Max XOR sum

Given four integers A, B, C, and D, the task is to find the maximum sum of the given numbers by using the below operations any number of times.

• Calculate X = A ^ B, and replace A or B with  X
• Calculate Y = C ^ D, and replace C or D with Y.

Examples:

Input: arr[] = {4, 5, 9, 12}
Output: 30
Explanation: X = 4 ^ 5 = 1
Y = 9 ^ 12 = 5 replace A with X and D with Y
X = 1 ^ 5 = 4
Y = 9 ^ 5 = 12 replace A with X and D with Y
Now, A + B + C + D = 30, it can be proved that this is the maximum sum that can be obtained

Input: arr[] = {0, 1, 0, 1}
Output: 4
Explanation: X = 0 ^ 1 = 1
Y = 1 ^ 0 = 1 replace A with X and C with Y
Now, A + B + C + D = 4, It can be proved that this is the maximum sum that can be proved.

Approach: This can be solved with the following idea:

This can be solved by some mathematical operations.

Below are the steps of implementation:

• Calculate the XOR of A and B, C and D.
• See maximum of (A + B), ( A + X), and (B + X).
• Similarly for C and D.
• Return the sum of the maximums found above.

Below is the implementation for the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

// Function to find maximum sum
int maxSumXor(int A, int B, int C, int D)
{
    // Calculate X
    int X = A ^ B;

    // Calculate Y
    int Y = C ^ D;

    // Calculating maximum value
    // A + b can gain
    int fpos = max(A + B, max(A + X, X + B));

    // Calculating maximum value
    // C + D can gain
    int spos = max(C + D, max(C + Y, Y + D));

    // Calculating (maximum of A + B) +
    // (maximum of C + D)
    int ans = fpos + spos;

    return ans;
}

// Driver code
int main()
{
    int a = 4, b = 5, c = 9, d = 12;

    // Function call
    cout << maxSumXor(a, b, c, d);
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;

class GFG {

  // Function to find maximum sum
  public static int maxSumXor(int A, int B, int C, int D)
  {
    // Calculate X
    int X = A ^ B;

    // Calculate Y
    int Y = C ^ D;

    // Calculating maximum value
    // A + b can gain
    int fpos = Math.max(A + B, Math.max(A + X, X + B));

    // Calculating maximum value
    // C + D can gain
    int spos = Math.max(C + D, Math.max(C + Y, Y + D));

    // Calculating (maximum of A + B) +
    // (maximum of C + D)
    int ans = fpos + spos;

    return ans;
  }

  // Driver code
  public static void main(String[] args)
  {
    int a = 4, b = 5, c = 9, d = 12;

    // Function call
    System.out.println(maxSumXor(a, b, c, d));
  }
}

// This code is contributed by Prasad Kandekar(prasad264)

Python3

# Python code for the above approach

# Function to find maximum sum


def maxSumXor(A, B, C, D):
    # Calculate X
    X = A ^ B

    # Calculate Y
    Y = C ^ D

    # Calculating maximum value
    # A + b can gain
    fpos = max(A + B, max(A + X, X + B))

    # Calculating maximum value
    # C + D can gain
    spos = max(C + D, max(C + Y, Y + D))

    # Calculating (maximum of A + B) +
    # (maximum of C + D)
    ans = fpos + spos

    return ans


# Driver code
a, b, c, d = 4, 5, 9, 12

# Function call
print(maxSumXor(a, b, c, d))

C#

// C# code for the above approach

using System;

class GFG {
    // Function to find maximum sum
    static int maxSumXor(int A, int B, int C, int D)
    {
        // Calculate X
        int X = A ^ B;

        // Calculate Y
        int Y = C ^ D;

        // Calculating maximum value
        // A + b can gain
        int fpos = Math.Max(A + B, Math.Max(A + X, X + B));

        // Calculating maximum value
        // C + D can gain
        int spos = Math.Max(C + D, Math.Max(C + Y, Y + D));

        // Calculating (maximum of A + B) +
        // (maximum of C + D)
        int ans = fpos + spos;

        return ans;
    }

    // Driver code
    public static void Main()
    {
        int a = 4, b = 5, c = 9, d = 12;

        // Function call
        Console.Write(maxSumXor(a, b, c, d));
    }
}

Javascript

<script>
  // Function to find maximum sum
  function maxSumXor(A, B, C, D) {
    // Calculate X
    let X = A ^ B;

    // Calculate Y
    let Y = C ^ D;

    // Calculating maximum value
    // A + b can gain
    let fpos = Math.max(A + B, Math.max(A + X, X + B));

    // Calculating maximum value
    // C + D can gain
    let spos = Math.max(C + D, Math.max(C + Y, Y + D));

    // Calculating (maximum of A + B) +
    // (maximum of C + D)
    let ans = fpos + spos;

    return ans;
  }

  // Driver code
  let a = 4,
    b = 5,
    c = 9,
    d = 12;

  // Function call
  console.log(maxSumXor(a, b, c, d));
</script>

Output

30

Time Complexity: O(1)
Auxiliary space: O(1)