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Maximize Array sum by subtracting absolute of odd and adding absolute of even elements

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  • Last Updated : 07 Jun, 2022

Given an array arr[] of size N, the task is to maximize the array sum by subtracting the absolute values of all odd and adding and absolute values of all even elements, with only at most one exceptional even-odd pair, i.e., only one even value can be subtracted and one odd element can be added.

Examples:

Input: arr[] = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Output: 4
Explanation: Total absolute value subtracted = 5 + 3 + 1 + 1 + 3 + 5 = 18.
Total absolute value added = 4 + 2 + 0 + 2 + 4 = 12
Sum achieved = -6. Make {5, 0} the exceptional elements
Add 5 to sum and subtract 0 from sum.
So total value subtracted = 13 and total value added = 17
New Sum = 17 – 3 = 4. This is the maximum possible sum.

Input: arr[] = {1, -2, 3}
Output: 0

 

Approach: The problem can be solved based on the following idea:

To maximize the sum, the maximum absolute value which is subtracted (say max) and the minimum absolute value which is added (say min) should be the exceptional elements. So the sum will increase by 2*(max – min).

Now, this should be performed only when min is less than max. Otherwise, the above term will become negative and will reduce the total sum.

Follow the below steps to solve the problem:

  • Initialize the variables Max, Min, Sum to store maximum absolute odd, minimum absolute even and total sum respectively.
  • Traverse the array from 0 to N – 1.
    • If element is odd then subtract its absolute value from the Sum and update Max.
    • If element is even then add its absolute value to the Sum and update Min.
  •  If Min is greater than Max then no need to update sum.
  • Else, update the Sum, Sum = Sum + 2*(Max – Min).

Below is the implementation of the above approach: 

C++




// C++ code for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return maximum sum
// after swapping
int maxSum(int* arr, int N)
{
    // Initialize the variables
    int Max = INT_MIN;
    int Min = INT_MAX;
    int X, Sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        X = arr[i];
 
        // If element is odd then subtract
        // it from the Sum and update maximum
        if (X & 1) {
            Max = max(Max, abs(X));
            Sum -= abs(X);
        }
 
        // Else add it to the Sum and
        // update minimum
        else {
            Min = min(Min, abs(X));
            Sum += abs(X);
        }
    }
 
    // If minimum even element is greater
    // than maximum odd element then no
    // need to swap
    if (Min >= Max)
        return Sum;
 
    // Else print the sum after swapping
    return (Sum + 2 * (Max - Min));
}
 
// Driver Code
int main()
{
    int arr[] = { -5, -4, -3, -2, -1,
                  0, 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxSum(arr, N);
    return 0;
}

Java




// Java code for above approach
import java.io.*;
 
class GFG
{
   
    // Function to return maximum sum
    // after swapping
    public static int maxSum(int arr[], int N)
    {
       
        // Initialize the variables
        int Max = Integer.MIN_VALUE;
        int Min = Integer.MAX_VALUE;
        int X, Sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
            X = arr[i];
 
            // If element is odd then subtract
            // it from the Sum and update maximum
            if ((X & 1) != 0) {
                Max = Math.max(Max, Math.abs(X));
                Sum -= Math.abs(X);
            }
 
            // Else add it to the Sum and
            // update minimum
            else {
                Min = Math.min(Min, Math.abs(X));
                Sum += Math.abs(X);
            }
        }
 
        // If minimum even element is greater
        // than maximum odd element then no
        // need to swap
        if (Min >= Max)
            return Sum;
 
        // Else print the sum after swapping
        return (Sum + (2 * (Max - Min)));
    }
    public static void main(String[] args)
    {
        int arr[]
            = { -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 };
        int N = arr.length;
 
        // Function call
        System.out.print(maxSum(arr, N));
    }
}
 
// This code is contributed by Rohit Pradhan.

Python3




# Python code to implement the approach
 
# Function to return maximum sum
# after swapping
import sys
 
def maxSum(arr, N):
   
    # Initialize the variables
    Max = -sys.maxsize -1
    Min = sys.maxsize
    X, Sum = 0,0
 
    # Traverse the array
    for i in range(N):
        X = arr[i]
 
        # If element is odd then subtract
        # it from the Sum and update maximum
        if (X & 1):
            Max = max(Max, abs(X))
            Sum -= abs(X)
 
        # Else add it to the Sum and
        # update minimum
        else:
            Min = min(Min, abs(X))
            Sum += abs(X)
 
    # If minimum even element is greater
    # than maximum odd element then no
    # need to swap
    if (Min >= Max):
        return Sum
 
    # Else print the sum after swapping
    return (Sum + 2 * (Max - Min))
 
# Driver Code
 
arr = [ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 ]
N = len(arr)
 
# Function call
print(maxSum(arr, N))
 
# This code is contributed by shinjanpatra

C#




// C# code for above approach
using System;
class GFG {
 
    // Function to return maximum sum
    // after swapping
    static int maxSum(int[] arr, int N)
    {
        // Initialize the variables
        int Max = Int32.MinValue;
        int Min = Int32.MaxValue;
        int X, Sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
            X = arr[i];
 
            // If element is odd then subtract
            // it from the Sum and update maximum
            if ((X & 1) != 0) {
                Max = Math.Max(Max, Math.Abs(X));
                Sum -= Math.Abs(X);
            }
 
            // Else add it to the Sum and
            // update minimum
            else {
                Min = Math.Min(Min, Math.Abs(X));
                Sum += Math.Abs(X);
            }
        }
 
        // If minimum even element is greater
        // than maximum odd element then no
        // need to swap
        if (Min >= Max)
            return Sum;
 
        // Else print the sum after swapping
        return (Sum + 2 * (Max - Min));
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr
            = { -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 };
        int N = arr.Length;
 
        // Function call
        Console.Write(maxSum(arr, N));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return maximum sum
// after swapping
function maxSum(arr,N)
{
    // Initialize the variables
    let Max = Number.MIN_VALUE;
    let Min = Number.MAX_VALUE;
    let X, Sum = 0;
 
    // Traverse the array
    for (let i = 0; i < N; i++) {
        X = arr[i];
 
        // If element is odd then subtract
        // it from the Sum and update maximum
        if (X & 1) {
            Max = Math.max(Max, Math.abs(X));
            Sum -= Math.abs(X);
        }
 
        // Else add it to the Sum and
        // update minimum
        else {
            Min = Math.min(Min, Math.abs(X));
            Sum += Math.abs(X);
        }
    }
 
    // If minimum even element is greater
    // than maximum odd element then no
    // need to swap
    if (Min >= Max)
        return Sum;
 
    // Else print the sum after swapping
    return (Sum + 2 * (Max - Min));
}
 
// Driver Code
 
let arr = [ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 ];
let N = arr.length;
 
// Function call
document.write(maxSum(arr, N));
 
// This code is contributed by shinjanpatraa
 
</script>

Output

4

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


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