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Maximum prefix sum after K reversals of a given array
• Last Updated : 27 Apr, 2021

Given an array arr[] of size N and a positive integer K, the task is to find the maximum prefix sum after K reversals of the given array.

Examples:

Input: arr[] = {1, 5, 8, 9, 11, 2}, K = 1
Output: 36
Explanation: Reverse the array once. Therefore, the array becomes {2, 11, 9, 8, 5, 1}. Maximum prefix sum = 2 + 11 + 9 + 8 + 5 + 1 = 36.

Input: arr[] = {5, 6, -4, 3, -2, -10}, K = 2
Output : 11
Explanation: Reverse the array twice. Therefore, the array becomes {5, 6, -4, 3, -2, -10}. Maximum prefix sum = 5 + 6=11.

Naive Approach: The simplest approach is to reverse the array K times and after K reversals, find the maximum prefix sum possible by traversing the array and print the maximum sum.
Time Complexity: O(N * K)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the observation that if K is odd, then the array gets reversed. Otherwise, the array remains unchanged. Therefore, if K is odd, find the maximum suffix sum. Otherwise, find the maximum prefix sum. Follow the steps below to solve the problem:

• Initialize sum as INT_MIN to store the required answer.
• If K is odd, reverse the array arr[].
• Initialize currSum as 0 to store the prefix sum of elements.
• Traverse the array arr[] using the variable i and perform the following:
• Add arr[i] to the variable currSum.
• If the value of currSum is greater than the sum, then update the sum as currSum.
• After the above steps, print the value of the sum as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum prefix``// sum after K reversals of the array``int` `maxSumAfterKReverse(``int` `arr[],``                        ``int` `K, ``int` `N)``{``    ``// Stores the required sum``    ``int` `sum = INT_MIN;` `    ``// If K is odd, reverse the array``    ``if` `(K & 1)``        ``reverse(arr, arr + N);` `    ``// Store current prefix sum of array``    ``int` `currsum = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Add arr[i] to currsum``        ``currsum += arr[i];` `        ``// Update maximum prefix sum``        ``// till now``        ``sum = max(sum, currsum);``    ``}` `    ``// Print the answer``    ``cout << sum;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 5, 8, 9, 11, 2 };``    ``int` `K = 1;``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``maxSumAfterKReverse(arr, K, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;``class` `GFG``{` `  ``// Function to find the maximum prefix``  ``// sum after K reversals of the array``  ``static` `void` `maxSumAfterKReverse(Integer arr[], ``int` `K, ``int` `N)``  ``{``    ` `    ``// Stores the required sum``    ``int` `sum = Integer.MIN_VALUE;` `    ``// If K is odd, reverse the array``    ``if` `(K % ``2` `!= ``0``)``      ``Collections.reverse(Arrays.asList(arr));` `    ``// Store current prefix sum of array``    ``int` `currsum = ``0``;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{` `      ``// Add arr[i] to currsum``      ``currsum += arr[i];` `      ``// Update maximum prefix sum``      ``// till now``      ``sum = Math.max(sum, currsum);``    ``}` `    ``// Print the answer``    ``System.out.print(sum);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``Integer[] arr = { ``1``, ``5``, ``8``, ``9``, ``11``, ``2` `};``    ``int` `K = ``1``;``    ``int` `N = arr.length;` `    ``// Function Call``    ``maxSumAfterKReverse(arr, K, N);``  ``}``}` `// This code is contributed by Dharanendra L V.`

## Python3

 `# Python3 program for the above approach``import` `sys` `# Function to find the maximum prefix``# sum after K reversals of the array``def` `maxSumAfterKReverse(arr, K, N) :``    ` `    ``# Stores the required sum``    ``sum` `=` `-``sys.maxsize ``-` `1` `    ``# If K is odd, reverse the array``    ``if` `(K & ``1``) :``        ``arr.reverse()` `    ``# Store current prefix sum of array``    ``currsum ``=` `0` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(N):` `        ``# Add arr[i] to currsum``        ``currsum ``+``=` `arr[i]` `        ``# Update maximum prefix sum``        ``# till now``        ``sum` `=` `max``(``sum``, currsum)``    ` `    ``# Prthe answer``    ``print``(``sum``)` `# Driver Code``arr ``=` `[ ``1``, ``5``, ``8``, ``9``, ``11``, ``2` `]``K ``=` `1``N ``=` `len``(arr)` `# Function Call``maxSumAfterKReverse(arr, K, N)` `# This code is contributed by code_hunt.`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `  ``// Function to find the maximum prefix``  ``// sum after K reversals of the array``  ``static` `void` `maxSumAfterKReverse(``int``[] arr, ``int` `K, ``int` `N)``  ``{` `    ``// Stores the required sum``    ``int` `sum = Int32.MinValue;` `    ``// If K is odd, reverse the array``    ``if` `(K % 2 != 0)``      ``Array.Reverse(arr);` `    ``// Store current prefix sum of array``    ``int` `currsum = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++)``    ``{` `      ``// Add arr[i] to currsum``      ``currsum += arr[i];` `      ``// Update maximum prefix sum``      ``// till now``      ``sum = Math.Max(sum, currsum);``    ``}` `    ``// Print the answer``    ``Console.Write(sum);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] arr = { 1, 5, 8, 9, 11, 2 };``    ``int` `K = 1;``    ``int` `N = arr.Length;` `    ``// Function Call``    ``maxSumAfterKReverse(arr, K, N);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``

Output:
`36`

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

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