Find Subarrays With Equal Sum

Find Subarrays With Equal Sum - Problem

Given a 0-indexed integer array nums, determine whether there exist two subarrays of length 2 with equal sum. Note that the two subarrays must begin at different indices.

Return true if these subarrays exist, and false otherwise.

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

Input & Output

Example 1 — Found Equal Sums

$ Input: nums = [4,2,4]

› Output: true

💡 Note: We can take subarrays [4,2] (sum=6) and [2,4] (sum=6). Both have the same sum of 6.

Example 2 — No Equal Sums

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

› Output: false

💡 Note: Subarrays are [1,2] (sum=3), [2,3] (sum=5), [3,4] (sum=7). All sums are different.

Example 3 — Too Short Array

$ Input: nums = [0,0]

› Output: false

💡 Note: Only one subarray [0,0] possible, need at least two different subarrays.

Constraints

2 ≤ nums.length ≤ 1000
-1000 ≤ nums[i] ≤ 1000

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use a hash set to track sums of adjacent pairs as we iterate through the array. The moment we find a sum that already exists in our set, we know there are two subarrays with equal sum. Best approach is the hash set method with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Compare All Pairs

⏱️ Time: O(n²) Space: O(1)

Generate all possible subarrays of length 2, calculate their sums, and compare each sum with all other sums to find duplicates. This involves nested loops to compare all pairs.

Hash Set - Track Seen Sums

⏱️ Time: O(n) Space: O(n)

Iterate through the array once, calculating the sum of each adjacent pair. Store each sum in a hash set. If we encounter a sum that's already in the set, we've found two subarrays with equal sum.

Brute Force - Compare All Pairs — Algorithm Steps

Generate all subarrays of length 2
For each subarray pair, compare their sums
Return true if any two sums are equal

Visualization

Tap to expand

Step-by-Step Walkthrough

Generate Pairs

Find all subarrays of length 2

Compare Sums

Check each pair against all others

Found Match

Return true when sums are equal

Code -

solution.c — C

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

bool solution(int* nums, int numsSize) {
    if (numsSize < 3) return false;
    
    for (int i = 0; i < numsSize - 1; i++) {
        int sum1 = nums[i] + nums[i + 1];
        for (int j = i + 1; j < numsSize - 1; j++) {
            int sum2 = nums[j] + nums[j + 1];
            if (sum1 == sum2) {
                return true;
            }
        }
    }
    
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    bool result = solution(nums, size);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops comparing each of n-1 pairs against all others

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to store sums and indices

✓ Linear Space

24.5K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Compare All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler