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 <iostream>#include <vector>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<Integer> 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<Integer> arr = new ArrayList<Integer>(); 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_sumarr = [1, 5, 8, 9, 11, 2]K = 1print(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 <bits/stdc++.h>using namespace std;// Function to find the maximum prefix// sum after K reversals of the arrayint 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 Codeint 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 approachimport 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 approachimport sys# Function to find the maximum prefix# sum after K reversals of the arraydef 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 Codearr = [ 1, 5, 8, 9, 11, 2 ]K = 1N = len(arr)# Function CallmaxSumAfterKReverse(arr, K, N)# This code is contributed by code_hunt. |
C#
// C# program for the above approachusing 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
<script>// Javascript program for the above approach // Function to find the maximum prefix // sum after K reversals of the array function maxSumAfterKReverse(arr, K, N) { // Stores the required sum let sum = Number.MIN_VALUE; // If K is odd, reverse the array if (K % 2 != 0) arr.reverse(); // Store current prefix sum of array let currsum = 0; // Traverse the array arr[] for (let 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 document.write(sum); }// Driver Code let arr = [ 1, 5, 8, 9, 11, 2 ]; let K = 1; let N = arr.length; // Function Call maxSumAfterKReverse(arr, K, N); </script> |
Output
36
Time Complexity: O(N)
Auxiliary Space: O(1)
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