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 = 10Input: 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 <bits/stdc++.h>using namespace std;// Function to maximize the sum after// performing atmost one operationint 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 Codeint 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 approachimport java.io.*;class GFG { // Function to maximize the sum after// performing atmost one operationstatic 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 operationdef 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 Codeif __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 approachusing 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
<script> // JavaScript program for the above approach // Function to maximize the sum after // performing atmost one operation function MaxSum(arr, n, M) { // Variable to store the sum let sum = 0; // Traverse the array and modify arr[] for (let i = 0; i < n; i++) { sum += arr[i]; arr[i] = (M - 2 * arr[i]); } let ans = 0, mx = 0; // Apply Kadane's algorithm for (let 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 // Given Input let arr = [2, 4, 3]; let N = arr.length; let M = 5; // Function Call document.write(MaxSum(arr, N, M)); // This code is contributed by Potta Lokesh </script> |
Output
10
Time Complexity: O(N)
Auxiliary Space: O(1)
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