Maximum equilibrium sum in an array
Last Updated :
28 Feb, 2025
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Given an array arr[]. Find the maximum value of prefix sum which is also suffix sum for index i in arr[].
Examples :
Input : arr[] = {-1, 2, 3, 0, 3, 2, -1}
Output : 4
Explanation : Prefix sum of arr[0..3] = Suffix sum of arr[3..6]
Input : arr[] = {-3, 5, 3, 1, 2, 6, -4, 2}
Output : 7
Explanation : Prefix sum of arr[0..3] = Suffix sum of arr[3..7]
[Naive Approach] – Two Nested Loops – O(n^2) Time and O(1) Space
The idea s to one by one check the given condition (prefix sum equal to suffix sum) for every element and returns the element that satisfies the given condition with maximum value.
// CPP program to find
// maximum equilibrium sum.
#include <bits/stdc++.h>
using namespace std;
// Function to find
// maximum equilibrium sum.
int findMaxSum(vector<int>& arr)
{
int res = INT_MIN;
int n = arr.size();
// Try every element one by one
for (int i = 0; i < n; i++)
{
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = max(res, prefix_sum);
}
return res;
}
// Driver Code
int main()
{
vector<int> arr = {-2, 5, 3, 1,
2, 6, -4, 2 };
cout << findMaxSum(arr);
return 0;
}
// Function to find
// maximum equilibrium sum.
class MaxEquilibriumSum {
public static int findMaxSum(int[] arr) {
int res = Integer.MIN_VALUE;
int n = arr.length;
// Try every element one by one
for (int i = 0; i < n; i++) {
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = Math.max(res, prefix_sum);
}
return res;
}
// Driver Code
public static void main(String[] args) {
int[] arr = {-2, 5, 3, 1, 2, 6, -4, 2};
System.out.println(findMaxSum(arr));
}
}
# Function to find
# maximum equilibrium sum.
def find_max_sum(arr):
res = float('-inf')
n = len(arr)
# Try every element one by one
for i in range(n):
prefix_sum = arr[i]
for j in range(i):
prefix_sum += arr[j]
suffix_sum = arr[i]
for j in range(n - 1, i, -1):
suffix_sum += arr[j]
if prefix_sum == suffix_sum:
res = max(res, prefix_sum)
return res
# Driver Code
arr = [-2, 5, 3, 1, 2, 6, -4, 2]
print(find_max_sum(arr))
// Function to find
// maximum equilibrium sum.
using System;
using System.Linq;
class MaxEquilibriumSum {
public static int FindMaxSum(int[] arr) {
int res = int.MinValue;
int n = arr.Length;
// Try every element one by one
for (int i = 0; i < n; i++) {
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = Math.Max(res, prefix_sum);
}
return res;
}
// Driver Code
public static void Main() {
int[] arr = {-2, 5, 3, 1, 2, 6, -4, 2};
Console.WriteLine(FindMaxSum(arr));
}
}
// Function to find
// maximum equilibrium sum.
function findMaxSum(arr) {
let res = Number.MIN_SAFE_INTEGER;
const n = arr.length;
// Try every element one by one
for (let i = 0; i < n; i++) {
let prefix_sum = arr[i];
for (let j = 0; j < i; j++)
prefix_sum += arr[j];
let suffix_sum = arr[i];
for (let j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum === suffix_sum)
res = Math.max(res, prefix_sum);
}
return res;
}
// Driver Code
const arr = [-2, 5, 3, 1, 2, 6, -4, 2];
console.log(findMaxSum(arr));
Output
7
[Better Approach] – Precompute Prefix and Suffix Sums – O(n) Time and O(n) Space
Traverse the array and store prefix sum for each index in array presum[], in which presum[i] stores sum of subarray arr[0..i]. Do another traversal of the array and store suffix sum in another array suffsum[], in which suffsum[i] stores sum of subarray arr[i..n-1]. After this for each index check if presum[i] is equal to suffsum[i] and if they are equal then compare their value with the overall maximum so far.
// CPP program to find
// maximum equilibrium sum.
#include <bits/stdc++.h>
using namespace std;
// Function to find maximum
// equilibrium sum.
int findMaxSum(vector<int>& arr)
{
int n = arr.size();
// Vector to store prefix sum.
vector<int> preSum(n);
// Vector to store suffix sum.
vector<int> suffSum(n);
// Variable to store maximum sum.
int res = INT_MIN;
// Calculate prefix sum.
preSum[0] = arr[0];
for (int i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
// Calculate suffix sum and compare
// it with prefix sum. Update res
// accordingly.
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
res = max(res, preSum[n - 1]);
for (int i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
res = max(res, preSum[i]);
}
return res;
}
// Driver Code
int main()
{
vector<int> arr = { -2, 5, 3, 1,
2, 6, -4, 2 };
cout << findMaxSum(arr);
return 0;
}
// Java program to find maximum equilibrium sum.
import java.util.Arrays;
public class MaxEquilibriumSum {
// Function to find maximum equilibrium sum.
public static int findMaxSum(int[] arr) {
int n = arr.length;
// Array to store prefix sum.
int[] preSum = new int[n];
// Array to store suffix sum.
int[] suffSum = new int[n];
// Variable to store maximum sum.
int res = Integer.MIN_VALUE;
// Calculate prefix sum.
preSum[0] = arr[0];
for (int i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
// Calculate suffix sum and compare it with
// prefix sum. Update res accordingly.
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
res = Math.max(res, preSum[n - 1]);
for (int i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
res = Math.max(res, preSum[i]);
}
return res;
}
// Driver Code
public static void main(String[] args) {
int[] arr = { -2, 5, 3, 1, 2, 6, -4, 2 };
System.out.println(findMaxSum(arr));
}
}
# Python program to find maximum equilibrium sum.
def findMaxSum(arr):
n = len(arr)
# List to store prefix sum.
preSum = [0] * n
# List to store suffix sum.
suffSum = [0] * n
# Variable to store maximum sum.
res = float('-inf')
# Calculate prefix sum.
preSum[0] = arr[0]
for i in range(1, n):
preSum[i] = preSum[i - 1] + arr[i]
# Calculate suffix sum and compare it with
# prefix sum. Update res accordingly.
suffSum[n - 1] = arr[n - 1]
if preSum[n - 1] == suffSum[n - 1]:
res = max(res, preSum[n - 1])
for i in range(n - 2, -1, -1):
suffSum[i] = suffSum[i + 1] + arr[i]
if suffSum[i] == preSum[i]:
res = max(res, preSum[i])
return res
# Driver Code
arr = [-2, 5, 3, 1, 2, 6, -4, 2]
print(findMaxSum(arr))
using System;
class Program {
static int FindMaxSum(int[] arr) {
int n = arr.Length;
// Array to store prefix sum.
int[] preSum = new int[n];
// Array to store suffix sum.
int[] suffSum = new int[n];
// Variable to store maximum sum.
int res = int.MinValue;
// Calculate prefix sum.
preSum[0] = arr[0];
for (int i = 1; i < n; i++) {
preSum[i] = preSum[i - 1] + arr[i];
}
// Calculate suffix sum and compare it with
// prefix sum. Update res accordingly.
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1]) {
res = Math.Max(res, preSum[n - 1]);
}
for (int i = n - 2; i >= 0; i--) {
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i]) {
res = Math.Max(res, preSum[i]);
}
}
return res;
}
static void Main() {
// Driver Code
int[] arr = { -2, 5, 3, 1, 2, 6, -4, 2 };
Console.WriteLine(FindMaxSum(arr));
}
}
function findMaxSum(arr) {
let n = arr.length;
// Array to store prefix sum.
let preSum = new Array(n).fill(0);
// Array to store suffix sum.
let suffSum = new Array(n).fill(0);
// Variable to store maximum sum.
let res = -Infinity;
// Calculate prefix sum.
preSum[0] = arr[0];
for (let i = 1; i < n; i++) {
preSum[i] = preSum[i - 1] + arr[i];
}
// Calculate suffix sum and compare it with
// prefix sum. Update res accordingly.
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] === suffSum[n - 1]) {
res = Math.max(res, preSum[n - 1]);
}
for (let i = n - 2; i >= 0; i--) {
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] === preSum[i]) {
res = Math.max(res, preSum[i]);
}
}
return res;
}
// Driver Code
let arr = [-2, 5, 3, 1, 2, 6, -4, 2];
console.log(findMaxSum(arr));
Output
7
[Expected Approach] – Only One Prefix Sum Variable – O(n) Time and O(1) Space
Now the above prefix sum array and suffix sum array approach can be further optimized in terms of space, by using prefix sum and suffix sum variables. The idea is that instead of storing the prefix sum and suffix sum for each element in an array, we can simply use the fact that
PrefixSum(arr[0 : pivot – 1]) + arr[pivot] + SuffixSum[pivot + 1: n – 1] = ArraySum
so, SuffixSum[pivot + 1: n – 1] = ArraySum – PrefixSum(arr[0 : pivot – 1])
Here, pivot refers to the index that we are currently checking for the equilibrium index.
#include <bits/stdc++.h>
using namespace std;
// Function to find
// maximum equilibrium sum.
int findMaxSum(vector<int>& arr)
{
int suffix_sum = accumulate(arr.begin(), arr.end(), 0);
int prefix_sum = 0, res = INT_MIN;
for (int i = 0; i < arr.size(); i++)
{
// Uppdate prefix sum
prefix_sum += arr[i];
if (prefix_sum == suffix_sum)
res = max(res, prefix_sum);
// Update suffix sum
suffix_sum -= arr[i];
}
return res;
}
// Driver Code
int main()
{
vector<int> arr = { -2, 5, 3, 1, 2, 6, -4, 2 };
cout << findMaxSum(arr);
return 0;
}
// Function to find
// maximum equilibrium sum.
import java.util.Arrays;
import java.util.List;
public class Main {
public static int findMaxSum(List<Integer> arr) {
int suffix_sum = arr.stream().mapToInt(Integer::intValue).sum();
int prefix_sum = 0, res = Integer.MIN_VALUE;
for (int i = 0; i < arr.size(); i++) {
// Update prefix sum
prefix_sum += arr.get(i);
if (prefix_sum == suffix_sum)
res = Math.max(res, prefix_sum);
// Update suffix sum
suffix_sum -= arr.get(i);
}
return res;
}
// Driver Code
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(-2, 5, 3, 1, 2, 6, -4, 2);
System.out.println(findMaxSum(arr));
}
}
# Function to find
# maximum equilibrium sum.
def findMaxSum(arr):
suffix_sum = sum(arr)
prefix_sum = 0
res = float('-inf')
for i in range(len(arr)):
# Update prefix sum
prefix_sum += arr[i]
if prefix_sum == suffix_sum:
res = max(res, prefix_sum)
# Update suffix sum
suffix_sum -= arr[i]
return res
# Driver Code
arr = [-2, 5, 3, 1, 2, 6, -4, 2]
print(findMaxSum(arr))
// Function to find
// maximum equilibrium sum.
using System;
using System.Linq;
using System.Collections.Generic;
class Program {
public static int FindMaxSum(List<int> arr) {
int suffix_sum = arr.Sum();
int prefix_sum = 0, res = int.MinValue;
for (int i = 0; i < arr.Count; i++) {
// Update prefix sum
prefix_sum += arr[i];
if (prefix_sum == suffix_sum)
res = Math.Max(res, prefix_sum);
// Update suffix sum
suffix_sum -= arr[i];
}
return res;
}
// Driver Code
static void Main() {
List<int> arr = new List<int> { -2, 5, 3, 1, 2, 6, -4, 2 };
Console.WriteLine(FindMaxSum(arr));
}
}
// Function to find
// maximum equilibrium sum.
function findMaxSum(arr) {
let suffix_sum = arr.reduce((a, b) => a + b, 0);
let prefix_sum = 0;
let res = Number.MIN_SAFE_INTEGER;
for (let i = 0; i < arr.length; i++) {
// Update prefix sum
prefix_sum += arr[i];
if (prefix_sum === suffix_sum)
res = Math.max(res, prefix_sum);
// Update suffix sum
suffix_sum -= arr[i];
}
return res;
}
// Driver Code
const arr = [-2, 5, 3, 1, 2, 6, -4, 2];
console.log(findMaxSum(arr));
Output
7
