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.
Java
// java program to find maximum// equilibrium sum.import java.io.*;class GFG { // Function to find maximum // equilibrium sum. static int findMaxSum(int []arr, int n) { int res = Integer.MIN_VALUE; 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 }; int n = arr.length; System.out.println(findMaxSum(arr, n)); }}// This code is contributed by anuj_67. |
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.
Java
// Java program to find maximum equilibrium sum.import java.io.*;public class GFG { // Function to find maximum // equilibrium sum. static int findMaxSum(int []arr, int n) { // 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 ans = 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 ans // accordingly. suffSum[n - 1] = arr[n - 1]; if (preSum[n - 1] == suffSum[n - 1]) ans = Math.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 = Math.max(ans, preSum[i]); } return ans; } // Driver Code static public void main (String[] args) { int []arr = { -2, 5, 3, 1, 2, 6, -4, 2 }; int n = arr.length; System.out.println( findMaxSum(arr, n)); }}// This code is contributed by anuj_67 |
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.
Java
// Java program to find maximum equilibrium// sum.import java.lang.Math.*;import java.util.stream.*;class GFG { // Function to find maximum equilibrium // sum. static int findMaxSum(int arr[], int n) { int sum = IntStream.of(arr).sum(); int prefix_sum = 0, res = Integer.MIN_VALUE; for (int 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 public static void main(String[] args) { int arr[] = { -2, 5, 3, 1, 2, 6, -4, 2 }; int n = arr.length; System.out.print(findMaxSum(arr, n)); }}// This code is contributed by Smitha. |
7
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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