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C++ Program for Maximum equilibrium sum in an array

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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]

A Simple Solution is 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. 

C++




// CPP program to find
// maximum equilibrium sum.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find
// maximum equilibrium sum.
int findMaxSum(int arr[], int n)
{
    int res = INT_MIN;
    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()
{
    int arr[] = {-2, 5, 3, 1,
                  2, 6, -4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMaxSum(arr, n);
    return 0;
}


Output : 

7

 

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

A Better Approach is to 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.

C++




// CPP program to find
// maximum equilibrium sum.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum
// equilibrium sum.
int findMaxSum(int arr[], int n)
{
    // Array to store prefix sum.
    int preSum[n];
 
    // Array to store suffix sum.
    int suffSum[n];
 
    // Variable to store maximum sum.
    int ans = 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 ans
    // accordingly.
    suffSum[n - 1] = arr[n - 1];
    if (preSum[n - 1] == suffSum[n - 1])
        ans = max(ans, preSum[n - 1]);
         
    for (int i = n - 2; i >= 0; i--)
    {
        suffSum[i] = suffSum[i + 1] + arr[i];
        if (suffSum[i] == preSum[i])
            ans = max(ans, preSum[i]);    
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { -2, 5, 3, 1,
                   2, 6, -4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMaxSum(arr, n);
    return 0;
}


Output: 

7

 

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.

C++




// CPP program to find
// maximum equilibrium sum.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find
// maximum equilibrium sum.
int findMaxSum(int arr[], int n)
{
    int sum = accumulate(arr, arr + n, 0);
    int prefix_sum = 0, res = INT_MIN;
    for (int i = 0; i < n; i++)
    {
    prefix_sum += arr[i];
    if (prefix_sum == sum)
        res = max(res, prefix_sum);
    sum -= arr[i];
    }
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { -2, 5, 3, 1,
                   2, 6, -4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMaxSum(arr, n);
    return 0;
}


Output : 

7

 

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

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



Last Updated : 31 Jan, 2022
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