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max-chunks-to-make-sorted-ii | O(n) beats 100% precomputing. | on-beats-100-precomputing-by-luudanhhieu-g9n6 | Intuition
Array arr can seperate to 2 chunks arr1, arr2 only if data in arr1 < any data in arr2
Approach
Precomputing calculate maxLeft and minRight of arr
Com | luudanhhieu | NORMAL | 2025-03-06T07:19:32.787943+00:00 | 2025-03-06T07:19:32.787943+00:00 | 66 | false | # Intuition
- Array `arr` can seperate to 2 chunks `arr1, arr2 only if data in arr1 < any data in arr2 `
# Approach
- Precomputing calculate `maxLeft` and `minRight` of `arr`
- Compare and choose the right index to seperate
- `i == len(arr)-1` or `maxLeft[i] == minRight[i]` or `maxLeft[i] <= minRight[i+1] `
# Com... | 1 | 0 | ['Go'] | 0 |
max-chunks-to-make-sorted-ii | Beats 100.00% || Used PrefixMax And SuffixMin Array || Very Easy Solution || C++ | beats-10000-used-prefixmax-and-suffixmin-q3e8 | Complexity -:
Time complexity -: O(n)
Space complexity -: O(n)
Code -: | ShivamShrivastav | NORMAL | 2024-12-30T06:04:57.545525+00:00 | 2024-12-30T06:04:57.545525+00:00 | 12 | false |
# Complexity -:
- Time complexity -: O(n)
- Space complexity -: O(n)
# Code -:
```cpp []
class Solution { // prefixMax, suffixMin -:
public:
int maxChunksToSorted(vector<int>& arr) {
int n=arr.size();
// find the prefix Max array
vector<int>pre(n);
pre[0]=arr[0];
f... | 1 | 0 | ['Array', 'Greedy', 'Sorting', 'C++'] | 0 |
max-chunks-to-make-sorted-ii | Easy way to solve this problem, easy to understand for beginner | easy-way-to-solve-this-problem-easy-to-u-wj4n | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Beshoy25 | NORMAL | 2024-12-19T22:59:16.468644+00:00 | 2024-12-19T22:59:16.468644+00:00 | 61 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Easy solution to solve this problem and easy to understand
# Approach
<!-- Describe your approach to solving the problem. -->
I used another array as a index.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g... | 1 | 0 | ['C++'] | 1 |
max-chunks-to-make-sorted-ii | Java Monotonic Stack O(n) Solution with Simple Explanation | java-monotonic-stack-on-solution-with-si-opym | IntuitionKeep the max number of each chunk in the stack in ascending order.If the current number is less than previous chunk's max number, then we should merge | _BugMaker_ | NORMAL | 2024-12-19T20:51:38.528490+00:00 | 2024-12-19T20:52:02.250660+00:00 | 145 | false | # Intuition
Keep the max number of each chunk in the stack in ascending order.
If the current number is less than previous chunk's max number, then we should merge current number into the previous chunk. Here ‘previous chunk’ means the chunk currently at the top of the stack, they are represented by their max numbers.... | 1 | 0 | ['Monotonic Stack', 'Java'] | 0 |
max-chunks-to-make-sorted-ii | O(n) -->TIME COMPLEXITY, PREFIX_MAXIMUM AND SUFFIX_MINIMUM | on-time-complexity-prefix_maximum-and-su-fjlt | IntuitionTo split the array into the maximum number of chunks such that each chunk, when sorted individually and concatenated, results in a fully sorted array, | Star5courage1436 | NORMAL | 2024-12-19T19:28:56.153051+00:00 | 2024-12-19T19:28:56.153051+00:00 | 60 | false | # Intuition
To split the array into the maximum number of chunks such that each chunk, when sorted individually and concatenated, results in a fully sorted array, we need to identify partition points. The key insight is that the maximum element on the left of a partition must be less than or equal to the minimum elemen... | 1 | 0 | ['Greedy', 'C++'] | 0 |
max-chunks-to-make-sorted-ii | Simple C++ code | O(N) | Monotonic Stack solution | | simple-c-code-on-monotonic-stack-solutio-f7eq | Approach -> Monotonic StackComplexity
Time complexity : O(N)
Space complexity : O(N)
Code | Akt_00 | NORMAL | 2024-12-19T14:03:14.791154+00:00 | 2024-12-19T14:03:14.791154+00:00 | 113 | false | # Approach -> Monotonic Stack
# Complexity
- Time complexity : O(N)
- Space complexity : O(N)
# Code
```cpp []
class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
int n = arr.size();
stack<int> st;
for(int i=0; i<n; i++){
int ele = arr[i];
while(!s... | 1 | 0 | ['Array', 'Stack', 'Monotonic Stack', 'C++'] | 0 |
max-chunks-to-make-sorted-ii | Easy Solution here. using the concept of leftMax and RightMin to check | easy-solution-here-using-the-concept-of-xlb8n | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | HYDRO2070 | NORMAL | 2024-12-19T11:36:18.985912+00:00 | 2024-12-19T11:36:18.985912+00:00 | 148 | 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 | ['C++'] | 1 |
max-chunks-to-make-sorted-ii | beats 100% fastest solution monotonic stack | beats-100-fastest-solution-monotonic-sta-f96b | IntuitionApproachmonotonic stack approach, code is commented for explanation, this is very similar to maxChunksToSorted medium questionComplexity
Time complexit | neilpant | NORMAL | 2024-12-19T11:19:08.749633+00:00 | 2024-12-19T11:19:08.749633+00:00 | 11 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
monotonic stack approach, code is commented for explanation, this is very similar to maxChunksToSorted medium question
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: N
<!-- Add your time... | 1 | 0 | ['Python3'] | 0 |
max-chunks-to-make-sorted-ii | beats 100% fastest solution monotonic stack | beats-100-fastest-solution-monotonic-sta-1z5l | IntuitionApproachmonotonic stack approach, code is commented for explanation, this is very similar to maxChunksToSorted medium questionComplexity
Time complexit | neilpant | NORMAL | 2024-12-19T11:19:02.480250+00:00 | 2024-12-19T11:19:02.480250+00:00 | 13 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
monotonic stack approach, code is commented for explanation, this is very similar to maxChunksToSorted medium question
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: N
<!-- Add your time... | 1 | 0 | ['Python3'] | 0 |
max-chunks-to-make-sorted-ii | Simple 10 line C++ O(N) Solution! | simple-10-line-c-on-solution-by-knightfu-zsgy | IntuitionJust sort the array and compare the prefix sums at each index!Complexity
Time complexity: O(N)
Space complexity: O(1)
Code | knightfury24 | NORMAL | 2024-12-19T06:41:24.762423+00:00 | 2024-12-19T06:41:24.762423+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Just sort the array and compare the prefix sums at each index!
# Complexity
- Time complexity: O(N)
- Space complexity: O(1)
# Code
```cpp []
class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
vector<int> nums ... | 1 | 1 | ['C++'] | 0 |
max-chunks-to-make-sorted-ii | Beats 100%🔥|| easy JAVA Solution✅ | beats-100-easy-java-solution-by-priyansh-mx31 | Code\n\nclass Solution {\n public int maxChunksToSorted(int[] arr) {\n int n=arr.length;\n int min[]=new int[n];\n min[n-1]=arr[n-1];\n | priyanshu1078 | NORMAL | 2023-12-05T10:03:41.074858+00:00 | 2023-12-05T10:03:41.074888+00:00 | 107 | false | # Code\n```\nclass Solution {\n public int maxChunksToSorted(int[] arr) {\n int n=arr.length;\n int min[]=new int[n];\n min[n-1]=arr[n-1];\n for(int i=n-2;i>=0;i--){\n min[i]=Math.min(min[i+1],arr[i]);\n }\n int ans=0,max=Integer.MIN_VALUE;\n for(int i=0;i<... | 1 | 0 | ['Java'] | 0 |
max-chunks-to-make-sorted-ii | Solution by using two pointers approach | solution-by-using-two-pointers-approach-kco3s | Intuition\n Describe your first thoughts on how to solve this problem. \nTwo pointers\n\n# Approach\nTwo Pointers Approach\n\n# Complexity\n- Time complexity:\n | naman05022002 | NORMAL | 2023-09-07T12:30:18.011047+00:00 | 2023-09-07T12:30:18.011071+00:00 | 17 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTwo pointers\n\n# Approach\nTwo Pointers Approach\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. $$O(n)$$ -->\nO(n)\n\n# C... | 1 | 0 | ['C++'] | 0 |
max-chunks-to-make-sorted-ii | C++ Easy Sorting | c-easy-sorting-by-xenox_grix-2n67 | Intuition\nThe last number that belongs to the chunk should be present at that index or before that index.\n\nbefore sorting [array, index] :{{2,0},{1,1},{3,2}, | xenox_grix | NORMAL | 2023-04-01T09:19:37.710474+00:00 | 2023-04-01T10:02:46.962297+00:00 | 27 | false | # Intuition\nThe last number that belongs to the chunk should be present at that index or before that index.\n\nbefore sorting [array, index] :{{2,0},{1,1},{3,2},{4,3},{4,4}}\nafter sorting.... [array, index] :{{1,1},{2,0},{3,2},{4,3},{4,4}}\n\nNow, while traversing the sorted array arr[n][2], we know that we need to i... | 1 | 0 | ['Sorting', 'C++'] | 0 |
max-chunks-to-make-sorted-ii | Python || Sorting+Hashmap || O(n logn) | python-sortinghashmap-on-logn-by-in_sidi-mo3h | We have 2 arrays. One sorted and one unsorted.Now traverse over both.For one array increment by one for every value and for other decrement.We delete an element | iN_siDious | NORMAL | 2023-01-13T16:32:01.606711+00:00 | 2023-01-13T16:38:01.798799+00:00 | 313 | false | We have 2 arrays. One sorted and one unsorted.Now traverse over both.For one array increment by one for every value and for other decrement.We delete an element from map when its count is reached 0, it means when one element who was in excess in one array has been compensated by the same value in other array.(basically... | 1 | 0 | ['Sorting', 'Python3'] | 0 |
max-chunks-to-make-sorted-ii | Easy and Fast Java Solution || Beats 100% | easy-and-fast-java-solution-beats-100-by-xu6v | \n\n# Code\n\nclass Solution {\n public int maxChunksToSorted(int[] arr) {\n int rightMin[]=new int[arr.length+1];\n rightMin[arr.length]=Integ | 20_Saurabh-kumar | NORMAL | 2023-01-12T20:33:37.289420+00:00 | 2023-01-12T20:33:37.289451+00:00 | 138 | false | \n\n# Code\n```\nclass Solution {\n public int maxChunksToSorted(int[] arr) {\n int rightMin[]=new int[arr.length+1];\n rightMin[arr.length]=Integer.MAX_VALUE;\n for(int i=arr.length-1; i>=0; i--){\n rightMin[i]=Math.min(rightMin[i+1], arr[i]);\n }\n int count=0;\n ... | 1 | 0 | ['Array', 'Stack', 'Greedy', 'Monotonic Stack', 'Java'] | 0 |
max-chunks-to-make-sorted-ii | Advance Sorting Technique | advance-sorting-technique-by-shrey0811-c3d2 | class Solution {\npublic:\n int maxChunksToSorted(vector& arr)\n {\n vector right_min(arr.size()+1,INT_MAX);\n int count = 0;\n int l | shrey0811 | NORMAL | 2022-11-16T18:19:05.230808+00:00 | 2022-11-16T18:19:05.230844+00:00 | 1,121 | false | class Solution {\npublic:\n int maxChunksToSorted(vector<int>& arr)\n {\n vector<int> right_min(arr.size()+1,INT_MAX);\n int count = 0;\n int left_max = INT_MIN;\n for(int i=arr.size()-1;i>=0;i--)\n {\n right_min[i] = min(right_min[i+1],arr[i]);\n }\n \n... | 1 | 0 | [] | 1 |
max-chunks-to-make-sorted-ii | C++ || DP || TLE (128 Passed out of 136) || O(n^2) || Partition based | c-dp-tle-128-passed-out-of-136-on2-parti-thm8 | \n bool check(vector<int> &nums , int l , int r , vector<int> &sorted_arr)\n {\n unordered_map< int , int > mp;\n \n for(int i=l;i<=r;i+ | KR_SK_01_In | NORMAL | 2022-10-08T14:04:31.429715+00:00 | 2022-10-08T14:04:31.429742+00:00 | 34 | false | ```\n bool check(vector<int> &nums , int l , int r , vector<int> &sorted_arr)\n {\n unordered_map< int , int > mp;\n \n for(int i=l;i<=r;i++)\n {\n mp[sorted_arr[i]]++;\n }\n \n for(int i=l;i<=r;i++)\n {\n if(mp.find(nums[i])==mp.end())\n ... | 1 | 0 | ['Array', 'Dynamic Programming', 'C', 'C++'] | 0 |
number-of-ways-to-split-array | Prefix Sum with O(1) space||C++beats 100% | prefix-sum-with-o1-spacebeats-100-by-anw-q7hw | IntuitionFind the number of ways splitting the array nums such that sum(nums[0...i])>=sum(nums[i+1...]).
How?
compute sum=sum(nums)
Use a loop to find whether i | anwendeng | NORMAL | 2025-01-03T00:15:05.380090+00:00 | 2025-01-03T11:20:33.931162+00:00 | 14,683 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Find the number of ways splitting the array `nums` such that `sum(nums[0...i])>=sum(nums[i+1...])`.
How?
1. compute `sum=sum(nums)`
2. Use a loop to find whether i satisfying `sum(nums[0..i])>=sum(nums[i+1..])=sum-sum(nums[0..i])`
2 pass s... | 56 | 2 | ['Math', 'Prefix Sum', 'C++', 'Python3'] | 14 |
number-of-ways-to-split-array | Detailed Explanation Total Sum | Prefix Sum 100%Beats | detailed-explanation-without-prefix-100b-ab3z | 🌟 IntuitionThe task requires finding the number of ways to split the array such that the sum of the left part is greater than or equal to the sum of the right p | Sumeet_Sharma-1 | NORMAL | 2025-01-03T00:34:59.684174+00:00 | 2025-01-03T09:25:53.931433+00:00 | 12,156 | false | # 🌟 **Intuition**
The task requires finding the number of ways to split the array such that the **sum of the left part is greater than or equal to the sum of the right part**.
To solve this, consider:
- **Brute force approach**: For every possible split, calculate the sum of the left and right parts. This is in... | 37 | 0 | ['Array', 'C', 'C++', 'Java', 'Python3', 'Ruby', 'JavaScript', 'C#'] | 12 |
number-of-ways-to-split-array | Prefix Sum | O(1) Space | prefix-sum-o1-space-by-kamisamaaaa-u22i | \nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n \n long long sumFromBack(0), sumFromFront(0);\n for (auto& i : nu | kamisamaaaa | NORMAL | 2022-05-14T16:01:28.805268+00:00 | 2022-05-14T16:27:16.862806+00:00 | 4,706 | false | ```\nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n \n long long sumFromBack(0), sumFromFront(0);\n for (auto& i : nums) sumFromBack += i;\n \n int n(size(nums)), res(0);\n for (auto i=0; i<n-1; i++) {\n \n sumFromFront += nums... | 35 | 1 | ['C', 'Prefix Sum'] | 15 |
number-of-ways-to-split-array | Python: Easily explained brute force to optimized | python-easily-explained-brute-force-to-o-1av8 | Here, they are asking to count such events/occurances where:\n the array can be splitted into two non-empty parts, such that \n the sum of elements in first par | meaditya70 | NORMAL | 2022-05-14T17:08:25.894911+00:00 | 2022-05-15T00:41:05.355797+00:00 | 1,496 | false | Here, they are asking to **count such events**/occurances where:\n* the array can be splitted into **two non-empty parts**, such that \n* the sum of elements in first part **>=** sum of elements in second parts.\n\nA prefix sum array is an array that stores at ith index, the sum of all elements encountered from 0th in... | 26 | 0 | ['Python'] | 4 |
number-of-ways-to-split-array | Beats 100% | Prefix Sum | Solution for LeetCode#2270 | beats-100-prefix-sum-solution-for-leetco-u70n | IntuitionThe problem asks us to split an array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part | samir023041 | NORMAL | 2025-01-03T01:12:00.074297+00:00 | 2025-01-03T01:12:00.074297+00:00 | 3,258 | false | 
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem asks us to split an array into two non-empty parts such that the sum of the left part is greater than or equal t... | 19 | 1 | ['Prefix Sum', 'Java'] | 6 |
number-of-ways-to-split-array | Prefix Sum | prefix-sum-by-votrubac-qh6v | Python 3\npython\nclass Solution:\n def waysToSplitArray(self, n: List[int]) -> int:\n n = list(accumulate(n))\n return sum(n[i] >= n[-1] - n[i | votrubac | NORMAL | 2022-05-14T17:16:29.055856+00:00 | 2022-05-14T18:13:37.894102+00:00 | 2,781 | false | **Python 3**\n```python\nclass Solution:\n def waysToSplitArray(self, n: List[int]) -> int:\n n = list(accumulate(n))\n return sum(n[i] >= n[-1] - n[i] for i in range(len(n) - 1))\n```\n\n**C++**\nWe could use a prefix sum, but in C++ we would need to store it in another `long long` vector (because of ... | 18 | 1 | ['C', 'Python3'] | 6 |
number-of-ways-to-split-array | ✅C++ || Easy to Understand || Simple Sum || O(N) | c-easy-to-understand-simple-sum-on-by-ma-9xiq | Please Upvote If It Helps\n\n store total sum of nums in right_sum and iterate over nums\n\n add nums[i] in left sum and subtract nums[i] from right sum\n* when | mayanksamadhiya12345 | NORMAL | 2022-05-14T16:44:30.701173+00:00 | 2022-05-14T16:44:30.701204+00:00 | 1,119 | false | **Please Upvote If It Helps**\n\n* store total sum of nums in right_sum and iterate over nums\n\n* add nums[i] in left sum and subtract nums[i] from right sum\n* when condition (left_sum>=right_sum) match increase valid_split by 1.\n\n```\nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) \n {\n... | 13 | 0 | [] | 1 |
number-of-ways-to-split-array | 🚀💻Efficient Solution for Counting Valid Array Splits Using Prefix Sums | JAVA | PYTHON🐍🚀 | efficient-solution-for-counting-valid-ar-xbur | IntuitionThe problem requires splitting an array into two non-empty parts such that the sum of the left part greater than or 🟰 sum of the right part. To solve t | manoj55802 | NORMAL | 2025-01-03T04:48:32.155985+00:00 | 2025-01-03T08:05:30.939595+00:00 | 1,081 | false | # Intuition
The problem requires splitting an array into two non-empty parts such that the sum of the left part greater than or 🟰 sum of the right part. To solve this efficiently, we can use a prefix sum 🤷♂️ to calculate the sums for both parts in constant time during each iteration.
# Approach
1. Prefix Sum Calcul... | 11 | 0 | ['Array', 'Prefix Sum', 'Python', 'Java', 'Python3'] | 6 |
number-of-ways-to-split-array | Easy Optimal Solution Beats💯|| with explanation||Java||JavaScript||Go||C++||C||Python✅✅ | java-easy-solution-with-explanation-by-j-m2o8 | Here’s the revised content including Java, Python, Python3, C, C++, JavaScript, and Go implementations:IntuitionThe problem is to determine how many ways an arr | jeril-johnson | NORMAL | 2025-01-03T04:27:39.060151+00:00 | 2025-01-03T04:54:00.587280+00:00 | 653 | false | Here’s the revised content including Java, Python, Python3, C, C++, JavaScript, and Go implementations:
---
### Intuition
The problem is to determine how many ways an array can be split such that the sum of elements in the left part is greater than or equal to the sum of elements in the right part. The main insight i... | 10 | 0 | ['Prefix Sum', 'Python', 'Java', 'Go', 'Python3', 'JavaScript'] | 2 |
number-of-ways-to-split-array | ✅C++ | ✅Prefix Sum | ✅Easy O(N) solution | c-prefix-sum-easy-on-solution-by-yash2ar-s7t8 | Please upvote if you find this solution helpful :)\n\nTC- O(N), SC- O(1)\n\nclass Solution {\npublic:\n \n //store total sum of nums in right_sum and iter | Yash2arma | NORMAL | 2022-05-14T16:10:36.742463+00:00 | 2022-05-14T16:12:06.426825+00:00 | 1,066 | false | **Please upvote if you find this solution helpful :)**\n\n**TC- O(N), SC- O(1)**\n```\nclass Solution {\npublic:\n \n //store total sum of nums in right_sum and iterate over nums\n //add nums[i] in left sum and subtract nums[i] from right sum\n //when condition match increase valid_split by 1.\n int ways... | 10 | 0 | ['C', 'Prefix Sum', 'C++'] | 2 |
number-of-ways-to-split-array | Simple and Easy Solution | ✅Beats 100% | C++| Java | Python3 | simple-and-easy-solution-beats-100-c-jav-ql8h | 🛠️ ApproachThe problem is to find the number of ways to split an array into two non-empty parts such that the sum of the left part is greater than or equal to t | adhamsafir | NORMAL | 2025-01-03T04:56:07.425956+00:00 | 2025-01-03T04:56:07.425956+00:00 | 1,380 | false | # **🛠️ Approach**
The problem is to find the number of ways to split an array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part. The approach involves:
1. **Prefix Sum Calculation:**
Compute the prefix sum for the array. This allows efficient calculatio... | 9 | 0 | ['Array', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3'] | 1 |
number-of-ways-to-split-array | ✅ One Line Solution | one-line-solution-by-mikposp-u8r5 | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)Code #1 - ShortTime complexity: O(n). Space complexi | MikPosp | NORMAL | 2025-01-03T09:54:58.176449+00:00 | 2025-01-03T10:09:46.234392+00:00 | 328 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)
# Code #1 - Short
Time complexity: $$O(n)$$. Space complexity: $$O(n)$$.
```python3
class Solution:
def waysToSplitArray(self, a: List[int]) -> int:
return sum(map(ge,accumulate(a),[*accumulate(a[:0... | 8 | 0 | ['Array', 'Prefix Sum', 'Python', 'Python3'] | 1 |
number-of-ways-to-split-array | python beginner friendly code must check it for simply code and neat explaination . | python-beginner-friendly-code-must-check-nk3t | Intuitionsoo basically what we want in this problem is that to find the number of pairs that can be formed with splitting the nums ( array) given through.thats | RAHULMOST | NORMAL | 2025-01-03T05:18:29.591618+00:00 | 2025-01-03T05:18:29.591618+00:00 | 204 | false | # Intuition
soo basically what we want in this problem is that to find the number of pairs that can be formed with splitting the nums ( array) given through.thats the basic idea of this problem except under few critereas for those who are new to solve python u totally need not to understand the whole logic of the code ... | 8 | 0 | ['Python3'] | 1 |
number-of-ways-to-split-array | Java Easy Solutions | 3ms | java-easy-solutions-3ms-by-swaythak-kpff | IntuitionThe problem involves finding the number of ways to split an array into two non-empty parts such that the sum of the left subarray is greater than or eq | SwaythaK | NORMAL | 2025-01-03T04:42:11.442521+00:00 | 2025-01-03T04:42:11.442521+00:00 | 179 | false | # Intuition
The problem involves finding the number of ways to split an array into two non-empty parts such that the sum of the left subarray is greater than or equal to the sum of the right subarray. This can be achieved efficiently by maintaining cumulative sums for both parts and updating them as we iterate through ... | 7 | 0 | ['Java'] | 1 |
number-of-ways-to-split-array | Simple Prefix Sum Approach✅✅ | simple-prefix-sum-approach-by-arunk_leet-8sft | IntuitionThe problem involves splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right p | arunk_leetcode | NORMAL | 2025-01-03T03:57:33.310744+00:00 | 2025-01-03T03:57:33.310744+00:00 | 764 | false | # Intuition
The problem involves splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part. To solve this, we need to efficiently calculate the prefix sum and compare the left and right sums for every possible split point.
# Approach
1. Comput... | 7 | 0 | ['Array', 'Prefix Sum', 'Python', 'C++', 'Java'] | 4 |
number-of-ways-to-split-array | ✅💯🔥Simple Code🚀📌|| 🔥✔️Easy to understand🎯 || 🎓🧠Beats 100%🔥|| Beginner friendly💀💯 | simple-code-easy-to-understand-beats-100-zsbj | 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 | atishayj4in | NORMAL | 2024-07-31T20:17:41.686081+00:00 | 2024-08-01T19:23:27.533032+00:00 | 326 | 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. $$O(n)$$ --... | 7 | 1 | ['Array', 'C', 'Prefix Sum', 'Python', 'C++', 'Java'] | 0 |
number-of-ways-to-split-array | JAVA | O(N) Time and O(1) Space | Easy | java-on-time-and-o1-space-easy-by-sambar-jmeo | \nclass Solution {\n public int waysToSplitArray(int[] nums) {\n long sum = 0;\n for(int num : nums) {\n sum += num;\n }\n | sambarannnn | NORMAL | 2022-05-14T16:26:00.778985+00:00 | 2022-05-14T16:36:44.972844+00:00 | 1,036 | false | ```\nclass Solution {\n public int waysToSplitArray(int[] nums) {\n long sum = 0;\n for(int num : nums) {\n sum += num;\n }\n \n long leftSum = 0;\n int res = 0;\n for(int i = 0; i < nums.length-1; i++) {\n leftSum += nums[i];\n long r... | 7 | 0 | ['Prefix Sum'] | 6 |
number-of-ways-to-split-array | JAVA || 2ms || 100% Beat || Easy to Understand for Beginners🙌 | java-2ms-100-beat-easy-to-understand-for-fyho | IntuitionApproachComplexity
Time complexity:o(n)
Space complexity:o(1)
Code | sathish-77 | NORMAL | 2025-01-03T04:45:55.145274+00:00 | 2025-01-03T04:45:55.145274+00:00 | 213 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
prefix sum, suffix sum
# Complexity
- Time complexity:o(n)
- Space complexity:o(1)
# Code
```java []
class Solution {
public int waysToSplitArray(int[] nums) {
long total_sum=0;
for(int i=0;i<nums.len... | 6 | 0 | ['Array', 'Prefix Sum', 'Java'] | 2 |
number-of-ways-to-split-array | 0ms || Beats 100% || Efficient & Easy C++ Solution || Full Explained | 0ms-beats-100-efficient-easy-c-solution-kfqyb | IntuitionThe goal is to find the number of ways to split the array into two non-empty parts such that the sum of the left part is greater than or equal to the s | HarshitSinha | NORMAL | 2025-01-03T01:27:54.058682+00:00 | 2025-01-03T01:27:54.058682+00:00 | 233 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The goal is to find the number of ways to split the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part.
If we know the total sum of the array, we can calculate the sum of... | 6 | 0 | ['Array', 'Prefix Sum', 'C++'] | 2 |
number-of-ways-to-split-array | Simple O(n) Time and O(1) Space Solution || Python | simple-on-time-and-o1-space-solution-pyt-ycvx | If you have any doubts or suggestion, put in comments.\n\nclass Solution(object):\n def waysToSplitArray(self, nums):\n lSum, rSum = 0, sum(nums)\n | krush_r_sonwane | NORMAL | 2022-11-05T12:08:40.262047+00:00 | 2022-11-05T12:08:40.262082+00:00 | 545 | false | If you have any ***doubts*** or ***suggestion***, put in comments.\n```\nclass Solution(object):\n def waysToSplitArray(self, nums):\n lSum, rSum = 0, sum(nums)\n ans = 0\n for i in range(len(nums) - 1):\n lSum += nums[i]\n rSum -= nums[i]\n if lSum >= rSum: ans ... | 6 | 0 | ['Python', 'Python3'] | 1 |
number-of-ways-to-split-array | 🌟🧩 One Loop, Many Splits:💡 The Optimized Prefix Sum Trick!⚡🚀 | one-loop-many-splits-the-optimized-prefi-5jt2 | Intuition 🤔We split the array into two 🧩 parts: left and right. The goal is to count how many splits make the left sum ≥ right sum. Easy, right? 🚀Approach 🛠️
To | mansimittal935 | NORMAL | 2025-01-03T06:07:40.779519+00:00 | 2025-01-03T06:07:40.779519+00:00 | 196 | false | # Intuition 🤔
We split the array into two 🧩 parts: left and right. The goal is to count how many splits make the left sum ≥ right sum. Easy, right? 🚀
---
# Approach 🛠️
1. Total Sum ➕: Compute the sum of all elements.
2. Iterate 🔄:
- Start with leftSum = 0.
- For each index i, add the current element to l... | 5 | 0 | ['Swift', 'Prefix Sum', 'Python', 'C++', 'Java', 'Go', 'Rust', 'Ruby', 'JavaScript', 'Dart'] | 1 |
number-of-ways-to-split-array | Explanation with Diagram 💨 | Prefix Sum | O(N) Approach | Beats 100% Users | Python | explanation-with-diagram-prefix-sum-on-a-13q2 | IntuitionThe task is to split the array into two parts such that the sum of the left subarray is greater than or equal to the sum of the right subarray. To achi | i_god_d_sanjai | NORMAL | 2025-01-03T04:45:09.032758+00:00 | 2025-01-03T04:47:21.475634+00:00 | 83 | false | # Intuition
The task is to split the array into two parts such that the sum of the left subarray is greater than or equal to the sum of the right subarray. To achieve this efficiently, we can use a prefix sum approach to compute the cumulative sums of the array and compare the left and right sums for potential splits.
... | 5 | 0 | ['Array', 'Prefix Sum', 'Python3'] | 1 |
number-of-ways-to-split-array | C++ | Prefix Sum | Number of Ways to Split Array | O(n) Time, O(1) Space | c-prefix-sum-number-of-ways-to-split-arr-1l1h | IntuitionThe problem involves finding the number of valid splits in an array where the sum of the left part is greater than or equal to the sum of the right par | AureliusPyron | NORMAL | 2025-01-03T02:53:26.565763+00:00 | 2025-01-03T02:53:26.565763+00:00 | 206 | false | # Intuition
The problem involves finding the number of valid splits in an array where the sum of the left part is greater than or equal to the sum of the right part. A brute-force approach would involve calculating sums for every split, which is inefficient. Instead, we use the **prefix sum** technique to optimize the ... | 5 | 0 | ['Array', 'Prefix Sum', 'C++'] | 1 |
number-of-ways-to-split-array | Easy Java Solution using Arrays - O(n) | easy-java-solution-using-arrays-on-by-de-sf3y | This is one of the easiest question, where we need to compare leftsum and rightsum\n\nNote that we have to take leftSum and rightSum as long type and not of int | devesh096 | NORMAL | 2022-12-04T09:54:24.597220+00:00 | 2022-12-04T09:54:24.597264+00:00 | 1,085 | false | This is one of the easiest question, where we need to compare leftsum and rightsum\n\nNote that we have to take leftSum and rightSum as long type and not of int type looking at the constraints given in the problem.\n\nSteps: \n1.First Calculate the sum of all elements present in the array.\n2.Then Calculate left sum an... | 5 | 0 | ['Array', 'Java'] | 2 |
number-of-ways-to-split-array | Python || Easy || Faster Than 99 % | python-easy-faster-than-99-by-workingpay-y3cn | \nclass Solution:\n def waysToSplitArray(self, nums: List[int]) -> int:\n sumtot= sum(nums)-nums[0]\n leftsum = nums[0]\n res = 0\n | workingpayload | NORMAL | 2022-09-02T19:50:03.954332+00:00 | 2022-09-02T19:50:03.954375+00:00 | 444 | false | ```\nclass Solution:\n def waysToSplitArray(self, nums: List[int]) -> int:\n sumtot= sum(nums)-nums[0]\n leftsum = nums[0]\n res = 0\n \n \n for i in range(1,len(nums)):\n if leftsum>=sumto:\n res+=1\n sumtot-=nums[i]\n leftsum... | 5 | 0 | ['Python'] | 1 |
number-of-ways-to-split-array | ✅Detailed Explanation | 100% Beats | O(N) Solution | Java | C++ | Python | detailed-explanation-100-beats-on-soluti-nyla | IntuitionTo split the array into two non-empty parts such that the left sum is greater than or equal to the right sum, we can efficiently calculate prefix and s | prashanth_d4 | NORMAL | 2025-03-11T01:45:01.401807+00:00 | 2025-03-11T01:45:01.401807+00:00 | 53 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To split the array into two non-empty parts such that the left sum is greater than or equal to the right sum, we can efficiently calculate prefix and suffix sums while iterating through the array.
# Approach
<!-- Describe your approach to ... | 4 | 0 | ['Array', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3'] | 0 |
number-of-ways-to-split-array | Best Solution for Arrays, Prefix Sum 🚀 in C++, Python and Java || 100% working | best-solution-for-arrays-prefix-sum-in-c-9kn8 | Intuition😊 First, calculate the total sum of the array. We will split the array at each index and compare the sum of the left and right parts.Approach👉 Use a pr | BladeRunner150 | NORMAL | 2025-01-04T08:49:53.718849+00:00 | 2025-01-04T08:49:53.718849+00:00 | 22 | false | # Intuition
😊 First, calculate the total sum of the array. We will split the array at each index and compare the sum of the left and right parts.
# Approach
👉 Use a prefix sum to calculate the left part's sum at each index. Subtract it from the total sum to get the right part's sum.
✨ Check if the left sum is grea... | 4 | 0 | ['Array', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3'] | 0 |
number-of-ways-to-split-array | Simple py,JAVA code expalined!! Prefixsum || arrays !! | simple-pyjava-code-expalined-prefixsum-a-4dr2 | Approach:Prefix sumIntuition
read the question twice and understand what they are asking!
Since we need to find the sum of all the previous prefixes and sum of | arjunprabhakar1910 | NORMAL | 2025-01-03T19:02:04.834289+00:00 | 2025-01-03T19:02:04.834289+00:00 | 24 | false |
# Approach:Prefix sum
<!-- Describe your approach to solving the problem. -->
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
> read the question twice and understand what they are asking!
- Since we need to find the sum of all the previous *prefixes* and sum of all values for *suffixes... | 4 | 0 | ['Array', 'Prefix Sum', 'Java', 'Python3'] | 0 |
number-of-ways-to-split-array | Beats 100% || O(1) space || C++ || Prefix Sum | beats-100-o1-space-c-prefix-sum-by-akash-jyg3 | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | akash92 | NORMAL | 2025-01-03T10:50:54.582963+00:00 | 2025-01-03T10:50:54.582963+00:00 | 79 | false | # Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
#define ll long long
class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
int n = nums.size();
... | 4 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | Easy to understand Prefix Sum with O(1) space || C++ beats 100% Time Complexity. | easy-to-understand-prefix-sum-with-o1-sp-veyb | IntuitionThe problem requires splitting an array into two parts such that the sum of the left part is greater than or equal to the sum of the right part. This c | RishubhChahar2001 | NORMAL | 2025-01-03T09:08:33.040274+00:00 | 2025-01-03T09:14:31.200970+00:00 | 128 | false | # Intuition
The problem requires splitting an array into two parts such that the sum of the left part is greater than or equal to the sum of the right part. This can be solved by maintaining a running sum or using a prefix sum array for efficient computation of sums.
# Approach 1: Using Running Sums
1. **Calculate Tot... | 4 | 0 | ['C++'] | 2 |
number-of-ways-to-split-array | 💢Faster✅💯 Lesser C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 💯 | faster-lesser-cpython3javacpythoncexplai-rk0w | IntuitionApproach
JavaScript Code --> https://leetcode.com/problems/number-of-ways-to-split-array/submissions/1495859295
C++ Code --> https://leetcode.com/probl | Edwards310 | NORMAL | 2025-01-03T05:26:31.211944+00:00 | 2025-01-03T16:23:26.718819+00:00 | 96 | false | 
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your first thoughts on how to solve this problem. -->
- ***JavaScript Code -->*** https:... | 4 | 0 | ['Array', 'C', 'Prefix Sum', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'C#'] | 0 |
number-of-ways-to-split-array | 💡 C++ | O(N) | Very Easy Sol | Simple Logic | With Full Explanation ✏ | c-on-very-easy-sol-simple-logic-with-ful-u40e | IntuitionWe have to find the count of the left subarrays having sum >= the right subarraysApproach
We will store the sum of the whole array in the rs
Then we wi | Tusharr2004 | NORMAL | 2025-01-03T05:11:32.762909+00:00 | 2025-01-03T05:11:32.762909+00:00 | 26 | false | # Intuition
We have to find the count of the left subarrays having sum >= the right subarrays
# Approach
- We will store the sum of the whole array in the `rs`
- Then we will traverse the array and use sliding window concept and store the element sum in the `ls` and subtract it from the `rs`
- Run the check if the `ls... | 4 | 0 | ['Array', 'Math', 'Sliding Window', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | 📌100% || C++ || Python || Prefix Sum ||🌟Beginner-Friendly | 100-c-python-prefix-sum-beginner-friendl-m0gp | IntuitionTo solve this problem, we need to find the valid split points in the array where the sum of the first part is greater than or equal to the sum of the s | Rohithaaishu16 | NORMAL | 2025-01-03T04:41:28.174535+00:00 | 2025-01-03T04:41:28.174535+00:00 | 71 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To solve this problem, we need to find the valid split points in the array where the sum of the first part is greater than or equal to the sum of the second part. By iterating through the array and maintaining a running total for both parts... | 4 | 0 | ['Array', 'Prefix Sum', 'C++', 'Python3'] | 0 |
number-of-ways-to-split-array | C# Solution for Number of Ways To Split Array Problem | c-solution-for-number-of-ways-to-split-a-zgrn | IntuitionThe goal is to determine how many valid splits exist in the array, where the sum of the left subarray is greater than or equal to the sum of the right | Aman_Raj_Sinha | NORMAL | 2025-01-03T04:30:09.846776+00:00 | 2025-01-03T04:30:09.846776+00:00 | 56 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The goal is to determine how many valid splits exist in the array, where the sum of the left subarray is greater than or equal to the sum of the right subarray. The intuition is to avoid recalculating subarray sums repeatedly by leveraging ... | 4 | 0 | ['C#'] | 0 |
number-of-ways-to-split-array | Easy Approach in O(n) | Python | easy-approach-in-on-python-by-omtejaswin-jcll | IntuitionWe are given a list nums of integers. The goal is to determine how many ways we can split the array nums into two non-empty subarrays such that the sum | OmTejaswini2802 | NORMAL | 2025-01-03T03:39:34.284559+00:00 | 2025-01-03T03:39:34.284559+00:00 | 111 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We are given a list nums of integers. The goal is to determine how many ways we can split the array nums into two non-empty subarrays such that the sum of the elements in the left subarray is greater than or equal to the sum of the elements... | 4 | 0 | ['Prefix Sum', 'Python3'] | 0 |
number-of-ways-to-split-array | [Rust / Elixir] 2 passes | rust-2-passes-by-minamikaze392-7ktt | Let the sum of the first i + 1 elements be left, and the sum of the others be right. Then left + right = total where total is the total sum.
If a cut is valid, | Minamikaze392 | NORMAL | 2025-01-03T02:19:50.184622+00:00 | 2025-01-03T02:51:03.341636+00:00 | 74 | false | Let the sum of the first `i + 1` elements be `left`, and the sum of the others be `right`. Then `left + right = total` where `total` is the total sum.
If a cut is valid, then:
```text
left >= right
left >= total - left
total - 2 * left <= 0
```
# Complexity
- Time complexity: $$O(n)$$
- Space complexity: $$O(1)$$
# C... | 4 | 0 | ['Rust', 'Elixir'] | 1 |
number-of-ways-to-split-array | Easiest C++ solution || Beginner-friendly approach || O(N) time complexity | easiest-c-solution-beginner-friendly-app-5gze | Code\n\nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n long long int sumLeft = 0, sumRight = 0, ans = 0;\n for(int i=0 | prathams29 | NORMAL | 2023-07-03T11:18:35.976422+00:00 | 2023-07-03T11:18:35.976457+00:00 | 216 | false | # Code\n```\nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n long long int sumLeft = 0, sumRight = 0, ans = 0;\n for(int i=0; i<nums.size(); i++)\n sumRight += nums[i];\n for(int i=0; i<nums.size()-1; i++){\n sumLeft += nums[i];\n sumRight... | 4 | 0 | ['C++'] | 0 |
number-of-ways-to-split-array | Python3 || beats 99% || simple solution | python3-beats-99-simple-solution-by-salt-xio0 | \nclass Solution:\n\tdef waysToSplitArray(self, nums: List[int]) -> int:\n s0 = sum(nums)\n s1, d = 0, 0\n for i in nums[:-1]:\n | Saltkroka | NORMAL | 2022-12-04T20:21:22.605776+00:00 | 2022-12-04T20:32:03.342748+00:00 | 375 | false | ```\nclass Solution:\n\tdef waysToSplitArray(self, nums: List[int]) -> int:\n s0 = sum(nums)\n s1, d = 0, 0\n for i in nums[:-1]:\n s1 += i\n if s1 >= (s0-s1):\n d += 1\n return d\n``` | 4 | 0 | ['Array', 'Prefix Sum', 'Python3'] | 0 |
number-of-ways-to-split-array | [C++/Python3] Very Concise Prefix Sum O(1) | cpython3-very-concise-prefix-sum-o1-by-s-z466 | Python\n\nclass Solution:\n def waysToSplitArray(self, nums: List[int]) -> int:\n lsum, rsum, ans = 0, sum(nums), 0\n for i in range(len(nums) | Suraj1199 | NORMAL | 2022-05-14T16:11:39.252074+00:00 | 2022-05-14T16:15:52.618272+00:00 | 377 | false | **Python**\n```\nclass Solution:\n def waysToSplitArray(self, nums: List[int]) -> int:\n lsum, rsum, ans = 0, sum(nums), 0\n for i in range(len(nums) - 1):\n lsum += nums[i]\n rsum -= nums[i]\n ans += (lsum >= rsum)\n return ans\n\t\t\n```\n\n**C++**\n```\nclass ... | 4 | 0 | ['Prefix Sum', 'Python3'] | 1 |
number-of-ways-to-split-array | Short and easy python solution using a queue. O(n) | short-and-easy-python-solution-using-a-q-skke | \nAlgorithm.\nInitialize 2 sums, Left and Right. where left = nums[0] and right will sum of the rest of the elements from 1 to n.\nNow create a queue with nums[ | vineeth_moturu | NORMAL | 2022-05-14T16:00:52.236751+00:00 | 2022-05-14T16:18:34.823612+00:00 | 447 | false | \n**Algorithm.**\nInitialize 2 sums, Left and Right. where `left = nums[0]` and right will sum of the rest of the elements from 1 to n.\nNow create a queue with `nums[1:]`\nPop elements from the queue one by one until none are left.\nFor each element\n\t - add this to left sum\n\t - subtract this from right sum\n\t - i... | 4 | 0 | ['Python'] | 2 |
number-of-ways-to-split-array | Easy Aproach O(N) - Time complexity only | easy-aproach-on-time-complexity-only-by-0dref | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | muhammednihas2218 | NORMAL | 2025-01-03T19:24:52.729383+00:00 | 2025-01-03T19:24:52.729383+00:00 | 12 | 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
`... | 3 | 0 | ['Array', 'Python3'] | 0 |
number-of-ways-to-split-array | Beginner Friendly Solution - Core Logic Building approach. Java | Python | C++ | beginner-friendly-solution-core-logic-bu-idmz | IntuitionThe problem involves splitting the array into two non-empty parts and comparing their sums. Instead of recalculating sums repeatedly for every split, w | v-athithyaramaa | NORMAL | 2025-01-03T18:44:21.219489+00:00 | 2025-01-03T18:44:21.219489+00:00 | 9 | false | # Intuition
The problem involves splitting the array into two non-empty parts and comparing their sums. Instead of recalculating sums repeatedly for every split, we can use a prefix sum array to efficiently compute the left and right sums for each split.
# Approach
1. Compute the prefix sum array, where `prefixSum[i]`... | 3 | 0 | ['Python3'] | 0 |
number-of-ways-to-split-array | ✨ Explained in Hindi!! 🤩 Prefix Ka Jaadu🔮 | explained-in-hindi-prefix-ka-jaadu-by-an-9v53 | Intuition"Yaar, problem yeh hai ki array ko do parts mein todna hai—ek left aur ek right. Aur left ka sum right se bada ya at least barabar hona chahiye. Matlab | ANJ_07 | NORMAL | 2025-01-03T09:15:49.326346+00:00 | 2025-01-03T09:15:49.326346+00:00 | 16 | false | # Intuition
<img src="https://assets.leetcode.com/users/images/e6e437b6-a4c0-4c91-beaf-540d635534d9_1735895579.0807717.png" width="200" height="30">
"Yaar, problem yeh hai ki array ko do parts mein todna hai—ek left aur ek right. Aur left ka sum right se bada ya at least barabar hona chahiye. Matlab, socho ek jagah c... | 3 | 0 | ['C++'] | 1 |
number-of-ways-to-split-array | 🧩 Ways to Split Array (⏱ Runtime: 0ms, 🚀 Beats 100%) | ways-to-split-array-runtime-0ms-beats-10-5woc | ✨ IntuitionWhen splitting an array, the goal is to count splits where the sum of the left part is greater than or equal to the sum of the right part. By leverag | rishisahu193 | NORMAL | 2025-01-03T06:15:46.544924+00:00 | 2025-01-03T06:15:46.544924+00:00 | 43 | false |
### ✨ Intuition
When splitting an array, the goal is to count splits where the sum of the left part is greater than or equal to the sum of the right part. By leveraging prefix sums, we can efficiently calculate the left and right segment sums during a single pass.
### 🔍 Approach
1️⃣ Compute the total sum of the... | 3 | 0 | ['Array', 'Prefix Sum', 'C++'] | 1 |
number-of-ways-to-split-array | Simple Python Solution || Prefix Sum || O(n) | simple-python-solution-prefix-sum-on-by-zjs6l | if it's help, please up ⬆ vote! ❤️Complexity
Time complexity: O(n)
Space complexity: O(n)
Code | shishirRsiam | NORMAL | 2025-01-03T06:04:47.292107+00:00 | 2025-01-03T06:04:47.292107+00:00 | 35 | false | # if it's help, please up ⬆ vote! ❤️
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def waysToSplitArray(self, nums: List[int]) -> int:
pref = ... | 3 | 0 | ['Array', 'Hash Table', 'Math', 'Prefix Sum', 'Python', 'Python3'] | 1 |
number-of-ways-to-split-array | Simple C++ Solution || Prefix Sum || O(n) | simple-c-solution-by-shishirrsiam-kf7v | if it's help, please up ⬆ vote! ❤️Complexity
Time complexity: O(n)
Space complexity: O(n)
Code | shishirRsiam | NORMAL | 2025-01-03T05:57:55.780760+00:00 | 2025-01-03T06:06:34.710163+00:00 | 85 | false | # if it's help, please up ⬆ vote! ❤️
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int waysToSplitArray(vector<int>& nums)
{
vector<... | 3 | 0 | ['Array', 'Hash Table', 'Math', 'Prefix Sum', 'C++'] | 4 |
number-of-ways-to-split-array | ✅ Beginner Friendly | Easy to Understand | 100% Faster | Prefix Sum | Java | Video Explanation 🔥 | beginner-friendly-easy-to-understand-100-nkim | IntuitionTo solve this problem, the key observation is that for any split of the array into two non-empty parts, the left part's sum (psum) must be greater than | sahilpcs | NORMAL | 2025-01-03T05:03:40.955133+00:00 | 2025-01-03T05:03:55.985726+00:00 | 53 | false | # Intuition
To solve this problem, the key observation is that for any split of the array into two non-empty parts, the left part's sum (`psum`) must be greater than or equal to the right part's sum (`sum - psum`). By iterating through the array and maintaining a running prefix sum, we can efficiently check this condit... | 3 | 0 | ['Array', 'Prefix Sum', 'Java'] | 1 |
number-of-ways-to-split-array | Easy approach for beginner || Prefix Sum Algorithm | easy-approach-for-beginner-prefix-sum-al-ats5 | IntuitionTo determine the number of valid splits, we need to divide the array into two non-empty parts at each possible index and compare their sums. The key in | minhle106cse | NORMAL | 2025-01-03T04:29:34.568311+00:00 | 2025-01-03T04:29:34.568311+00:00 | 91 | false | # Intuition
To determine the number of valid splits, we need to divide the array into two non-empty parts at each possible index and compare their sums. The key insight is to precompute the prefix sums of the array, allowing efficient calculation of the sum for any subarray.
# Approach
1. **Prefix Sum Calculation**: ... | 3 | 0 | ['Prefix Sum', 'JavaScript'] | 1 |
number-of-ways-to-split-array | Simple Approach with Efficient time and space complexity :- | simple-approach-with-efficient-time-and-i2y4l | IntuitionThe problem involves splitting an array such that the left sum is greater than or equal to the right sum.
By precomputing the total sum (rSum) and dyna | Ayush_ranjan00 | NORMAL | 2025-01-03T04:29:16.539481+00:00 | 2025-01-03T04:29:16.539481+00:00 | 45 | false | # Intuition
The problem involves splitting an array such that the left sum is greater than or equal to the right sum.
By precomputing the total sum (rSum) and dynamically adjusting it during iteration, we avoid recomputation, making the solution efficient.
## At every step, the split condition
lSum≥rSum ensures only v... | 3 | 0 | ['Array', 'Java'] | 1 |
number-of-ways-to-split-array | Simple and Easy Solution | ✅Beats 100% | C++| Java | Python | JavaScript | simple-and-easy-solution-beats-100-c-jav-90p0 | ⬆️Upvote if it helps ⬆️Connect with me on Linkedin [Bijoy Sing]Solution in C++, Python, Java, and JavaScriptIntuitionThe goal is to split the array such that th | BijoySingh7 | NORMAL | 2025-01-03T04:04:20.841743+00:00 | 2025-01-03T04:04:20.841743+00:00 | 231 | false | # ⬆️Upvote if it helps ⬆️
---
## Connect with me on Linkedin [Bijoy Sing]
---
### Solution in C++, Python, Java, and JavaScript
```cpp []
class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
long long tot = 0, left = 0;
int ans = 0;
for (int num : nums) {
tot ... | 3 | 2 | ['Array', 'C', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 9 |
number-of-ways-to-split-array | Java🍵|Prefix Sum✅|Easy🔥 | javaprefix-sumeasy-by-mani-26-t5nt | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | mani-26 | NORMAL | 2025-01-03T03:27:23.053176+00:00 | 2025-01-03T03:27:23.053176+00:00 | 19 | false | # Complexity
- Time complexity: **O(n)**
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: **O(1)**
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int waysToSplitArray(int[] nums) {
long right = 0;
long left = 0;
for... | 3 | 0 | ['Array', 'Prefix Sum', 'Java'] | 0 |
number-of-ways-to-split-array | beats 100% || very easy solution! | beats-100-very-easy-solution-by-ishwarya-v1lt | IntuitionThe intuition is to find the number of ways the array can be splitted such the left subarray is always maximum then the right subarray.Note: The array | Ishwaryaice | NORMAL | 2025-01-03T03:25:01.266799+00:00 | 2025-01-03T03:25:01.266799+00:00 | 11 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The intuition is to find the number of ways the array can be splitted such the left subarray is always maximum then the right subarray.
**Note:** The array also consists of negative integers and should be split such that atleast one eleme... | 3 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | Beats 100% of users with C++ | beats-100-of-users-with-c-by-sahilsaw-cb63 | IntuitionTo solve the problem, I first considered the idea of iterating through the array and trying to split it into two non-empty parts. The challenge is to f | SahilSaw | NORMAL | 2025-01-03T00:07:22.897358+00:00 | 2025-01-03T00:07:22.897358+00:00 | 168 | false |
# Intuition
To solve the problem, I first considered the idea of iterating through the array and trying to split it into two non-empty parts. The challenge is to find a point where the sum of the left part is greater than or equal to the sum of the right part. This is where the comparison between the left and right su... | 3 | 0 | ['C++'] | 1 |
number-of-ways-to-split-array | Beats 100% of users with C++|| Using Presum and Postsum || Faster Solution || Easy to Understand || | beats-100-of-users-with-c-using-presum-a-j6mg | Intuition\n Describe your first thoughts on how to solve this problem. \n\nif you like the approach please upvote it\n\n\n\n\n\n\nif you like the approach pleas | abhirajpratapsingh | NORMAL | 2023-12-21T10:53:27.863465+00:00 | 2023-12-21T10:53:27.863504+00:00 | 80 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nif you like the approach please upvote it\n\n\n\n\n\n\nif you like the approach please upvote it\n\n\n\n\n\n\n\n# Approach... | 3 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | 7 Lines Python Code || Beats 82 % | 7-lines-python-code-beats-82-by-jigglyyp-wght | Intuition\nso you have to count the number of ways we can split array basically in two parts such that sum of left Part >= right Part\nso first find the total | JigglyyPuff | NORMAL | 2023-07-12T06:56:47.543697+00:00 | 2023-07-12T06:56:47.543724+00:00 | 127 | false | # Intuition\nso you have to count the number of ways we can split array basically in two parts such that `sum of left Part >= right Part`\nso first find the total sum of the array\nand loop through your given array and add each element to left Part and subtract it from your totalSum (which is now your right Part)\nif ... | 3 | 0 | ['Python3'] | 0 |
number-of-ways-to-split-array | Simple JAVA Solution | simple-java-solution-by-kumarabhinav88-8alh | ```\nclass Solution {\n public int waysToSplitArray(int[] nums) {\n long sum = 0;\n for(int i : nums){\n sum+=i;\n }\n | kumarabhinav88 | NORMAL | 2022-05-14T16:31:50.529683+00:00 | 2022-05-14T16:31:50.529723+00:00 | 941 | false | ```\nclass Solution {\n public int waysToSplitArray(int[] nums) {\n long sum = 0;\n for(int i : nums){\n sum+=i;\n }\n int sol = 0;\n long localSum = 0;\n for(int i=0; i<nums.length-1;i++){\n localSum += nums[i];\n if(localSum >= sum-localSum... | 3 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | javascript 112ms | javascript-112ms-by-growthfromnewbie-fvv5 | \n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar waysToSplitArray = function(nums) {\n let result = 0;\n let letsum = 0;\n let rightsum | growthfromnewbie | NORMAL | 2022-05-14T16:22:04.276737+00:00 | 2022-07-17T11:18:52.635200+00:00 | 185 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar waysToSplitArray = function(nums) {\n let result = 0;\n let letsum = 0;\n let rightsum = nums.reduce((a,b)=> a+b);\n let end = nums.length-1;\n for (let i = 0;i<end;i++) {\n letsum+=nums[i];\n rightsum-=nums[i];\n if ... | 3 | 1 | ['JavaScript'] | 0 |
number-of-ways-to-split-array | C++ || O(n) solution || With comments | c-on-solution-with-comments-by-hrishilam-knz3 | \nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n \n long long sum=0, lsum=0,rsum=0;\n \n //calculate sum | hrishilamdade | NORMAL | 2022-05-14T16:21:52.858185+00:00 | 2022-05-14T16:21:52.858212+00:00 | 258 | false | ```\nclass Solution {\npublic:\n int waysToSplitArray(vector<int>& nums) {\n \n long long sum=0, lsum=0,rsum=0;\n \n //calculate sum of all values in nums\n for(auto i:nums){\n sum+=i;\n }\n \n int n=nums.size(),res=0;\n \n for(int i=0;i<n;... | 3 | 0 | ['C', 'Prefix Sum'] | 1 |
number-of-ways-to-split-array | CPP | Short and Simple | Prefix Sum | cpp-short-and-simple-prefix-sum-by-amirk-2cvz | \nclass Solution {\npublic:\n typedef long ll;\n int waysToSplitArray(vector<int>& v,ll ss=0,ll ans=0) {\n ll s=accumulate(v.begin(),v.end(),0LL);\ | amirkpatna | NORMAL | 2022-05-14T16:01:33.739303+00:00 | 2022-05-14T16:03:30.366534+00:00 | 260 | false | ```\nclass Solution {\npublic:\n typedef long ll;\n int waysToSplitArray(vector<int>& v,ll ss=0,ll ans=0) {\n ll s=accumulate(v.begin(),v.end(),0LL);\n for(int i=0;i<v.size()-1;i++){\n ss+=v[i];\n ans+=ss>=s-ss;\n }\n return ans;\n }\n};\n``` | 3 | 0 | ['Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | beginner friendly solution beats 99.9% user in JAVA | beginner-friendly-solution-beats-999-use-5oek | IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(1)
Code | soumya_kumar_gupta | NORMAL | 2025-01-21T06:38:26.246389+00:00 | 2025-01-21T06:38:26.246389+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | Count Valid Array Splits Based on Prefix and Suffix Sums | count-valid-array-splits-based-on-prefix-l22z | IntuitionThe problem seems to involve checking if the prefix sum of an array up to a certain index is greater than or equal to the sum of the remaining elements | abhivatsa1185 | NORMAL | 2025-01-04T14:08:45.208267+00:00 | 2025-01-04T14:08:45.208267+00:00 | 11 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem seems to involve checking if the prefix sum of an array up to a certain index is greater than or equal to the sum of the remaining elements of the array. This makes the problem a good fit for a prefix sum approach.
# Approach
<... | 2 | 0 | ['Java'] | 1 |
number-of-ways-to-split-array | ✅ Java ✅ Beats 100% ✅ Simple ✅ | java-beats-100-simple-by-abdullohmaraimo-8lby | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | AbdullohMaraimov | NORMAL | 2025-01-03T17:54:40.711325+00:00 | 2025-01-03T17:55:49.459419+00:00 | 21 | false |
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int waysToSplitArray(int[] nums) {
long leftSum = 0;
int res = 0;
long right... | 2 | 0 | ['Java'] | 2 |
number-of-ways-to-split-array | Beats 100% ✨ | beats-100-by-hriii11-hq3f | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(1)
Code | Hriii11 | NORMAL | 2025-01-03T16:56:39.858133+00:00 | 2025-01-03T16:56:39.858133+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
$$O(n)$$
- Space complexity:
$$O(1)$$
# Code
```cpp []
class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
lon... | 2 | 0 | ['Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | Simple Java Solution with O(n) tc | Beats 100%. | simple-java-solution-with-on-tc-beats-10-ipse | IntuitionApproachTo solve this problem, first calculate the total sum of the array. Then, iterate through the array while keeping a running sum for the left par | AnaghaBharadwaj | NORMAL | 2025-01-03T15:42:29.304744+00:00 | 2025-01-03T15:42:29.304744+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
To solve this problem, first calculate the total sum of the array. Then, iterate through the array while keeping a running sum for the left part (leftsum). At each step, check if leftsum is greater than or equal to the remaining... | 2 | 0 | ['Prefix Sum', 'Java'] | 0 |
number-of-ways-to-split-array | 6 lines logic || very simple | 6-lines-logic-very-simple-by-shashwat191-g3m6 | Code | shashwat1915 | NORMAL | 2025-01-03T14:37:06.323869+00:00 | 2025-01-03T14:37:06.323869+00:00 | 12 | false | # Code
```cpp []
class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
int ans=0;
long long total_sum=accumulate(nums.begin(),nums.end(),0L),last_sum=0;
for(int i=0;i<nums.size()-1;i++){
total_sum-=nums[i];
last_sum+=nums[i];
if(total_sum<=las... | 2 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | C++ || Prefix Sum || Number of Ways to Split Array - Solution Explanation || O(n) Time | c-prefix-sum-number-of-ways-to-split-arr-7dor | IntuitionTo determine the number of valid splits in the array, we need to check whether the sum of elements in the left part (up to index i) is greater than or | lokeshsingh07 | NORMAL | 2025-01-03T14:11:05.549350+00:00 | 2025-01-03T14:11:05.549350+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To determine the number of valid splits in the array, we need to check whether the sum of elements in the left part (up to index i) is greater than or equal to the sum of elements in the right part (from index i+1 onward). Using prefix sums... | 2 | 0 | ['C++'] | 0 |
number-of-ways-to-split-array | O(n) Solution using Prefix Sum with explaination! | 96.04% \ 87.64% | on-solution-using-prefix-sum-with-explai-ghgd | Intuition
The problem involves finding the number of ways to split an array such that the sum of the left part is greater than or equal to the sum of the right | Delta7Actual | NORMAL | 2025-01-03T13:00:04.378276+00:00 | 2025-01-03T13:00:04.378276+00:00 | 9 | false | # Intuition
- The problem involves finding the number of ways to split an array such that the sum of the left part is greater than or equal to the sum of the right part.
- Initially, it may seem like calculating the sums of both parts for every split is necessary, but this can be optimized by using cumulative sums.
---... | 2 | 0 | ['Array', 'Prefix Sum', 'Python3'] | 0 |
number-of-ways-to-split-array | Easy JavaScript Solution ❤️⭐ | easy-javascript-solution-by-priyanshuson-8f0k | IntuitionSo Here We need to observe one things which is that , at every index we need to verify if sum of left side is** >=** sum of right side , that's why i w | priyanshusoniii | NORMAL | 2025-01-03T12:42:42.805621+00:00 | 2025-01-03T12:42:42.805621+00:00 | 20 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
So Here We need to observe one things which is that , at every index we need to verify if sum of left side is** >=** sum of right side , that's why i will always point to last index-1 (at max.) to satisfy given question constraint , Hence *... | 2 | 0 | ['JavaScript'] | 0 |
number-of-ways-to-split-array | ✅✅Beats 100%🔥C++🔥Python|| 🚀🚀Super Simple and Efficient Solution🚀🚀||🔥Python🔥C++✅✅ | beats-100cpython-super-simple-and-effici-jf2r | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | shobhit_yadav | NORMAL | 2025-01-03T12:41:29.094735+00:00 | 2025-01-03T12:41:29.094735+00:00 | 9 | false | # Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int waysToSplitArray(vector<int>& nums) {
int ans = 0;
long prefix = 0;
long suffi... | 2 | 0 | ['Array', 'Prefix Sum', 'C++', 'Python3'] | 0 |
number-of-ways-to-split-array | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-m5ka | IntuitionTo split an array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part, we can leverage pr | CodeWithSparsh | NORMAL | 2025-01-03T12:35:50.831541+00:00 | 2025-01-03T12:35:50.831541+00:00 | 51 | false | 
---
### Intuition
To split an array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part, we can leverage prefix and suffix sums. We maintain... | 2 | 0 | ['Array', 'C', 'Prefix Sum', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 2 |
number-of-ways-to-split-array | Easy C++ solution || Beats 100% | easy-c-solution-beats-100-by-prashant_71-ozs6 | IntuitionWhen splitting an array into two non-empty subarrays, the sum of elements in the left subarray and the right subarray determines whether the split is v | prashant_71200 | NORMAL | 2025-01-03T11:54:33.840576+00:00 | 2025-01-03T11:54:33.840576+00:00 | 20 | false | # Intuition
When splitting an array into two non-empty subarrays, the sum of elements in the left subarray and the right subarray determines whether the split is valid. A split is valid if the left subarray's sum is greater than or equal to the right subarray's sum. The challenge is to efficiently compute these sums fo... | 2 | 0 | ['C++'] | 0 |
number-of-ways-to-split-array | easy solution prefix sum | easy-solution-prefix-sum-by-harshulgarg-7xf9 | IntuitionThe problem involves splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right p | harshulgarg | NORMAL | 2025-01-03T11:32:03.723342+00:00 | 2025-01-03T11:32:03.723342+00:00 | 10 | false | # Intuition
The problem involves splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part. To solve this, we need a way to calculate the sums of both parts efficiently at each potential split point.
# Approach
Suffix Sums: Precompute an array... | 2 | 0 | ['Python3'] | 0 |
number-of-ways-to-split-array | JAVA CODE | java-code-by-ayeshakhan7-d6qv | ApproachUsing cummulativeSum array/Prefix Sum ArrayComplexity
Time complexity: O(N)
Space complexity: O(1)
Code | ayeshakhan7 | NORMAL | 2025-01-03T11:19:35.984418+00:00 | 2025-01-03T11:19:35.984418+00:00 | 13 | false |
# Approach
<!-- Describe your approach to solving the problem. -->
Using cummulativeSum array/Prefix Sum Array
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | 🎉BEATS 100%🏆| O(n)🕒 O(1)🚀 Shortest most easiest optimal code Explained💻 | Running Sum 📈 PHP🌟 | beats-100-on-o1-shortest-most-easiest-op-86im | CodeIntuition:The function waysToSplitArray calculates the number of ways to split an array of integers nums into two non-empty parts such that the sum of the e | webdevanjali | NORMAL | 2025-01-03T11:08:05.458178+00:00 | 2025-01-03T11:38:35.554574+00:00 | 12 | false |
# Code
```php []
class Solution {
/**
* @param Integer[] $nums
* @return Integer
*/
function waysToSplitArray($nums) {
$total = array_sum($nums);
$len = count($nums);
$count = $leftSum = 0;
for ($i = 0; $i < $len - 1; $i++) {
$leftSum += $nums[$i];
... | 2 | 0 | ['Array', 'PHP'] | 0 |
number-of-ways-to-split-array | ✅✅ easy prefix sum approch 🔥🔥 || o(n) Space Complexity || o(n) Time Complexity | easy-prefix-sum-approch-on-space-complex-b3fd | IntuitionThe problem requires dividing the array into two non-overlapping parts such that the sum of the left part is greater than or equal to the sum of the ri | ishanbagra | NORMAL | 2025-01-03T11:07:19.034599+00:00 | 2025-01-03T11:07:19.034599+00:00 | 8 | false | Intuition
The problem requires dividing the array into two non-overlapping parts such that the sum of the left part is greater than or equal to the sum of the right part.
To efficiently solve this, we use prefix sums to compute the total sum of any part of the array in constant time.
Approach
Calculate Prefix Sums fro... | 2 | 0 | ['Prefix Sum', 'C++'] | 2 |
number-of-ways-to-split-array | Kotlin | Rust | kotlin-rust-by-samoylenkodmitry-rpz3 | Join me on Telegramhttps://t.me/leetcode_daily_unstoppable/853Problem TLDRCount splits left_sum >= right_sum #medium #prefix_sumIntuitionPrefix sum can help sol | SamoylenkoDmitry | NORMAL | 2025-01-03T11:05:48.140308+00:00 | 2025-01-03T11:06:12.961092+00:00 | 28 | false | 
https://youtu.be/MjsYNhMxHnM
#### Join me on Telegram
https://t.me/leetcode_daily_unstoppable/853
#### Problem TLDR
Count splits left_sum >= right_sum #medium #prefix_sum
#### Intuition
Prefix sum can... | 2 | 0 | ['Prefix Sum', 'C++', 'Rust', 'Kotlin'] | 0 |
number-of-ways-to-split-array | Easy approach ✅✅✅ || Prefix Sum || Simple Explanation 💯 💯 | easy-approach-prefix-sum-simple-explanat-va5s | IntuitionThe Problem aims to count the number of "valid split points" in an array nums, where the prefix sum of the left subarray (from index 0 to i) is greater | prabhas_rakurthi | NORMAL | 2025-01-03T08:49:47.134218+00:00 | 2025-01-03T08:49:47.134218+00:00 | 15 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The Problem aims to count the number of **"valid split points"** in an array nums, where the prefix sum of the left subarray (from index 0 to i) is greater than or equal to the sum of the right subarray (from index i+1 to the end of the arr... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | 0ms(100.00%) 10.12 MB(95.00%) | Go solution | 0ms10000-1012-mb9500-go-solution-by-kita-79qy | null | kitanoyoru_ | NORMAL | 2025-01-03T08:47:10.647176+00:00 | 2025-01-03T08:47:10.647176+00:00 | 17 | false | ```golang
func waysToSplitArray(nums []int) int {
var result, prefixSum, arraySum int
for _, num := range nums {
arraySum += num
}
for i := 0; i < len(nums)-1; i++ {
prefixSum += nums[i]
if prefixSum >= arraySum-prefixSum {
result++
}
}
return result
}
``` | 2 | 0 | ['Go'] | 1 |
number-of-ways-to-split-array | EASIEST SOLUTION | C/C++ | JavaScript | Python | easiest-solution-cc-javascript-python-by-cg7h | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Nurliaidin | NORMAL | 2025-01-03T08:05:23.894554+00:00 | 2025-01-03T08:05:23.894554+00:00 | 34 | 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
`... | 2 | 0 | ['Array', 'C', 'Prefix Sum', 'C++', 'Python3', 'JavaScript'] | 1 |
number-of-ways-to-split-array | BEATS 100% EASY TO UNDERSTAND | beats-100-easy-to-understand-by-shubhtro-ixcs | IntuitionWe have to split the array in such a way that left hand side is equal to right hand side and atmost 1 element each side
OK
then we make left hand side | ShubhTron | NORMAL | 2025-01-03T08:00:42.208078+00:00 | 2025-01-03T08:00:42.208078+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We have to split the array in such a way that left hand side is equal to right hand side and atmost 1 element each side
OK
then we make left hand side and right hand side total whenever lhs sum is greater than rhs sum we increment the ans... | 2 | 0 | ['Prefix Sum', 'Java'] | 0 |
number-of-ways-to-split-array | Easy Solution || Explained | easy-solution-explained-by-namandas918-bl14 | IntuitionThe idea is to iterate through the array while keeping track of the sum of the elements to the left and right of the current split. If the sum of the l | Hexagon_6_ | NORMAL | 2025-01-03T07:55:54.867785+00:00 | 2025-01-03T07:55:54.867785+00:00 | 13 | false | # Intuition
The idea is to iterate through the array while keeping track of the sum of the elements to the left and right of the current split. If the sum of the left part is greater than or equal to the sum of the right part, it counts as a valid split.
# Approach
1. **Calculate Total Sum:** First, compute the total ... | 2 | 0 | ['Array', 'Math', 'C++'] | 0 |
number-of-ways-to-split-array | BEATS 100%❤️🔥DRY-RUM❤️🔥SLIGHTLY AROUSING🤤 | beats-100dry-rumslightly-arousing-by-int-l7xy | IntuitionThe problem requires splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right p | intbliss | NORMAL | 2025-01-03T07:40:16.366782+00:00 | 2025-01-03T07:40:16.366782+00:00 | 56 | false |
#### **Intuition**
The problem requires splitting the array into two non-empty parts such that the sum of the left part is greater than or equal to the sum of the right part. To achieve this, we can use prefix sums to efficiently calculate the sum of the left and right parts as we iterate through the array.
- As... | 2 | 1 | ['Array', 'Suffix Array', 'Prefix Sum', 'Java'] | 2 |
number-of-ways-to-split-array | 🔥Space optimized Solution for Beginners! Beats 100%!🔥 | space-optimized-solution-for-beginners-b-szjr | ApproachComplexity
Time complexity:O(N)
Space complexity:O(1)
Code | Ansh_Ghawri | NORMAL | 2025-01-03T07:38:57.983172+00:00 | 2025-01-03T07:38:57.983172+00:00 | 4 | false | # Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space c... | 2 | 0 | ['C++'] | 0 |
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