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.
Javascript
<script>// Javascript program to find // maximum equilibrium sum.// Function to find // maximum equilibrium sum.function findMaxSum(arr, n){ var res = -1000000000; for (var i = 0; i < n; i++) { var prefix_sum = arr[i]; for (var j = 0; j < i; j++) prefix_sum += arr[j]; var suffix_sum = arr[i]; for (var j = n - 1; j > i; j--) suffix_sum += arr[j]; if (prefix_sum == suffix_sum) res = Math.max(res, prefix_sum); } return res;}// Driver Codevar arr = [-2, 5, 3, 1, 2, 6, -4, 2 ];var n = arr.length;document.write( findMaxSum(arr, n));</script> |
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.
Javascript
<script>// 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 codelet arr = [ -2, 5, 3, 1, 2, 6, -4, 2 ];let n = arr.length;document.write(findMaxSum(arr, n));// This code is contributed by rameshtravel07</script> |
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.
Javascript
<script> // 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; document.write(findMaxSum(arr, n)); // This code is contributed by vaibhavrabadiya117.</script> |
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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