Find Subarrays With Equal Sum - Practice Coding Problems

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

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Understanding the Visualization

Input Array

Array with adjacent pairs to check

Calculate Sums

Find sum of each length-2 subarray

Check for Duplicates

Return true if any two sums match

Key Takeaway

🎯 Key Insight: Use a hash set to detect duplicate sums in O(n) time instead of comparing all pairs

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

Approach	Time	Space	Notes
✓ Hash Set - Track Seen Sums	O(n)	O(n)	Use a hash set to track sums we've seen and detect duplicates immediately
Brute Force - Compare All Pairs	O(n²)	O(1)	Check every possible pair of adjacent elements against every other pair

Hash Set - Track Seen Sums — Algorithm Steps

Initialize an empty hash set
For each pair of adjacent elements, calculate sum
If sum exists in set, return true; otherwise add to set

Visualization

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Step-by-Step Walkthrough

Process Pairs

Calculate sum for each adjacent pair

Check & Store

Look for sum in set, add if new

Found Duplicate

Return true when sum already exists

Code -

solution.c — C

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

#define MAX_SUMS 1000

bool solution(int* nums, int numsSize) {
    if (numsSize < 4) return false;
    
    int seenSums[MAX_SUMS];
    int seenCount = 0;
    
    for (int i = 0; i < numsSize - 1; i++) {
        int currentSum = nums[i] + nums[i + 1];
        
        // Check if sum already seen
        for (int j = 0; j < seenCount; j++) {
            if (seenSums[j] == currentSum) {
                return true;
            }
        }
        
        // Add to seen sums
        seenSums[seenCount++] = currentSum;
    }
    
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    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(n)

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

✓ Linear Growth

Space Complexity

O(n)

Hash set stores up to n-1 sums in worst case

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Set - Track Seen Sums — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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