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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

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  • Last Updated : 06 Jul, 2021
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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 <bits/stdc++.h>
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




<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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