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Javascript Program for Maximum equilibrium sum in an array

Last Updated : 23 Jul, 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
Prefix sum of arr[0..3] = 
            Suffix sum of arr[3..6]

Input : arr[] = {-2, 5, 3, 1, 2, 6, -4, 2}
Output : 7
Prefix sum of arr[0..3] = 
              Suffix sum of arr[3..7]

Native Approach:

Checks the one-by-one given condition (prefix sum equal to suffix sum) for every element and returns the element that satisfies the given condition with maximum value. 

JavaScript 
// Javascript program to find 
// maximum equilibrium sum.

// Function to find 
// maximum equilibrium sum.
function findMaxSum(arr, n) {
    let res = -1000000000;
    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
let arr = [-2, 5, 3, 1,
    2, 6, -4, 2];
let n = arr.length;
console.log(findMaxSum(arr, n));

Output
7

Complexity Analysis:

  • Time Complexity: O(n2) 
  • Auxiliary Space: O(n)

Better Approach:

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.

JavaScript 
// Javascript program to find 
// maximum equilibrium sum.

// Function to find maximum
// equilibrium sum.
function findMaxSum(arr, n) {

    // Array to store prefix sum.
    let preSum = new Array(n);
    preSum.fill(0);

    // Array to store suffix sum.
    let suffSum = new Array(n);
    suffSum.fill(0);

    // Variable to store maximum sum.
    let ans = Number.MIN_VALUE;

    // 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 ans
    // accordingly.
    suffSum[n - 1] = arr[n - 1];

    if (preSum[n - 1] == suffSum[n - 1])
        ans = Math.max(ans, preSum[n - 1]);

    for (let i = n - 2; i >= 0; i--) {
        suffSum[i] = suffSum[i + 1] + arr[i];

        if (suffSum[i] == preSum[i])
            ans = Math.max(ans, preSum[i]);
    }
    return ans;
}

// Driver code
let arr = [-2, 5, 3, 1, 2, 6, -4, 2];
let n = arr.length;

console.log(findMaxSum(arr, n));

Output
7

Complexity Analysis:

  • Time Complexity: O(n) 
  • Auxiliary Space: O(n)

Further Optimization : 

We can avoid the use of extra space by first computing the total sum, then using it to find the current prefix and suffix sums.

JavaScript 
// javascript program to find
// maximum equilibrium sum.

// Function to find
// maximum equilibrium sum.
function findMaxSum(arr, n) {
    let sum = 0;
    for (let i = 0; i < n; i++) {
        sum = sum + arr[i];
    }
    let prefix_sum = 0, res = Number.MIN_VALUE;
    for (let i = 0; i < n; i++) {
        prefix_sum += arr[i];
        if (prefix_sum == sum)
            res = Math.max(res, prefix_sum);
        sum -= arr[i];
    }
    return res;
}

// Driver Code

let arr = [-2, 5, 3, 1, 2, 6, -4, 2];
let n = arr.length;
console.log(findMaxSum(arr, n));

Output
7

Complexity Analysis:

  • Time Complexity: O(n) 
  • Auxiliary Space: O(1)

Please refer complete article on Maximum equilibrium sum in an array for more details!


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