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Find Balance Index — Problem Statement & Solution Guide

ArraysEasyPrefix Sum

TimeO(N)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Find Balance Index problem optimally.

TopicArrays

PatternPrefix Sum

TimeO(N)

SpaceO(1)

Problem Description

Given an array of integers named elements, determine the lowest 0-based index i such that the aggregate sum of all values strictly to the left of i is equal to the aggregate sum of all values strictly to the right of i.

By definition, if the index i is at position 0, the sum of values to its left is considered 0. Similarly, if i is at the terminal index elements.length - 1, the sum of values to its right is considered 0.

Return the smallest index i that satisfies this equilibrium condition. If no such index exists within the array, return -1.

DSA Pattern Breakdown

"Find Balance Index"

easy

WHY DOES IT MATTER?

This problem illustrates the Prefix Sum pattern and algebraic reduction, which are fundamental techniques for converting nested dynamic window problems into linear single-pass scans.

OPTIMIZATION CHALLENGE

The core challenge lies in avoiding redundant array scans by expressing the right-hand sum dynamically using the total sum and the left-hand sum.

REAL-WORLD CONNECTION

In distributed database sharding and data partitioning, finding a balance index represents finding a split point in ordered keys so that read/write storage workloads are distributed equally across two storage nodes.

Pay strict attention to the order of operations in your loop: always test the balance condition BEFORE adding the current element to the running left sum, otherwise you will include the current pivot element in the left side calculation.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The Find Balance Index problem (commonly known as finding the Equilibrium Point or Pivot Index) evaluates a candidate's mastery over basic array transformations and state preservation. The objective is to identify an index where the sum of elements on its left equals the sum of elements on its right. A naive approach recalculates both sums from scratch for every index, leading to repeated subarray traversals and quadratic runtime. To optimize this, we must recognize the mathematical relationship linking the left sum, the current element, and the right sum.

In any array, the sum of all elements can be expressed as: Total Sum = Left Sum + Current Element + Right Sum. Rearranging this equation reveals that Right Sum = Total Sum - Left Sum - Current Element. This key identity eliminates the need to calculate the right sum via explicit iteration. Instead, if we compute the total sum of the array beforehand, we can derive the right sum in O(1) time at any step.

By leveraging this invariant, the problem reduces to a two-pass algorithm. The first pass computes the aggregate sum of the array. The second pass iterates through the array sequentially while maintaining a running accumulator for the left sum. At each step, we check if the running left sum equals the derived right sum. This optimal strategy guarantees a linear execution time and uses constant extra space.

Interview Questions on This Problem

Q1How would you modify your solution if the input array contained extremely large values that might cause integer overflow?

To prevent integer overflow, we can use 64-bit integer types (like 'long' in Java/C++ or 64-bit signed integers in C#) for storing the total sum and left sum. If values exceed 64-bit limits, we can utilize arbitrary-precision arithmetic libraries (such as BigInteger in Java or Python's native unbounded integers). Alternatively, in streaming scenarios where total sum precomputation is impossible due to memory or numeric limits, we could maintain dynamic balance counters or use safe overflow-checked arithmetic methods.

Q2Does this algorithm work correctly when the array contains negative numbers or zero?

Yes, the algorithm works seamlessly with negative numbers and zeros. Because the mathematical identity 'Right Sum = Total Sum - Left Sum - Current Element' is algebraically valid across all real numbers and integers, positive and negative cancellation does not invalidate the invariant. The running sum correctly handles negative additions and subtractions without requiring special boundary logic.

Q3How would you handle a variant where you need to return ALL valid balance indices instead of just the lowest one?

Instead of returning immediately upon finding the first index where 'left_sum == right_sum', we would append the current index 'i' to a result list. The loop continues traversing the array to completion, updating 'left_sum' at each step. Finally, we return the list of collected indices. The time complexity remains O(N) and the auxiliary space complexity becomes O(K), where K is the number of balance indices found.

Examples

Example 1

Input

elements = [2, 4, -2, 1, 3, 1]

Output

Explanation: Evaluating index 3: The sum of elements to the left is elements[0] + elements[1] + elements[2] = 2 + 4 + (-2) = 4. The sum of elements to the right is elements[4] + elements[5] = 3 + 1 = 4. Since the left sum (4) equals the right sum (4), index 3 is a valid balance point.

Example 2

Input

elements = [5, 1, 2, -3]

Output

Explanation: Evaluating index 0: The sum of elements to the left is 0 by definition. The sum of elements to the right is elements[1] + elements[2] + elements[3] = 1 + 2 + (-3) = 0. Since both sums equal 0, index 0 is the smallest balance index.

Example 3

Input

elements = [1, 2, 3]

Output

-1

Explanation: No index satisfies the condition where the sum of elements on the left equals the sum of elements on the right. Therefore, the function returns -1.

Example 4

Input

elements = [-1, 2, 0, -2, 1]

Output

Explanation: Evaluating index 1: The sum of elements to the left is elements[0] = -1. The sum of elements to the right is elements[2] + elements[3] + elements[4] = 0 + (-2) + 1 = -1. Since both left and right sums equal -1, index 1 is a valid balance index.

Constraints

1 <= elements.length <= 10^5
-10^4 <= elements[i] <= 10^4

Optimal Approach & Strategy

First compute the total sum of the array in a single pass. Then scan through the array maintaining a running left sum, calculating the right sum on the fly as total_sum - left_sum - element[i], returning the first matching index.

Brute Force Approach

For every index i, use nested loops to sum all elements from index 0 to i - 1, and another nested loop to sum elements from i + 1 to N - 1. Compare the two sums at each step and return the first index where they are equal.

Code Solutions

Language:

JavaScript Solution

Time: O(N)

function findBalanceIndex(elements) {
    let totalSum = elements.reduce((acc, curr) => acc + curr, 0);
    let leftSum = 0;
    for (let i = 0; i < elements.length; i++) {
        if (leftSum === totalSum - leftSum - elements[i]) {
            return i;
        }
        leftSum += elements[i];
    }
    return -1;
}

#include <iostream>
#include <vector>
#include <numeric>

int findBalanceIndex(const std::vector<int>& elements) {
    long long totalSum = 0;
    for (int num : elements) {
        totalSum += num;
    }
    long long leftSum = 0;
    for (int i = 0; i < static_cast<int>(elements.size()); ++i) {
        if (leftSum == totalSum - leftSum - elements[i]) {
            return i;
        }
        leftSum += elements[i];
    }
    return -1;
}

public class Solution {
    public static int findBalanceIndex(int[] elements) {
        long totalSum = 0;
        for (int num : elements) {
            totalSum += num;
        }
        long leftSum = 0;
        for (int i = 0; i < elements.length; i++) {
            if (leftSum == totalSum - leftSum - elements[i]) {
                return i;
            }
            leftSum += elements[i];
        }
        return -1;
    }
}

def find_balance_index(elements: list[int]) -> int:
    total_sum = sum(elements)
    left_sum = 0
    for i, num in enumerate(elements):
        if left_sum == total_sum - left_sum - num:
            return i
        left_sum += num
    return -1

function findBalanceIndex(elements) {
    let totalSum = elements.reduce((acc, curr) => acc + curr, 0);
    let leftSum = 0;
    for (let i = 0; i < elements.length; i++) {
        if (leftSum === totalSum - leftSum - elements[i]) {
            return i;
        }
        leftSum += elements[i];
    }
    return -1;
}

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