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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.
Java logic follows the same pattern. Increment zeroEndsHere to count zero subarrays, resetting it whenever a non-zero element appears in the sequence.