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# Maximize sum by choosing a subsegment from array [l, r] and convert arr[i] to (M–arr[i]) at most once

• Last Updated : 06 Jul, 2021

Given an array, arr[] consisting of N positive integers and a positive integer M, the task is to maximize the sum of the array after performing at most one operation. In one operation, choose a subsegment from the array [l, r] and convert arr[i] to M – arr[i] where l≤i≤r.

Examples:

Input: arr[] = {2, 4, 3}, M = 5
Output: 10
Explanation: Replace numbers in the subarray from index 0 to index 0, so the new array becomes [5-2, 4, 3], so the maximum sum will be (5-2) + 4 + 3 = 10

Input: arr[] = {4, 3, 4}, M = 5
Output: 11

Naive Approach: The simplest approach to solve the problem is to apply the operation on all the subarrays of the array and find the maximum sum applying the given operation.

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized further by using Kadane’s Algorithm. Follow the steps below to solve the problem:

• Initialize a variable, say sum as 0 to store the sum of the array.
• Iterate in the range [0, N-1] using the variable i and add arr[i] to the variable sum and modify the value of arr[i] as M – 2×arr[i].
• Now apply Kadane’s algorithm to the array arr[].
• Initialize two variables, say mx and ans as 0.
• Iterate in the range [0, N-1] using the variable i and perform the following steps:
• Add arr[i] to the ans.
• If ans<0, then modify the value of ans as 0.
• Modify the value of mx as max(mx, ans).
• Print the value of sum + mx as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to maximize the sum after``// performing atmost one operation``int` `MaxSum(``int` `arr[], ``int` `n, ``int` `M)``{``    ``// Variable to store the sum``    ``int` `sum = 0;` `    ``// Traverse the array and modify arr[]``    ``for` `(``int` `i = 0; i < n; i++) {``        ``sum += arr[i];``        ``arr[i] = (M - 2 * arr[i]);``    ``}` `    ``int` `ans = 0, mx = 0;` `    ``// Apply Kadane's algorithm``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// Add a[i] to ans``        ``ans += arr[i];` `        ``// If ans<0, modify the value``        ``// of ans as 0``        ``if` `(ans < 0)``            ``ans = 0;` `        ``// Update the value of mx``        ``mx = max(mx, ans);``    ``}` `    ``// Return the maximum sum``    ``return` `mx + sum;``}` `// Driver Code``int` `main()``{``    ``// Given Input``    ``int` `arr[] = { 2, 4, 3 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `M = 5;` `    ``// Function Call``    ``cout << MaxSum(arr, N, M);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG``{``  ` `  ``// Function to maximize the sum after``// performing atmost one operation``static` `int` `MaxSum(``int` `arr[], ``int` `n, ``int` `M)``{``  ` `    ``// Variable to store the sum``    ``int` `sum = ``0``;` `    ``// Traverse the array and modify arr[]``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``sum += arr[i];``        ``arr[i] = (M - ``2` `* arr[i]);``    ``}` `    ``int` `ans = ``0``, mx = ``0``;` `    ``// Apply Kadane's algorithm``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``      ` `        ``// Add a[i] to ans``        ``ans += arr[i];` `        ``// If ans<0, modify the value``        ``// of ans as 0``        ``if` `(ans < ``0``)``            ``ans = ``0``;` `        ``// Update the value of mx``        ``mx = Math.max(mx, ans);``    ``}` `    ``// Return the maximum sum``    ``return` `mx + sum;``}` `// Driver Code``    ``public` `static` `void` `main (String[] args) {``       ``// Given Input``    ``int` `arr[] = { ``2``, ``4``, ``3` `};``    ``int` `N = arr.length;``    ``int` `M = ``5``;` `    ``// Function Call``     ` `        ``System.out.println(MaxSum(arr, N, M));``    ``}``}` `// This code is contributed by potta lokesh.`

## Python3

 `# python 3 program for the above approach` `# Function to maximize the sum after``# performing atmost one operation``def` `MaxSum(arr, n, M):``    ``# Variable to store the sum``    ``sum` `=` `0` `    ``# Traverse the array and modify arr[]``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]``        ``arr[i] ``=` `(M ``-` `2` `*` `arr[i])` `    ``ans ``=` `0``    ``mx ``=` `0` `    ``# Apply Kadane's algorithm``    ``for` `i ``in` `range``(n):``      ` `        ``# Add a[i] to ans``        ``ans ``+``=` `arr[i]` `        ``# If ans<0, modify the value``        ``# of ans as 0``        ``if` `(ans < ``0``):``            ``ans ``=` `0` `        ``# Update the value of mx``        ``mx ``=` `max``(mx, ans)` `    ``# Return the maximum sum``    ``return` `mx ``+` `sum` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given Input``    ``arr ``=` `[``2``, ``4``, ``3``]``    ``N ``=` `len``(arr)``    ``M ``=` `5` `    ``# Function Call``    ``print``(MaxSum(arr, N, M))``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {` `    ``// Function to maximize the sum after``    ``// performing atmost one operation``    ``static` `int` `MaxSum(``int``[] arr, ``int` `n, ``int` `M)``    ``{` `        ``// Variable to store the sum``        ``int` `sum = 0;` `        ``// Traverse the array and modify arr[]``        ``for` `(``int` `i = 0; i < n; i++) {``            ``sum += arr[i];``            ``arr[i] = (M - 2 * arr[i]);``        ``}` `        ``int` `ans = 0, mx = 0;` `        ``// Apply Kadane's algorithm``        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// Add a[i] to ans``            ``ans += arr[i];` `            ``// If ans<0, modify the value``            ``// of ans as 0``            ``if` `(ans < 0)``                ``ans = 0;` `            ``// Update the value of mx``            ``mx = Math.Max(mx, ans);``        ``}` `        ``// Return the maximum sum``        ``return` `mx + sum;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``// Given Input``        ``int``[] arr = { 2, 4, 3 };``        ``int` `N = arr.Length;``        ``int` `M = 5;` `        ``// Function Call` `        ``Console.WriteLine(MaxSum(arr, N, M));``    ``}``}` `// This code is contributed by ukasp.`

## Javascript

 ``
Output
`10`

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

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