Maximum sum of Bitwise XOR of all elements of two equal length subsets

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Given an array arr[] of N integers, where N is an even number. The task is to divide the given N integers into two equal subsets such that the sum of Bitwise XOR of all elements of two subsets is maximum.

Examples:

Input: N= 4, arr[] = {1, 2, 3, 4}
Output: 10
Explanation:
There are 3 ways possible:
(1, 2)(3, 4) = (1^2)+(3^4) = 10
(1, 3)(2, 4) = (1^3)+(2^3) = 8
(1, 4)(2, 3) = (1^4)+(2^3) = 6
Hence, the maximum sum = 10

Input: N= 6, arr[] = {4, 5, 3, 2, 5, 6}
Output: 17

Naive Approach: The idea is to check every possible distribution of N/2 pairs. Print the Bitwise XOR of the sum of all elements of two subsets which is maximum.

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

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming Using Bit Masking. Follow the below steps to solve the problem:

Initially, the bitmask is 0, if the bit is set then the pair is already picked.
Iterate through all the possible pairs and check if it is possible to pick a pair i.e., the bits of i and j are not set in the mask:
- If it is possible to take the pair then find the Bitwise XOR sum for the current pair and check for the next pair recursively.
- Else check for the next pair of elements.
Keep updating the maximum XOR pair sum in the above step for each recursive call.
Print the maximum value of all possible pairs stored in dp[mask].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function that finds the maximum
// Bitwise XOR sum of the two subset
int xorSum(int a[], int n, 
           int mask, int dp[])
{
  // Check if the current state is
  // already computed
  if (dp[mask] != -1) 
  {
    return dp[mask];
  }
 
  // Initialize answer to minimum value
  int max_value = 0;
 
  // Iterate through all possible pairs
  for (int i = 0; i < n; i++)
  {
    for (int j = i + 1; j < n; j++) 
    {
 
      // Check whether ith bit and
      // jth bit of mask is not
      // set then pick the pair
      if (i != j && 
         (mask & (1 << i)) == 0 && 
         (mask & (1 << j)) == 0) 
      {
 
        // For all possible pairs
        // find maximum value pick
        // current a[i], a[j] and
        // set i, j th bits in mask
        max_value = max(max_value, (a[i] ^ a[j]) + 
                        xorSum(a, n, (mask | (1 << i) | 
                                               (1 << j)), dp));
      }
    }
  }
 
  // Store the maximum value
  // and return the answer
  return dp[mask] = max_value;
}
 
// Driver Code
int main()
{
   int n = 4;
 
   // Given array arr[]
   int arr[] = { 1, 2, 3, 4 };
 
   // Declare Initialize the dp states
   int dp[(1 << n) + 5];
   memset(dp, -1, sizeof(dp));
 
   // Function Call
   cout << (xorSum(arr, n, 0, dp));
}
 
// This code is contributed by Rohit_ranjan

Java

// Java program for the above approach
import java.util.*;
import java.io.*;
 
public class GFG {
 
    // Function that finds the maximum
    // Bitwise XOR sum of the two subset
    public static int xorSum(int a[], int n,
                             int mask, int[] dp)
    {
        // Check if the current state is
        // already computed
        if (dp[mask] != -1) {
            return dp[mask];
        }
 
        // Initialize answer to minimum value
        int max_value = 0;
 
        // Iterate through all possible pairs
        for (int i = 0; i < n; i++) {
 
            for (int j = i + 1; j < n; j++) {
 
                // Check whether ith bit and
                // jth bit of mask is not
                // set then pick the pair
                if (i != j
                    && (mask & (1 << i)) == 0
                    && (mask & (1 << j)) == 0) {
 
                    // For all possible pairs
                    // find maximum value pick
                    // current a[i], a[j] and
                    // set i, j th bits in mask
                    max_value = Math.max(
                        max_value,
                        (a[i] ^ a[j])
                            + xorSum(a, n,
                                     (mask | (1 << i)
                                      | (1 << j)),
                                     dp));
                }
            }
        }
 
        // Store the maximum value
        // and return the answer
        return dp[mask] = max_value;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 4;
 
        // Given array arr[]
        int arr[] = { 1, 2, 3, 4 };
 
        // Declare Initialize the dp states
        int dp[] = new int[(1 << n) + 5];
        Arrays.fill(dp, -1);
 
        // Function Call
        System.out.println(xorSum(arr, n, 0, dp));
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function that finds the maximum 
# Bitwise XOR sum of the two subset
def xorSum(a, n, mask, dp):
 
    # Check if the current state is
    # already computed
    if(dp[mask] != -1):
        return dp[mask]
 
    # Initialize answer to minimum value
    max_value = 0
 
    # Iterate through all possible pairs
    for i in range(n):
        for j in range(i + 1, n):
 
            # Check whether ith bit and
            # jth bit of mask is not
            # set then pick the pair
            if(i != j and
              (mask & (1 << i)) == 0 and
              (mask & (1 << j)) == 0):
 
                # For all possible pairs
                # find maximum value pick
                # current a[i], a[j] and
                # set i, j th bits in mask
                max_value = max(max_value,
                               (a[i] ^ a[j]) +
                                xorSum(a, n, 
                                (mask | (1 << i) |
                                (1 << j)), dp))
 
    # Store the maximum value
    # and return the answer
    dp[mask] = max_value
 
    return dp[mask]
 
# Driver Code
n = 4
 
# Given array arr[]
arr = [ 1, 2, 3, 4 ]
 
# Declare Initialize the dp states
dp = [-1] * ((1 << n) + 5)
 
# Function call
print(xorSum(arr, n, 0, dp))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function that finds the maximum
// Bitwise XOR sum of the two subset
public static int xorSum(int []a, int n,
                         int mask, int[] dp)
{
     
    // Check if the current state is
    // already computed
    if (dp[mask] != -1)
    {
        return dp[mask];
    }
 
    // Initialize answer to minimum value
    int max_value = 0;
 
    // Iterate through all possible pairs
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
             
            // Check whether ith bit and
            // jth bit of mask is not
            // set then pick the pair
            if (i != j &&
               (mask & (1 << i)) == 0 && 
               (mask & (1 << j)) == 0) 
            {
 
                // For all possible pairs
                // find maximum value pick
                // current a[i], a[j] and
                // set i, j th bits in mask
                max_value = Math.Max(
                            max_value,
                            (a[i] ^ a[j]) + 
                            xorSum(a, n, (mask |
                            (1 << i) | (1 << j)), dp));
            }
        }
    }
 
    // Store the maximum value
    // and return the answer
    return dp[mask] = max_value;
}
 
// Driver Code
public static void Main(String []args)
{
    int n = 4;
 
    // Given array []arr
    int []arr = { 1, 2, 3, 4 };
 
    // Declare Initialize the dp states
    int []dp = new int[(1 << n) + 5];
    for(int i = 0; i < dp.Length; i++)
        dp[i] = -1;
 
    // Function call
    Console.WriteLine(xorSum(arr, n, 0, dp));
}
}
 
// This code is contributed by amal kumar choubey

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function that finds the maximum
// Bitwise XOR sum of the two subset
function xorSum(a, n, mask, dp)
{
  // Check if the current state is
  // already computed
  if (dp[mask] != -1) 
  {
    return dp[mask];
  }
 
  // Initialize answer to minimum value
  var max_value = 0;
 
  // Iterate through all possible pairs
  for (var i = 0; i < n; i++)
  {
    for (var j = i + 1; j < n; j++) 
    {
 
      // Check whether ith bit and
      // jth bit of mask is not
      // set then pick the pair
      if (i != j && 
         (mask & (1 << i)) == 0 && 
         (mask & (1 << j)) == 0) 
      {
 
        // For all possible pairs
        // find maximum value pick
        // current a[i], a[j] and
        // set i, j th bits in mask
        max_value = Math.max(max_value, (a[i] ^ a[j]) + 
                        xorSum(a, n, (mask | (1 << i) | 
                                        (1 << j)), dp));
      }
    }
  }
 
  // Store the maximum value
  // and return the answer
  return dp[mask] = max_value;
}
 
// Driver Code
 
var n = 4;
 
// Given array arr[]
var arr = [1, 2, 3, 4];
 
// Declare Initialize the dp states
var dp = Array((1 << n) + 5).fill(-1);
 
// Function Call
document.write(xorSum(arr, n, 0, dp));
 
</script>

Output

Time Complexity: O(N²*2^N), where N is the size of the given array
Auxiliary Space: O(N)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.

Steps to solve this problem :

Create a table to store the solution of the subproblems.
Initialize the table with base cases
Fill up the table iteratively
Return the final solution

Implementation:

C++

// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function that finds the maximum
// Bitwise XOR sum of the two subset
int xorSum(int a[], int n)
{
    // Declare dp table and initialize
    // with all elements as 0
    int dp[1 << n];
    memset(dp, 0, sizeof(dp));
 
    // Fill the dp table
    for (int mask = 0; mask < (1 << n); mask++)
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                // Check whether ith bit and
                // jth bit of mask is not set
                // then pick the pair
                if ((mask & (1 << i)) == 0 &&
                    (mask & (1 << j)) == 0)
                {
                    // Update dp table with
                    // maximum value pick
                    // current a[i], a[j] and
                    // set i, j th bits in mask
                    dp[mask | (1 << i) | (1 << j)] =
                        max(dp[mask | (1 << i) | (1 << j)],
                            dp[mask] + (a[i] ^ a[j]));
                }
            }
        }
    }
 
    // Return the maximum value
    return dp[(1 << n) - 1];
}
 
// Driver Code
int main()
{
    int n = 4;
 
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
 
    // Function Call
    cout << (xorSum(arr, n));
}
 
// this code is contributed by bhardwajji

Java

// java code addition 
import java.io.*;
import java.util.*;
 
public class Main {
 
    // Function that finds the maximum
    // Bitwise XOR sum of the two subset
    static int xorSum(int a[], int n) 
    {
       
        // Declare dp table and initialize
        // with all elements as 0
        int dp[] = new int[1 << n];
        Arrays.fill(dp, 0);
 
        // Fill the dp table
        for (int mask = 0; mask < (1 << n); mask++) {
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) 
                {
                   
                    // Check whether ith bit and
                    // jth bit of mask is not set
                    // then pick the pair
                    if ((mask & (1 << i)) == 0 && (mask & (1 << j)) == 0) 
                    {
                       
                        // Update dp table with
                        // maximum value pick
                        // current a[i], a[j] and
                        // set i, j th bits in mask
                        dp[mask | (1 << i) | (1 << j)] =
                                Math.max(dp[mask | (1 << i) | (1 << j)],
                                        dp[mask] + (a[i] ^ a[j]));
                    }
                }
            }
        }
 
        // Return the maximum value
        return dp[(1 << n) - 1];
    }
 
    // Driver Code
    public static void main(String[] args) {
        int n = 4;
 
        // Given array arr[]
        int arr[] = { 1, 2, 3, 4 };
 
        // Function Call
        System.out.println(xorSum(arr, n));
    }
}
 
// The code is contributed by Nidhi goel. 

Python

# Python program for above approach
import math
 
# Function that finds the maximum
# Bitwise XOR sum of the two subset
 
 
def xorSum(a, n):
    # Declare dp table and initialize
    # with all elements as 0
    dp = [0]*(1 << n)
 
    # Fill the dp table
    for mask in range(1 << n):
        for i in range(n):
            for j in range(i+1, n):
 
                # Check whether ith bit and
                # jth bit of mask is not set
                # then pick the pair
                if (mask & (1 << i)) == 0 and (mask & (1 << j)) == 0:
 
                    # Update dp table with
                    # maximum value pick
                    # current a[i], a[j] and
                    # set i, j th bits in mask
                    dp[mask | (1 << i) | (1 << j)] = max(
                        dp[mask | (1 << i) | (1 << j)], dp[mask] + (a[i] ^ a[j]))
 
    # Return the maximum value
    return dp[(1 << n) - 1]
 
 
# Driver Code
if __name__ == '__main__':
    n = 4
    # Given array arr[]
    arr = [1, 2, 3, 4]
 
    # Function Call
    print(xorSum(arr, n))
# This code is contributed by user_dtewbxkn77n

C#

using System;
 
public class Program {
     
    // Function that finds the maximum
    // Bitwise XOR sum of the two subset
    static int xorSum(int[] a, int n)
    {
        // Declare dp table and initialize
        // with all elements as 0
        int[] dp = new int[1 << n];
        Array.Fill(dp, 0);
 
        // Fill the dp table
        for (int mask = 0; mask < (1 << n); mask++)
        {
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    // Check whether ith bit and
                    // jth bit of mask is not set
                    // then pick the pair
                    if ((mask & (1 << i)) == 0 &&
                        (mask & (1 << j)) == 0)
                    {
                        // Update dp table with
                        // maximum value pick
                        // current a[i], a[j] and
                        // set i, j th bits in mask
                        dp[mask | (1 << i) | (1 << j)] =
                            Math.Max(dp[mask | (1 << i) | (1 << j)],
                                     dp[mask] + (a[i] ^ a[j]));
                    }
                }
            }
        }
 
        // Return the maximum value
        return dp[(1 << n) - 1];
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        int n = 4;
 
        // Given array arr[]
        int[] arr = { 1, 2, 3, 4 };
 
        // Function Call
        Console.WriteLine(xorSum(arr, n));
    }
}

Javascript

// JavaScript code equivalent of the Java code above
 
function xorSum(a, n) {
// Declare dp table and initialize
// with all elements as 0
let dp = new Array(1 << n).fill(0);
 
// Fill the dp table
for (let mask = 0; mask < (1 << n); mask++) {
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
// Check whether ith bit and jth bit of mask is not set
// then pick the pair
if ((mask & (1 << i)) == 0 && (mask & (1 << j)) == 0) {
// Update dp table with
// maximum value pick
// current a[i], a[j] and
// set i, j th bits in mask
dp[mask | (1 << i) | (1 << j)] =
Math.max(
dp[mask | (1 << i) | (1 << j)],
dp[mask] + (a[i] ^ a[j])
);
}
}
}
}
 
// Return the maximum value
return dp[(1 << n) - 1];
}
 
// Driver Code
let n = 4;
 
// Given array arr[]
let arr = [1, 2, 3, 4];
 
// Function Call
console.log(xorSum(arr, n));

Output

Time Complexity: O(N²*2^N)
Auxiliary Space: O(N)

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