LeetCode #2395 — EASY

Find Subarrays With Equal Sum

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: nums = [4,2,4]
Output: true
Explanation: The subarrays with elements [4,2] and [2,4] have the same sum of 6.

Example 2:

Input: nums = [1,2,3,4,5]
Output: false
Explanation: No two subarrays of size 2 have the same sum.

Example 3:

Input: nums = [0,0,0]
Output: true
Explanation: The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0. 
Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.

Constraints:

2 <= nums.length <= 1000
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[4,2,4]

Example 2

[1,2,3,4,5]

Example 3

[0,0,0]

Core Insight

What unlocks the optimal approach

Use a counter to keep track of the subarray sums.
Use a hashset to check if any two sums are equal.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2395: Find Subarrays With Equal Sum
class Solution {
    public boolean findSubarrays(int[] nums) {
        Set<Integer> vis = new HashSet<>();
        for (int i = 1; i < nums.length; ++i) {
            if (!vis.add(nums[i - 1] + nums[i])) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #2395: Find Subarrays With Equal Sum
func findSubarrays(nums []int) bool {
	vis := map[int]bool{}
	for i, b := range nums[1:] {
		x := nums[i] + b
		if vis[x] {
			return true
		}
		vis[x] = true
	}
	return false
}

# Accepted solution for LeetCode #2395: Find Subarrays With Equal Sum
class Solution:
    def findSubarrays(self, nums: List[int]) -> bool:
        vis = set()
        for a, b in pairwise(nums):
            if (x := a + b) in vis:
                return True
            vis.add(x)
        return False

// Accepted solution for LeetCode #2395: Find Subarrays With Equal Sum
use std::collections::HashSet;
impl Solution {
    pub fn find_subarrays(nums: Vec<i32>) -> bool {
        let n = nums.len();
        let mut set = HashSet::new();
        for i in 1..n {
            if !set.insert(nums[i - 1] + nums[i]) {
                return true;
            }
        }
        false
    }
}

// Accepted solution for LeetCode #2395: Find Subarrays With Equal Sum
function findSubarrays(nums: number[]): boolean {
    const vis: Set<number> = new Set<number>();
    for (let i = 1; i < nums.length; ++i) {
        const x = nums[i - 1] + nums[i];
        if (vis.has(x)) {
            return true;
        }
        vis.add(x);
    }
    return false;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.