Number of Zero-Filled Subarrays - Practice Coding Problems

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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Understanding the Visualization

Input Array

Array with mixed zeros and non-zeros

Find Zero Segments

Identify consecutive zero groups

Count Subarrays

Apply formula n*(n+1)/2 per segment

Key Takeaway

🎯 Key Insight: For n consecutive zeros, there are exactly n×(n+1)/2 possible zero-filled subarrays

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

Approach	Time	Space	Notes
✓ One Pass with Math Formula	O(n)	O(1)	Count consecutive zeros and use formula n*(n+1)/2 for each segment
Brute Force - Check All Subarrays	O(n³)	O(1)	Check every possible subarray to see if it contains only zeros

One Pass with Math Formula — Algorithm Steps

Iterate through the array
Count consecutive zeros in each segment
For each segment of length n, add n*(n+1)/2 to result
Reset count when non-zero is encountered

Visualization

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Step-by-Step Walkthrough

Find Segments

Identify consecutive zero segments

Apply Formula

For each segment of n zeros, add n*(n+1)/2

Sum Results

Total count of 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 result = 0;
    int currentZeros = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 0) {
            currentZeros++;
            // Each new zero adds currentZeros new subarrays
            result += currentZeros;
        } else {
            currentZeros = 0;
        }
    }
    
    return result;
}

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(n)

Single pass through the array

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track count and result

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One Pass with Math Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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