# Maximum prefix sum after K reversals of a given array

Last Updated : 11 Oct, 2023

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.

Algorithm to implement this approach:

`Reverse the entire array K times using a loop.We can implement the reversal operation by swapping the first and last elements, then the second and second-last elements, and so on, until we reach the middle of the array.Traverse the array and calculate the prefix sum at each index.Keep track of the maximum prefix sum seen so far.Return the maximum prefix sum.`

Implementation

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `int` `max_prefix_sum(vector<``int``>& arr, ``int` `K) {` `    ``int` `n = arr.size();` `    ``for` `(``int` `i = 0; i < K; i++) {` `        ``for` `(``int` `j = 0; j < n/2; j++) {` `            ``swap(arr[j], arr[n-j-1]);` `        ``}` `    ``}` `    ``int` `max_prefix_sum = 0;` `    ``int` `prefix_sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``prefix_sum += arr[i];` `        ``if` `(prefix_sum > max_prefix_sum) {` `            ``max_prefix_sum = prefix_sum;` `        ``}` `    ``}` `    ``return` `max_prefix_sum;` `}`   `int` `main() {` `    ``vector<``int``> arr = {1, 5, 8, 9, 11, 2};` `    ``int` `K = 1;` `    ``cout << max_prefix_sum(arr, K) << endl;` `    ``return` `0;` `}`

## Java

 `import` `java.util.ArrayList;`   `public` `class` `Main {` `    ``public` `static` `int` `maxPrefixSum(ArrayList arr, ``int` `K) {` `        ``int` `n = arr.size();`   `        ``// Reverse the first n/2 elements of the array K times` `        ``for` `(``int` `i = ``0``; i < K; i++) {` `            ``for` `(``int` `j = ``0``; j < n/``2``; j++) {` `                ``int` `temp = arr.get(j);` `                ``arr.set(j, arr.get(n-j-``1``));` `                ``arr.set(n-j-``1``, temp);` `            ``}` `        ``}`   `        ``// Compute the maximum prefix sum` `        ``int` `max_prefix_sum = ``0``;` `        ``int` `prefix_sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``prefix_sum += arr.get(i);` `            ``if` `(prefix_sum > max_prefix_sum) {` `                ``max_prefix_sum = prefix_sum;` `            ``}` `        ``}` `        ``return` `max_prefix_sum;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``ArrayList arr = ``new` `ArrayList();` `        ``arr.add(``1``);` `        ``arr.add(``5``);` `        ``arr.add(``8``);` `        ``arr.add(``9``);` `        ``arr.add(``11``);` `        ``arr.add(``2``);` `        ``int` `K = ``1``;` `        ``System.out.println(maxPrefixSum(arr, K));` `    ``}` `}`

## Python

 `def` `max_prefix_sum(arr, K):` `    ``n ``=` `len``(arr)` `    ``for` `i ``in` `range``(K):` `        ``for` `j ``in` `range``(n``/``/``2``):` `            ``arr[j], arr[n``-``j``-``1``] ``=` `arr[n``-``j``-``1``], arr[j]` `    ``max_prefix_sum ``=` `0` `    ``prefix_sum ``=` `0` `    ``for` `i ``in` `range``(n):` `        ``prefix_sum ``+``=` `arr[i]` `        ``if` `prefix_sum > max_prefix_sum:` `            ``max_prefix_sum ``=` `prefix_sum` `    ``return` `max_prefix_sum` `arr ``=` `[``1``, ``5``, ``8``, ``9``, ``11``, ``2``]` `K ``=` `1` `print``(max_prefix_sum(arr, K))`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {` `    ``public` `static` `int` `MaxPrefixSum(List<``int``> arr, ``int` `K) {` `        ``int` `n = arr.Count;`   `        ``// Reverse the first n/2 elements of the array K times` `        ``for` `(``int` `i = 0; i < K; i++) {` `            ``for` `(``int` `j = 0; j < n / 2; j++) {` `                ``int` `temp = arr[j];` `                ``arr[j] = arr[n - j - 1];` `                ``arr[n - j - 1] = temp;` `            ``}` `        ``}`   `        ``// Compute the maximum prefix sum` `        ``int` `maxPrefixSum = 0;` `        ``int` `prefixSum = 0;` `        ``foreach` `(``int` `num ``in` `arr) {` `            ``prefixSum += num;` `            ``if` `(prefixSum > maxPrefixSum) {` `                ``maxPrefixSum = prefixSum;` `            ``}` `        ``}` `        ``return` `maxPrefixSum;` `    ``}`   `    ``public` `static` `void` `Main(``string``[] args) {` `        ``List<``int``> arr = ``new` `List<``int``> { 1, 5, 8, 9, 11, 2 };` `        ``int` `K = 1;` `        ``Console.WriteLine(MaxPrefixSum(arr, K));` `    ``}` `}`

## Javascript

 `function` `maxPrefixSum(arr, K) {` `    ``const n = arr.length;` `    `  `    ``// Reverse the first K subarrays` `    ``for` `(let i = 0; i < K; i++) {` `        ``for` `(let j = 0; j < Math.floor(n / 2); j++) {` `            ``[arr[j], arr[n - j - 1]] = [arr[n - j - 1], arr[j]]; ``// Swap elements` `        ``}` `    ``}` `    `  `    ``let maxPrefixSum = 0;` `    ``let prefixSum = 0;` `    `  `    ``// Calculate the maximum prefix sum` `    ``for` `(let i = 0; i < n; i++) {` `        ``prefixSum += arr[i];` `        ``if` `(prefixSum > maxPrefixSum) {` `            ``maxPrefixSum = prefixSum;` `        ``}` `    ``}` `    `  `    ``return` `maxPrefixSum;` `}`   `const arr = [1, 5, 8, 9, 11, 2];` `const K = 1;` `console.log(maxPrefixSum(arr, K));`

Output

```36

```

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[0]);`   `    ``// 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)` `    `  `    ``# Print the 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)