\n\nWith each new bit, we can determine the new remainder by left-shifting the new bit onto the existing remainder a
SquirtleHerder
NORMAL
2024-02-23T22:22:53.859215+00:00
2024-04-13T04:44:12.891124+00:00
631
false
# Solution 1 | Bit Manipulation:\n## Intuition\n\nWith each new bit, we can determine the new remainder by left-shifting the new bit onto the existing remainder and then computing the modulo of the resultant value. \n\nThis approach ensures that the length of the resulting value would remain at most 4 bits vs other bit...
2
0
['Bit Manipulation', 'C', 'Simulation', 'C++', 'Python3']
0
binary-prefix-divisible-by-5
Solution
solution-by-deleted_user-n77d
C++ []\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool> r;\n int n=0;\n for(int b : nums){\n
deleted_user
NORMAL
2023-05-21T07:25:16.874924+00:00
2023-05-21T07:49:00.002614+00:00
712
false
```C++ []\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool> r;\n int n=0;\n for(int b : nums){\n n = ((n<<1) | b) % 5;\n r.push_back(n==0);\n }\n return r;\n }\n};\n```\n\n```Python3 []\nclass Solution:\n def pr...
2
0
['C++', 'Java', 'Python3']
0
binary-prefix-divisible-by-5
Java && JavaScript Solution does not need to convert anything
java-javascript-solution-does-not-need-t-mncq
There are many solutions for solving this problem and most of them are awesome. This solution based on that and just modified it to easier and cleaner\n=> Here
tonybuidn
NORMAL
2022-11-26T06:16:58.187901+00:00
2023-02-26T15:26:01.924719+00:00
713
false
There are many solutions for solving this problem and most of them are awesome. This solution based on that and just modified it to easier and cleaner\n=> Here using loop enhancement instead original loop => because we always to get the value for each index start the first to the end of the array.\n=> so we use loop en...
2
0
['Java']
1
binary-prefix-divisible-by-5
C++ || easy and efficient
c-easy-and-efficient-by-prakhar_20-i3xo
\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool> v1;\n int sum=0;\n for(int i=0;i
prakhar_20
NORMAL
2022-09-01T19:12:26.420036+00:00
2022-09-01T19:12:26.420077+00:00
580
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n vector<bool> v1;\n int sum=0;\n for(int i=0;i<nums.size();i++)\n { \n sum+=nums[i]; //to check for each element\n sum*=2; // to conve...
2
0
['C++']
1
binary-prefix-divisible-by-5
Least Runtime Solution
least-runtime-solution-by-day_tripper-06m3
\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n c,l=0,[]\n for i in nums:\n c=(c*2+i)%5\n
Day_Tripper
NORMAL
2022-07-19T09:39:49.620194+00:00
2022-07-19T09:39:49.620231+00:00
159
false
```\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n c,l=0,[]\n for i in nums:\n c=(c*2+i)%5\n l.append(c==0)\n return l\n```
2
0
[]
0
binary-prefix-divisible-by-5
𝗗𝗲𝘁𝗮𝗶𝗹𝗲𝗱 𝗘𝘅𝗽𝗹𝗮𝗻𝗮𝘁𝗶𝗼𝗻 | Python | C | 𝗢(𝗻) | DFA | Theory | Thought Process
detailed-explanation-python-c-on-dfa-the-gp10
\u2714 \uD835\uDDD7\uD835\uDDD8\uD835\uDDE7\uD835\uDDD4\uD835\uDDDC\uD835\uDDDF\uD835\uDDD8\uD835\uDDD7 \uD835\uDDD8\uD835\uDDEB\uD835\uDDE3\uD835\uDDDF\uD835\u
RohitSgh
NORMAL
2022-06-02T01:50:35.934185+00:00
2023-01-10T20:39:42.305959+00:00
377
false
# \u2714 \uD835\uDDD7\uD835\uDDD8\uD835\uDDE7\uD835\uDDD4\uD835\uDDDC\uD835\uDDDF\uD835\uDDD8\uD835\uDDD7 \uD835\uDDD8\uD835\uDDEB\uD835\uDDE3\uD835\uDDDF\uD835\uDDD4\uD835\uDDE1\uD835\uDDD4\uD835\uDDE7\uD835\uDDDC\uD835\uDDE2\uD835\uDDE1 \n\nThis **Solution** will use the concept of **Finite Automaton**, and is simila...
2
0
['C', 'Python']
0
binary-prefix-divisible-by-5
Simple C++ Solution
simple-c-solution-by-nikhilgupta25-spd6
\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool>ans;\n int dec=0;\n for(int i=0;i<nums.size
NikhilGupta25
NORMAL
2022-04-14T10:30:42.975446+00:00
2022-04-14T10:30:42.975486+00:00
332
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool>ans;\n int dec=0;\n for(int i=0;i<nums.size();i++){\n dec = dec * 2 + nums[i];\n if(dec%5==0){\n ans.push_back(true);\n }\n else{\n ...
2
0
['C', 'C++']
0
binary-prefix-divisible-by-5
Easy Python code for beginners
easy-python-code-for-beginners-by-vistri-elgf
\ndef prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n l=[]\n su=""\n for i in nums:\n su+=str(i)\n if int(su,2
vistrit
NORMAL
2021-11-24T07:23:06.602663+00:00
2021-11-24T07:23:06.602699+00:00
195
false
```\ndef prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n l=[]\n su=""\n for i in nums:\n su+=str(i)\n if int(su,2)%5==0:\n l.append(True)\n else:\n l.append(False)\n return l\n```
2
0
['Python', 'Python3']
0
binary-prefix-divisible-by-5
Java O(n) solution with explaination
java-on-solution-with-explaination-by-ak-1jr0
\n\n/* \n * First thing first, note that by adding a zero bit digit to any binary\n * number, we get a new binary number that represent double of the original b
aknuclear
NORMAL
2021-06-12T03:43:20.411419+00:00
2021-06-12T03:43:33.765006+00:00
132
false
```\n\n/* \n * First thing first, note that by adding a zero bit digit to any binary\n * number, we get a new binary number that represent double of the original binary value.\n * Similarly, adding a 1 bit digit yields original value * 2 + 1.\n * So we can walk the binary array nums from left to right,\n * starting fro...
2
0
[]
0
binary-prefix-divisible-by-5
Python3 simple solution
python3-simple-solution-by-eklavyajoshi-lnvl
\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n res = []\n n = 0\n for i in nums:\n n *= 2\n
EklavyaJoshi
NORMAL
2021-05-26T03:36:50.010714+00:00
2021-05-26T03:36:50.010749+00:00
96
false
```\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n res = []\n n = 0\n for i in nums:\n n *= 2\n if i == 1:\n n += 1\n res.append(n % 5 == 0)\n return res\n```\n**If you like this solution, please upvote for thi...
2
0
['Python3']
0
binary-prefix-divisible-by-5
Java solution with given hint
java-solution-with-given-hint-by-ayansam-m7d7
class Solution {\n\n\n public List prefixesDivBy5(int[] a) {\n List l = new ArrayList<>();\n long num = a[0];\n l.add(num%5 == 0?true:fa
ayansam
NORMAL
2021-04-22T08:11:30.521418+00:00
2021-04-22T08:11:30.521471+00:00
100
false
class Solution {\n\n\n public List<Boolean> prefixesDivBy5(int[] a) {\n List<Boolean> l = new ArrayList<>();\n long num = a[0];\n l.add(num%5 == 0?true:false);\n for(int i=1;i<a.length;i++){\n num = 2*num +a[i];\n l.add(num%5 == 0?true:false);\n num = num%...
2
0
[]
0
binary-prefix-divisible-by-5
java-solution
java-solution-by-meteor18-talj
\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n int temp = 0;\n List<Boolean> result = new ArrayList<Boolean>(A.length);\
meteor18
NORMAL
2019-12-27T02:58:47.958033+00:00
2019-12-27T02:58:47.958070+00:00
190
false
```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] A) {\n int temp = 0;\n List<Boolean> result = new ArrayList<Boolean>(A.length);\n for (int i = 0; i < A.length; i++) {\n temp = (temp * 2 + A[i]) % 5;\n if (temp == 0)\n result.add(i, Boolean.T...
2
0
['Java']
1
binary-prefix-divisible-by-5
[JavaScript] Beats 100% Time and Memory
javascript-beats-100-time-and-memory-by-2jw93
Bit shift only to least significant digit to avoid overflow. Current number is divisible by 5 if LSD is 0 or 5.\n\njavascript\nvar prefixesDivBy5 = function(A)
ethan-marsh
NORMAL
2019-09-27T04:09:33.993199+00:00
2019-09-27T04:09:33.993242+00:00
373
false
Bit shift only to least significant digit to avoid overflow. Current number is divisible by 5 if LSD is 0 or 5.\n\n```javascript\nvar prefixesDivBy5 = function(A) {\n let lsd = 0;\n return A.map((bit) => {\n lsd <<= 1;\n lsd += bit;\n lsd %= 10;\n return lsd === 0 || lsd === 5;\n });\n};\n```
2
1
['JavaScript']
1
binary-prefix-divisible-by-5
Go golang solution
go-golang-solution-by-leaf_peng-nvhb
Runtime: 332 ms, faster than 71.93% of Go online submissions for Binary Prefix Divisible By 5.\nMemory Usage: 8.2 MB, less than 40.00% of Go online submissions
leaf_peng
NORMAL
2019-09-22T02:28:27.270286+00:00
2019-09-22T02:28:27.270335+00:00
88
false
>Runtime: 332 ms, faster than 71.93% of Go online submissions for Binary Prefix Divisible By 5.\nMemory Usage: 8.2 MB, less than 40.00% of Go online submissions for Binary Prefix Divisible By 5.\n\n```go\nfunc prefixesDivBy5(A []int) []bool {\n\n\tans := []bool{}\n\tn := 0\n\n\tfor _, v := range A {\n\t\tn = (n<<1 + v)...
2
0
[]
0
binary-prefix-divisible-by-5
javascript
javascript-by-huihancarl188-a9fv
\nvar prefixesDivBy5 = function(A) {\n // i => current binary number;\n let cur = 0;\n let res = [];\n for(let i = 0 ; i < A.length; i++) {\n
huihancarl188
NORMAL
2019-05-26T04:28:19.159410+00:00
2019-05-26T04:28:19.159459+00:00
164
false
```\nvar prefixesDivBy5 = function(A) {\n // i => current binary number;\n let cur = 0;\n let res = [];\n for(let i = 0 ; i < A.length; i++) {\n cur = cur << 1;\n cur = cur % 5;\n cur += A[i];\n res.push(cur % 5 === 0);\n }\n return res;\n};\n```
2
0
[]
0
binary-prefix-divisible-by-5
Weird Javascript Question, can anyone explain? (Solved)
weird-javascript-question-can-anyone-exp-ehug
If I write my code like this\n\nvar prefixesDivBy5 = function(A) {\n let res = [];\n let remainder = 0;\n for(let i=0; i<A.length; i++){\n remai
0618
NORMAL
2019-03-31T14:47:53.906920+00:00
2019-03-31T14:47:53.906967+00:00
302
false
If I write my code like this\n```\nvar prefixesDivBy5 = function(A) {\n let res = [];\n let remainder = 0;\n for(let i=0; i<A.length; i++){\n remainder = remainder*2 + A[i];\n res.push(remainder%5===0);\n\t\tif(remainder%5===0) remainder = 0;\n }\n\n return res;\n};\n```\nI can\'t pass test...
2
0
[]
2
binary-prefix-divisible-by-5
Python straightforward bit manipulation
python-straightforward-bit-manipulation-0jke7
On each iteration, num bits shift 1 index to left while least-significant bit is A[i].\n\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bo
cenkay
NORMAL
2019-03-31T04:27:57.284956+00:00
2019-03-31T04:27:57.284989+00:00
188
false
* On each iteration, num bits shift 1 index to left while least-significant bit is A[i].\n```\nclass Solution:\n def prefixesDivBy5(self, A: List[int]) -> List[bool]:\n num = 0\n for i in range(len(A)):\n num = (num << 1) + A[i]\n A[i] = num % 5 == 0\n return A\n```\nTime c...
2
2
[]
0
binary-prefix-divisible-by-5
Simple C++ Solution || Beats 100%
simple-c-solution-beats-100-by-devanshv1-ynbn
IntuitionThe problem requires checking whether the binary number formed by each prefix of the given array (nums) is divisible by 5. Instead of constructing larg
Devanshv123
NORMAL
2025-03-01T14:26:10.961546+00:00
2025-03-01T14:26:10.961546+00:00
43
false
# Intuition The problem requires checking whether the binary number formed by each prefix of the given array (nums) is divisible by 5. Instead of constructing large numbers explicitly (which could lead to overflow), we can track only the remainder when divided by 5. <!-- Describe your first thoughts on how to solve thi...
1
0
['Array', 'Bit Manipulation', 'C++']
0
binary-prefix-divisible-by-5
Simple C++ Solution || Beats 100%
simple-c-solution-beats-100-by-devanshv1-xt6a
IntuitionThe problem requires checking whether the binary number formed by each prefix of the given array (nums) is divisible by 5. Instead of constructing larg
Devanshv123
NORMAL
2025-03-01T14:26:07.802732+00:00
2025-03-01T14:26:07.802732+00:00
30
false
# Intuition The problem requires checking whether the binary number formed by each prefix of the given array (nums) is divisible by 5. Instead of constructing large numbers explicitly (which could lead to overflow), we can track only the remainder when divided by 5. <!-- Describe your first thoughts on how to solve thi...
1
0
['Array', 'Bit Manipulation', 'C++']
0
binary-prefix-divisible-by-5
Java Solution
java-solution-by-hksuthar12-t6fv
IntuitionApproachComplexity Time complexity: Space complexity: Code
Hksuthar12
NORMAL
2025-02-03T12:14:14.144809+00:00
2025-02-03T12:14:14.144809+00:00
79
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code `...
1
0
['Java']
0
binary-prefix-divisible-by-5
EASY SOLUTION IN JAVA BEATING 98.16% IN 3ms
easy-solution-in-java-beating-9816-in-3m-66ac
IntuitionThe problem asks us to determine whether the number represented by the binary subarray nums[0..i] (where i ranges from 0 to n-1) is divisible by 5. The
arshi_bansal
NORMAL
2025-02-03T06:47:04.671421+00:00
2025-02-03T06:47:04.671421+00:00
68
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> The problem asks us to determine whether the number represented by the binary subarray nums[0..i] (where i ranges from 0 to n-1) is divisible by 5. The binary number formed from nums[0..i] is built progressively as we move through the array...
1
0
['Array', 'Bit Manipulation', 'Java']
0
binary-prefix-divisible-by-5
O(n)
on-by-bhavanabharatisingh-w2fh
Intuition\n Describe your first thoughts on how to solve this problem. \nTake each number from the array nums and convert into decimal and check its divisiabli
bhavanabharatisingh
NORMAL
2024-10-01T04:15:54.151698+00:00
2024-10-01T04:15:54.151730+00:00
33
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTake each number from the array nums and convert into decimal and check its divisiablity with 5\n \n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\n- O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n...
1
0
['Python3']
0
binary-prefix-divisible-by-5
easy-peasy solution | O(n) time complexity
easy-peasy-solution-on-time-complexity-b-vgie
Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n# Co
ilovevectors
NORMAL
2024-09-09T18:48:01.472457+00:00
2024-09-09T18:48:01.472495+00:00
274
false
# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool> ret;\n int x...
1
0
['C++']
0
binary-prefix-divisible-by-5
EASY SOLUTION USING ONLY ONE WHILE LOOP IN C++
easy-solution-using-only-one-while-loop-7gi5b
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
vasu049
NORMAL
2024-08-23T16:26:56.841103+00:00
2024-08-23T16:26:56.841133+00:00
20
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --...
1
0
['C++']
0
binary-prefix-divisible-by-5
Python Easy Approach
python-easy-approach-by-_ss_official_006-sgon
Code\n\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n Integer=[]\n output=[]\n Final=[]\n for i in
_ss_official_006
NORMAL
2024-08-10T08:17:56.846253+00:00
2024-08-10T08:17:56.846278+00:00
168
false
# Code\n```\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n Integer=[]\n output=[]\n Final=[]\n for i in nums:\n Integer.append(str(i))\n for i in range(len(Integer)):\n output.append(int("".join(Integer[:i+1]),2))\n for i ...
1
0
['Array', 'Bit Manipulation', 'Python3']
1
binary-prefix-divisible-by-5
Simple java code for beginners
simple-java-code-for-beginners-by-arobh-o0yc
\n# Complexity\n- \n\n# Code\n\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n ArrayList<Boolean> ans = new ArrayList<>();\n
Arobh
NORMAL
2024-01-07T03:36:22.099830+00:00
2024-01-07T03:36:22.099852+00:00
35
false
\n# Complexity\n- \n![image.png](https://assets.leetcode.com/users/images/c0d0819d-7fda-4b6b-bd52-d59bbd45c2e0_1704598577.6489384.png)\n# Code\n```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n ArrayList<Boolean> ans = new ArrayList<>();\n int num = 0;\n for(int i = 0 ;...
1
0
['Java']
0
binary-prefix-divisible-by-5
Simple C# solution
simple-c-solution-by-dmitriy-maksimov-aled
Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n# Code\n\npublic class Solution\n{\n public IList<bool> PrefixesDivBy5(int[] nums)\n {\n
dmitriy-maksimov
NORMAL
2023-09-06T01:59:33.491451+00:00
2023-09-06T01:59:33.491478+00:00
28
false
# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```\npublic class Solution\n{\n public IList<bool> PrefixesDivBy5(int[] nums)\n {\n var result = new List<bool>();\n var num = 0;\n foreach (var n in nums)\n {\n num = (num * 2 + n) % 5;\n ...
1
0
['C#']
0
binary-prefix-divisible-by-5
Simple Javascript solution
simple-javascript-solution-by-kamalbhera-zpfh
\n# Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n
kamalbhera
NORMAL
2023-05-16T01:37:30.366102+00:00
2023-05-16T01:37:30.366151+00:00
161
false
\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * @param {number[]} nums\n * @return {boolean[]}\n */\nvar prefixesDivBy5 = function(nums) {\n let result = ...
1
0
['JavaScript']
0
binary-prefix-divisible-by-5
Python3 Solution || Beats 99.30% || Prefix Approach
python3-solution-beats-9930-prefix-appro-rgmw
python\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n currBinary, result = 0, []\n for bit in nums:\n
Heber_Alturria
NORMAL
2023-02-24T02:41:39.625254+00:00
2023-02-24T02:42:31.996141+00:00
656
false
```python\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n currBinary, result = 0, []\n for bit in nums:\n currBinary = (currBinary * 2 + bit) % 5 \n result.append(currBinary == 0)\n return result\n```
1
0
['Array', 'Prefix Sum', 'Python']
0
binary-prefix-divisible-by-5
Java and Python Solution
java-and-python-solution-by-a_shekhar-ymag
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
a_shekhar
NORMAL
2022-10-11T16:46:41.474064+00:00
2023-10-03T16:27:46.843363+00:00
482
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n...
1
0
['Python', 'Java', 'Python3']
1
binary-prefix-divisible-by-5
[C++] Faster than 98%
c-faster-than-98-by-nil24-ok90
\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n int n=0;\n vector<bool> res;\n\t\t\n for(int i=0;i<nums
Nil24
NORMAL
2022-08-17T08:55:54.330319+00:00
2022-08-17T08:55:54.330367+00:00
347
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n int n=0;\n vector<bool> res;\n\t\t\n for(int i=0;i<nums.size();i++){\n n=(n<<1)+nums[i];\n res.push_back(n%5==0);\n n=n%5;\n }\n\t\t\n return res;\n }\n};\n```
1
0
['C']
1
binary-prefix-divisible-by-5
java solution | 100% efficient
java-solution-100-efficient-by-aayushkum-tno9
\nclass Solution {\n func prefixesDivBy5(_ nums: [Int]) -> [Bool] {\n var divisibleByFive = [Bool]()\n var n = 0\n for bit in nums {\n
aayushkumar20
NORMAL
2022-07-19T18:59:55.692192+00:00
2022-07-19T18:59:55.692238+00:00
147
false
```\nclass Solution {\n func prefixesDivBy5(_ nums: [Int]) -> [Bool] {\n var divisibleByFive = [Bool]()\n var n = 0\n for bit in nums {\n n = (n * 2 + bit) % 10\n divisibleByFive.append(n == 5 || n == 0)\n }\n return divisibleByFive\n }\n}\n```
1
0
['Swift']
1
binary-prefix-divisible-by-5
C++ simple solution
c-simple-solution-by-osmanay-pedp
\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n int n = nums.size();\n vector<bool> res(n);\n
osmanay
NORMAL
2022-07-02T23:30:06.654306+00:00
2022-07-02T23:30:06.654344+00:00
92
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n \n int n = nums.size();\n vector<bool> res(n);\n \n int num = 0;\n \n for(int i=0; i<n; i++)\n {\n num = num*2 + nums[i];\n num %= 5;\n res[i] =...
1
0
[]
0
binary-prefix-divisible-by-5
Java | 3 liner | Explained
java-3-liner-explained-by-prashant404-c81j
Idea: Iterate over the array, form the new number and check for divisibility by 5. Mod by 5 at every iteration to avoid overflows\n>T/S: O(n)/O(1), where n = si
prashant404
NORMAL
2022-03-07T02:08:31.188053+00:00
2022-03-07T02:09:41.834256+00:00
194
false
**Idea:** Iterate over the array, form the new number and check for divisibility by 5. Mod by 5 at every iteration to avoid overflows\n>**T/S:** O(n)/O(1), where n = size(nums). Ignoring space for output\n```\npublic List<Boolean> prefixesDivBy5(int[] nums) {\n\tvar answer = new ArrayList<Boolean>(nums.length);\n\tfor ...
1
0
['Java']
0
binary-prefix-divisible-by-5
[ Problem No. 1018 ] Easy solution for beginners | C++
problem-no-1018-easy-solution-for-beginn-bikh
\nclass Solution {\npublic:\n\tvector<bool> prefixesDivBy5(vector<int>& nums) {\n\t\tvector<bool> res;\n\t\tint t = 0;\n\t\tfor(int i = 0; i < nums.size(); i++)
rakibhasan1030
NORMAL
2022-03-01T18:35:13.335117+00:00
2022-03-01T18:35:13.335158+00:00
141
false
```\nclass Solution {\npublic:\n\tvector<bool> prefixesDivBy5(vector<int>& nums) {\n\t\tvector<bool> res;\n\t\tint t = 0;\n\t\tfor(int i = 0; i < nums.size(); i++){\n\t\t\tt = (t * 2 + nums[i]) % 5;\n\t\t\tres.push_back(t == 0);\n\t\t}\n\t\treturn res;\n\t}\n};\n```\n***Please upvote if it was helpful for you, thank ...
1
0
['Array', 'C', 'C++']
0
binary-prefix-divisible-by-5
Simple Python Solution
simple-python-solution-by-idebbarh-x4io
\n def prefixesDivBy5(self, nums):\n output = []\n nums = "".join([str(i) for i in nums])\n for i in range(0,len(nums)) :\n i
idebbarh
NORMAL
2022-02-28T11:27:13.890599+00:00
2022-02-28T11:27:13.890639+00:00
48
false
```\n def prefixesDivBy5(self, nums):\n output = []\n nums = "".join([str(i) for i in nums])\n for i in range(0,len(nums)) :\n if int(nums[0:i+1], 2)%5==0 :\n output.append(True)\n else :\n output.append(False)\n return output\n```
1
0
[]
0
binary-prefix-divisible-by-5
(C++) 1018. Binary Prefix Divisible By 5
c-1018-binary-prefix-divisible-by-5-by-q-uh13
\n\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool> ans; \n int prefix = 0; \n for (auto& x
qeetcode
NORMAL
2022-02-02T21:17:45.986963+00:00
2022-02-02T21:17:45.986995+00:00
166
false
\n```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool> ans; \n int prefix = 0; \n for (auto& x : nums) {\n prefix = (2*prefix + x) % 5; \n ans.push_back(!prefix); \n }\n return ans; \n }\n};\n```
1
0
['C']
0
binary-prefix-divisible-by-5
Golang easy solution 8ms 100%
golang-easy-solution-8ms-100-by-fshiori-1rct
\nfunc prefixesDivBy5(nums []int) []bool {\n ans := make([]bool, len(nums))\n var t int\n for i, num := range nums {\n t = t << 1 + num\n
fshiori
NORMAL
2021-12-26T13:15:40.404183+00:00
2021-12-26T13:15:40.404223+00:00
136
false
```\nfunc prefixesDivBy5(nums []int) []bool {\n ans := make([]bool, len(nums))\n var t int\n for i, num := range nums {\n t = t << 1 + num\n if t % 5 == 0 {\n ans[i] = true\n t = 0\n } else {\n t %= 5\n }\n }\n return ans\n}\n```
1
0
['Go']
0
binary-prefix-divisible-by-5
Binary Prefix Divisible By 5: Combining a bit shift with modulo
binary-prefix-divisible-by-5-combining-a-7mzn
This solution has a space complexity of O(n) because the length of the list returned is n. But if we use a generator instead, this becomes O(1). We iterate thro
jodawill
NORMAL
2021-12-07T20:40:38.882876+00:00
2021-12-07T20:40:38.882908+00:00
65
false
This solution has a space complexity of `O(n`) because the length of the list returned is `n`. But if we use a generator instead, this becomes `O(1)`. We iterate through each bit in the list, shifting as we go, and adding true or false to the results, based on whether `n % 5 == 0`. So the time complexity is also `O(n)`...
1
0
[]
0
binary-prefix-divisible-by-5
Python: 5 lines of code with bad explanation
python-5-lines-of-code-with-bad-explanat-e7ol
python\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n n = 0\n for i, bit in enumerate(nums):\n n = (n
ctonic
NORMAL
2021-12-06T20:04:31.252818+00:00
2021-12-06T20:04:31.252856+00:00
100
false
```python\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n n = 0\n for i, bit in enumerate(nums):\n n = (n << 1) + bit\n nums[i] = not (n % 5)\n return nums`\n```\n\nReuse the list to save space\nIDK if (n << 1) or (n * 2) is faster
1
0
['Python3']
0
binary-prefix-divisible-by-5
O(n) time, O(n) space.
on-time-on-space-by-yayarokya-7byy
\npublic class Solution {\n public IList<bool> PrefixesDivBy5(int[] nums) {\n var res = new List<bool>();\n int curr = 0;\n foreach(var
yayarokya
NORMAL
2021-09-10T06:14:27.890789+00:00
2021-09-10T06:19:16.431115+00:00
99
false
```\npublic class Solution {\n public IList<bool> PrefixesDivBy5(int[] nums) {\n var res = new List<bool>();\n int curr = 0;\n foreach(var n in nums) {\n curr = (2*curr + n) % 5;\n res.Add(curr == 0);\n }\n return res;\n }\n}\n```
1
1
[]
0
binary-prefix-divisible-by-5
Python simple solution O(n)
python-simple-solution-on-by-doya1-gc1o
\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n answer = []\n start = 1\n val = 0\n for i in nums:
doya1
NORMAL
2021-07-26T05:01:55.450139+00:00
2021-07-26T05:01:55.450188+00:00
68
false
```\nclass Solution:\n def prefixesDivBy5(self, nums: List[int]) -> List[bool]:\n answer = []\n start = 1\n val = 0\n for i in nums:\n val += i\n if val%5==0:\n answer.append(True)\n else:\n answer.append(False)\n v...
1
0
[]
0
binary-prefix-divisible-by-5
Java Simple Solution
java-simple-solution-by-kunalojha1999-yn13
\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int cur = 0;\n ArrayList<Boolean> res = new ArrayList<>();\n \n
kunalojha1999
NORMAL
2021-07-22T15:45:35.357586+00:00
2021-07-22T15:45:35.357638+00:00
178
false
```\nclass Solution {\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int cur = 0;\n ArrayList<Boolean> res = new ArrayList<>();\n \n for(int i = 0 ; i < nums.length ; i++){\n cur = (2 * cur + nums[i]) % 5;\n if(cur == 0)\n res.add(true);\n ...
1
0
[]
0
binary-prefix-divisible-by-5
Java | Bit-manipulation
java-bit-manipulation-by-tangjiawei18-0s32
\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int a = 0;\n List<Boolean> res = new ArrayList<>();\n for (int i :nums) {\n
tangjiawei18
NORMAL
2021-06-29T07:05:10.178541+00:00
2021-06-29T07:05:10.178600+00:00
97
false
```\n public List<Boolean> prefixesDivBy5(int[] nums) {\n int a = 0;\n List<Boolean> res = new ArrayList<>();\n for (int i :nums) {\n a = (a << 1 | i) % 5;\n res.add(a == 0);\n }\n return res;\n }\n```
1
0
['Bit Manipulation']
0
binary-prefix-divisible-by-5
C++ ||Clean Code|| Easy explanation||Short| T: 90%, Mem:100%
c-clean-code-easy-explanationshort-t-90-udlzx
the intution is based on storing prefix sum, but instead of sum, i am storing the number upto that point.\nIt would be easier to understand if we consider decim
DivyanshuCS
NORMAL
2021-06-21T19:23:54.502129+00:00
2021-06-21T19:23:54.502171+00:00
108
false
the intution is based on storing prefix sum, but instead of sum, i am storing the number upto that point.\nIt would be easier to understand if we consider decimal number system.\nImagine nums represet a decimal number array instead of binary and you have an array like\n1 0 2 3 8 9 0 1\nso at index 0, the number it will...
1
0
['C++']
1
binary-prefix-divisible-by-5
100% FASTER || c++ SOLUTION || SIMPLE 5 LINES SOLUTION
100-faster-c-solution-simple-5-lines-sol-ili8
```\nclass Solution {\npublic:\n vector prefixesDivBy5(vector& nums) {\n vectorans;\n int sum=0;\n \n \n for(int i=0;i<num
SJ4u
NORMAL
2021-05-30T19:01:41.494150+00:00
2021-05-30T19:01:41.494185+00:00
124
false
```\nclass Solution {\npublic:\n vector<bool> prefixesDivBy5(vector<int>& nums) {\n vector<bool>ans;\n int sum=0;\n \n \n for(int i=0;i<nums.size();i++)\n {\n sum*=2;\n sum+=nums[i];\n sum=sum%5;\n ans.push_back(sum%5==0);\n ...
1
1
['C']
0