Number of Zero-Filled Subarrays

Number of Zero-Filled Subarrays - Problem

Array Math Medium

Given an integer array nums, return the number of subarrays filled with 0.

A subarray is a contiguous non-empty sequence of elements within an array.

Input & Output

Example 1 — Mixed Array

$ Input: nums = [1,3,0,0,2,0,0,4]

› Output: 6

💡 Note: Two segments of zeros: [0,0] contributes 3 subarrays, [0,0] contributes 3 subarrays. Total: 3+3=6

Example 2 — All Zeros

$ Input: nums = [0,0,0,0,0]

› Output: 15

💡 Note: One segment of 5 zeros: 5×(5+1)/2 = 15 subarrays total

Example 3 — No Zeros

$ Input: nums = [2,10,2019]

› Output: 0

💡 Note: No zeros in the array, so no zero-filled subarrays exist

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

M Microsoft 15 a Amazon 12

The key insight is recognizing that for a segment of n consecutive zeros, the number of subarrays is n×(n+1)/2. The optimal approach scans the array once, counting consecutive zeros and applying this formula. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n³) Space: O(1)

Generate all possible subarrays and count how many contain only zeros. For each starting position, expand the subarray and check if all elements are zero.

One Pass with Math Formula

⏱️ Time: O(n) Space: O(1)

Scan the array once. When we encounter consecutive zeros, count the length of the segment. For a segment of length n, it contributes n*(n+1)/2 subarrays.

Brute Force - Check All Subarrays — Algorithm Steps

For each starting index i
For each ending index j ≥ i
Check if subarray nums[i:j+1] contains only zeros
If yes, increment counter

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subarrays

Check Each

Verify if subarray contains only zeros

Count Valid

Increment counter for zero-filled subarrays

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long long solution(int* nums, int numsSize) {
    long long count = 0;
    
    // Check all possible subarrays
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            // Check if subarray nums[i:j+1] contains only zeros
            int allZeros = 1;
            for (int k = i; k <= j; k++) {
                if (nums[k] != 0) {
                    allZeros = 0;
                    break;
                }
            }
            
            if (allZeros) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[100000];
    int count = 0;
    
    char* token = strtok(line, "[,]\n");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, "[,]\n");
    }
    
    printf("%lld\n", solution(nums, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Nested loops to generate O(n²) subarrays, each taking O(n) time to check

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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