Widest Pair of Indices With Equal Range Sum

Widest Pair of Indices With Equal Range Sum - Problem

You are given two 0-indexed binary arrays nums1 and nums2.

Find the widest pair of indices (i, j) such that i <= j and nums1[i] + nums1[i+1] + ... + nums1[j] == nums2[i] + nums2[i+1] + ... + nums2[j].

The widest pair of indices is the pair with the largest distance between i and j. The distance between a pair of indices is defined as j - i + 1.

Return the distance of the widest pair of indices. If no pair of indices meets the conditions, return 0.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [1,0,1,1,0], nums2 = [0,1,1,1,0]

› Output: 3

💡 Note: The widest pair is (2,4) where both arrays have sum 2 in range [2,4]. Distance = 4-2+1 = 3.

Example 2 — Equal Arrays

$ Input: nums1 = [1,1,0], nums2 = [1,1,0]

› Output: 3

💡 Note: Arrays are identical, so entire range [0,2] has equal sums. Distance = 2-0+1 = 3.

Example 3 — No Valid Range

$ Input: nums1 = [1,0], nums2 = [0,1]

› Output: 0

💡 Note: No range has equal sums: [0,0] has 1≠0, [1,1] has 0≠1, [0,1] has 1≠1.

Constraints

1 ≤ nums1.length, nums2.length ≤ 10⁵
nums1.length == nums2.length
nums1[i], nums2[i] ∈ {0, 1}

Visualization

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Asked in

G Google 35 f Meta 28 a Amazon 22 M Microsoft 18

The key insight is to transform this into a zero-sum subarray problem by creating a difference array where diff[i] = nums1[i] - nums2[i]. Equal range sums mean their difference is zero. Best approach uses prefix sum with hash map to find the longest zero-sum subarray efficiently. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n³) Space: O(1)

For each possible starting index i and ending index j, calculate the sum of both arrays in range [i,j] and check if they are equal. Track the maximum distance found.

Prefix Sum with Hash Map

⏱️ Time: O(n) Space: O(n)

Create a difference array where diff[i] = nums1[i] - nums2[i]. The problem becomes finding the longest subarray with sum 0. Use prefix sums and hash map to track first occurrence of each sum.

Brute Force - Check All Pairs — Algorithm Steps

Iterate through all pairs (i,j) where i <= j
Calculate sum of nums1[i...j] and nums2[i...j]
If sums are equal, update maximum distance
Return the maximum distance found

Visualization

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Step-by-Step Walkthrough

Try All Pairs

Check every (i,j) where i ≤ j

Calculate Sums

Sum both arrays in range [i,j]

Track Maximum

Keep largest distance where sums match

Code -

solution.c — C

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

int solution(int* nums1, int* nums2, int n) {
    int max_distance = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int sum1 = 0;
            int sum2 = 0;
            for (int k = i; k <= j; k++) {
                sum1 += nums1[k];
                sum2 += nums2[k];
            }
            
            if (sum1 == sum2) {
                int distance = j - i + 1;
                if (distance > max_distance) {
                    max_distance = distance;
                }
            }
        }
    }
    
    return max_distance;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line1[1000];
    char line2[1000];
    
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int nums1[1000], nums2[1000];
    int n1, n2;
    
    parseArray(line1, nums1, &n1);
    parseArray(line2, nums2, &n2);
    
    printf("%d\n", solution(nums1, nums2, n1));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Nested loops for i,j pairs (O(n²)) and sum calculation for each pair (O(n))

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for tracking indices and sums

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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