Leetcode 2750: Ways to Split Array Into Good Subarrays
You are given a binary array ’nums’ containing only 0s and 1s. A good subarray is defined as a contiguous subarray that contains exactly one 1. Your task is to find the number of ways to split the array into good subarrays. Return the result modulo (10^9 + 7).
Problem
Approach
Steps
Complexity
Input: You will be given a binary array 'nums' where each element is either 0 or 1.
Example: nums = [0, 1, 0, 0, 1]
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 1
Output: Return the number of ways to split the array 'nums' into good subarrays, modulo (10^9 + 7).
Example: For nums = [0, 1, 0, 0, 1], the output is 4.
Constraints:
• The result should be returned modulo (10^9 + 7).
Goal: Count the number of ways to split 'nums' into good subarrays where each subarray contains exactly one 1.
Steps:
• Iterate through the array and count contiguous zeros between 1s.
• For each segment of zeros, calculate the number of ways to split the segment into good subarrays.
• Multiply the results for all segments together to get the final answer.
Goal: The length of the binary array is at least 1 and can be up to 100,000. Each element of the array is either 0 or 1.
Steps:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 1
Assumptions:
• The array 'nums' will always contain at least one element.
• Input: For nums = [0, 1, 0, 0, 1]
• Explanation: You need to count how many ways you can split the array into subarrays where each subarray contains exactly one 1. In this case, there are 4 ways.
Approach: We need to find contiguous segments of zeros between the ones and calculate the number of possible ways to split those segments into subarrays containing exactly one 1.
Observations:
• The problem involves finding segments of zeros between ones in the array.
• For each segment of zeros, the number of ways to split it depends on the length of the segment.
• We can iterate through the array, identify the segments between 1s, and count the number of ways to split them.
Steps:
• Initialize a variable to store the result.
• Loop through the array, and for each 1 encountered, calculate the number of ways to split the zeros before and after it.
• Multiply the results of all the splits together and return the result modulo (10^9 + 7).
Empty Inputs:
• The input will always have at least one element, so no empty input case.
Large Inputs:
• If the array is large (close to 100,000 elements), the solution needs to be efficient enough to handle this within the time limit.
Special Values:
• If the array only contains 1s or only 0s, no valid splits are possible.
Constraints:
• Ensure that the result is computed modulo (10^9 + 7).
int numberOfGoodSubarraySplits(vector<int>& nums) {
int mod = (int) 1e9 + 7;
long ans = 1, cnt = 0;
int i = 0, n = nums.size();
while(i < n && nums[i] == 0) cnt++, i++;
if(cnt == n) return 0;
cnt = 0;
for(; i < n; i++) {
if(nums[i]) {
ans = (ans * (cnt + 1)) % mod;
cnt = 0;
} else cnt++;
}
return ans;
}
1 : Function Definition
int numberOfGoodSubarraySplits(vector<int>& nums) {
Defines the function `numberOfGoodSubarraySplits` which takes a vector `nums` containing integers and returns the number of good subarray splits.
2 : Modular Arithmetic
int mod = (int) 1e9 + 7;
Defines the modulus `mod` as (10^9 + 7), which is commonly used in competitive programming to prevent integer overflow.
3 : Variable Initialization
long ans = 1, cnt = 0;
Initializes the result variable `ans` to 1, and the counter `cnt` to 0. `cnt` will track the number of consecutive zeroes in the array.
4 : Size Calculation
int i = 0, n = nums.size();
Initializes the index `i` to 0, and `n` to the size of the `nums` array. These will be used to traverse the array.
5 : Count Leading Zeroes
while(i < n && nums[i] == 0) cnt++, i++;
Counts the number of leading zeroes in the array by iterating through the array until a non-zero element is found.
6 : Edge Case Check
if(cnt == n) return 0;
If the entire array consists of zeroes, there are no valid subarray splits, so the function returns 0.
7 : Reset Counter
cnt = 0;
Resets the `cnt` variable to 0, as we are starting to process non-zero elements in the array.
8 : Iterate Through Array
for(; i < n; i++) {
Starts a loop to iterate over the remaining elements in the `nums` array, starting from the first non-zero element.
9 : Non-zero Element Check
if(nums[i]) {
Checks if the current element `nums[i]` is non-zero.
10 : Update Result
ans = (ans * (cnt + 1)) % mod;
If the current element is non-zero, the result `ans` is updated by multiplying it with `cnt + 1` (the number of subarrays that can be formed from the zeroes), then taking the result modulo `mod`.
11 : Reset Counter After Non-zero Element
cnt = 0;
Resets the counter `cnt` to 0 after processing a non-zero element.
12 : Increment Zero Counter
} else cnt++;
If the current element is zero, the counter `cnt` is incremented to track the consecutive zeroes.
13 : Return Statement
return ans;
Returns the result `ans`, which represents the total number of good subarray splits modulo (10^9 + 7).
Best Case: O(n) where n is the length of nums.
Average Case: O(n).
Worst Case: O(n).
Description: We iterate through the array once to find the segments and compute the result, which is linear in time complexity.
Best Case: O(1).
Worst Case: O(1).
Description: We only use a few variables to store intermediate results, so the space complexity is constant.
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