Sponsored
Sponsored
This approach involves iterating through the array and counting the length of contiguous segments of zeros. For each segment of zeros with length n, the total number of zero-filled subarrays is the sum of the first n natural numbers.
For a segment with length n, the number of zero-filled subarrays is (n * (n + 1)) / 2. This formula is derived from the well-known sum formula for the first n integers.
Time Complexity: O(n), where n is the size of the input array. This is because we only traverse the array once.
Space Complexity: O(1), since we only use a constant number of extra variables.
1#include <stdio.h>
2
3int zeroFilledSubarray(int* nums, int numsSize) {
4 int count = 0;
5 long The C solution iterates over the array, increasing the count for each contiguous zero encountered and adds the current count to the total subarrayCount anytime zeros[i] != 0 is encountered.
This approach is slightly variation. We keep track of the number of zero subarrays as we proceed through the array. Whenever a zero is encountered, the count of zero subarrays ending at that position is incremented by the number of zero subarrays ending at the previous position plus one. When a non-zero is encountered, we reset our counter.
Time Complexity: O(n) - traversing each element once to compute the solution.
Space Complexity: O(1) - using only a few extra integer variables for tracking.
using namespace std;
class Solution {
public:
long long zeroFilledSubarray(vector<int>& nums) {
long long count = 0, zeroEndsHere = 0;
for (int num : nums) {
if (num == 0) {
zeroEndsHere++;
count += zeroEndsHere;
} else {
zeroEndsHere = 0;
}
}
return count;
}
};In C++, the concept of cumulative counting with zeroEndsHere is used to build up the count variable. Whenever zeros continue, we add their potential subarrays incrementally.