Contiguous Array - Practice Coding Problems

Contiguous Array - Problem

Given a binary array nums, return the maximum length of a contiguous subarray with an equal number of 0s and 1s.

A contiguous subarray is a sequence of adjacent elements in the array.

Example: For array [0,1,0,0,1,1,0], the subarray [0,1,0,0,1,1] has 3 zeros and 3 ones, so the maximum length is 6.

Input & Output

Example 1 — Basic Case

$ Input: nums = [0,1]

› Output: 2

💡 Note: The entire array has 1 zero and 1 one, so the maximum length is 2

Example 2 — Longer Array

$ Input: nums = [0,1,0,0,1,1,0]

› Output: 6

💡 Note: The subarray [0,1,0,0,1,1] from index 0 to 5 has 3 zeros and 3 ones, length = 6

Example 3 — All Same

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

› Output: 0

💡 Note: No subarray has equal number of 0s and 1s, so return 0

Constraints

1 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1

Visualization

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Understanding the Visualization

Input

Binary array with 0s and 1s

Process

Find longest subarray with equal count

Output

Maximum length found

Key Takeaway

🎯 Key Insight: Transform 0s to -1s and find longest zero-sum subarray using prefix sums

Asked in

f Facebook 45 G Google 35 a Amazon 30

The key insight is to transform the problem: convert 0s to -1s and find the longest subarray with sum 0. Best approach uses prefix sum with hash map to track first occurrence of each cumulative sum. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Prefix Sum with Hash Map	O(n)	O(n)	Transform 0s to -1s and find longest subarray with sum 0
Brute Force	O(n²)	O(1)	Check all possible subarrays and count 0s and 1s

Prefix Sum with Hash Map — Algorithm Steps

Transform: replace 0s with -1s
Calculate running sum at each position
Use hash map to store first occurrence of each sum
When same sum appears again, we found a zero-sum subarray

Visualization

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Step-by-Step Walkthrough

Transform

Convert 0s to -1s

Cumulative Sum

Track running sum

Hash Map

Store first occurrence of each sum

Code -

solution.c — C

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

#define MAX_SIZE 50000

typedef struct {
    int sum;
    int index;
} SumEntry;

int solution(int* nums, int numsSize) {
    SumEntry sumMap[MAX_SIZE];
    int mapSize = 0;
    
    // Add initial entry: sum 0 at index -1
    sumMap[mapSize++] = (SumEntry){0, -1};
    
    int cumulativeSum = 0;
    int maxLength = 0;
    
    for (int i = 0; i < numsSize; i++) {
        // Transform 0 to -1
        cumulativeSum += nums[i] == 1 ? 1 : -1;
        
        // Check if sum exists
        int found = -1;
        for (int j = 0; j < mapSize; j++) {
            if (sumMap[j].sum == cumulativeSum) {
                found = sumMap[j].index;
                break;
            }
        }
        
        if (found != -1) {
            // Found subarray with sum 0
            int length = i - found;
            if (length > maxLength) {
                maxLength = length;
            }
        } else {
            // Store first occurrence of this sum
            sumMap[mapSize++] = (SumEntry){cumulativeSum, i};
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[10000];
    int numsSize = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array with O(1) hash map operations

✓ Linear Growth

Space Complexity

O(n)

Hash map stores up to n different cumulative sums

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Prefix Sum with Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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