Maximize sum of XOR of each element of Array with partition number

Given an array arr of positive integers of size N, the task is to split the array into 3 partitions, such that the sum of bitwise XOR of each element of the array with its partition number is maximum.

Examples:

Input: arr[] = { 2, 4, 7, 1, 8, 7, 2 }
Output: First partition: 2 4 7 1 8
Second partition: 7
Third partition: 2
Sum: 244

Input: arr[] = {95, 2, 86, 12, 9, 14, 45, 11}
Output: First partition: 95 2 86 12 9 14
Second partition: 45
Third partition: 11
Sum: 1994

Approach: The idea is to use nested loops for three partitions. 

• Compute the XOR sum of each element of each partition with its partition number.
• Find the global maximum sum of all three partitions’ XOR sum.
• Return and print the three partitions and their maximum XOR sum.

Below is the implementation of the above approach:

C++

// C++ program for maximize the sum of
// bitwise XOR of each element of the array
// with it's partition number

#include <bits/stdc++.h>
using namespace std;

// Utility function to print the partitions
void ShowPartition(vector<int> sum, vector<int> arr)
{
    cout << "First partition: ";
    for (int i = 0; i <= sum[0]; i++)
        cout << arr[i] << " ";

    cout << "\nSecond partition: ";
    for (int i = sum[0] + 1; i <= sum[1]; i++)
        cout << arr[i] << " ";

    cout << "\nThird partition: ";
    for (int i = sum[1] + 1; i <= sum[2]; i++)
        cout << arr[i] << " ";

    cout << "\nSum: ";
    cout << sum[3];
}

// Function to maximise the partitions sum
vector<int> MaximumSumPartition(vector<int> arr)
{
    int i, j, k;
    int n = arr.size();
    vector<int> sum(4, 0);

    // initialise the dummy sum values.
    int s1 = 0, s2 = 0, s3 = 0, s = INT_MIN;
    int x, y, z;

    // nested for  loop
    for (i = 0; i <= n - 3; i++) {

        // XOR sum of first partition.
        s1 += 1 ^ arr[i];
        x = i;

        for (j = i + 1; j <= n - 2; j++) {

            // XOR sum of second partition.
            s2 += 2 ^ arr[j];
            y = j;

            for (k = j + 1; k <= n - 1; k++) {

                // XOR sum of third partition.
                s3 += 3 ^ arr[k];
                z = k;

                // XOR sum of all three partition.
                if (s1 + s2 + s3 > s) {

                    s = s1 + s2 + s3;
                    sum[0] = x;
                    sum[1] = y;
                    sum[2] = z;
                    sum[3] = s;
                }
            }
        }
    }

    // return the vector.
    return sum;
}

// Driver code
int main()
{
    vector<int> sum, arr{ 2, 4, 7, 1, 8, 7, 2 };

    sum = MaximumSumPartition(arr);
    ShowPartition(sum, arr);
    return 0;
}

Java

// Java program for maximize the sum of
// bitwise XOR of each element of the array
// with it's partition number
import java.io.*;

class GFG {

  // Utility function to print the partitions
  static void ShowPartition(int []sum, int []arr)
  {
    System.out.print("First partition: ");
    for (int i = 0; i <= sum[0]; i++)
      System.out.print(arr[i] + " ");

    System.out.print("\nSecond partition: ");
    for (int i = sum[0] + 1; i <= sum[1]; i++)
      System.out.print(arr[i] + " ");

    System.out.print("\nThird partition: ");
    for (int i = sum[1] + 1; i <= sum[2]; i++)
      System.out.print(arr[i] + " ");

    System.out.print("\nSum: ");
    System.out.print(sum[3]);
  }

  // Function to maximise the partitions sum
  static int[] MaximumSumPartition(int []arr)
  {
    int i = 0, j = 0, k = 0;
    int n = arr.length;
    int []sum = new int[4];
    for(i = 0; i < 4; i++) {
      sum[i] = 0;
    }

    // initialise the dummy sum values.
    int s1 = 0, s2 = 0, s3 = 0, s = Integer.MIN_VALUE;
    int x = 0, y = 0, z = 0;

    // nested for  loop
    for (i = 0; i <= n - 3; i++) {

      // XOR sum of first partition.
      s1 += 1 ^ arr[i];
      x = i;

      for (j = i + 1; j <= n - 2; j++) {

        // XOR sum of second partition.
        s2 += 2 ^ arr[j];
        y = j;

        for (k = j + 1; k <= n - 1; k++) {

          // XOR sum of third partition.
          s3 += 3 ^ arr[k];
          z = k;

          // XOR sum of all three partition.
          if (s1 + s2 + s3 > s) {

            s = s1 + s2 + s3;
            sum[0] = x;
            sum[1] = y;
            sum[2] = z;
            sum[3] = s;
          }
        }
      }
    }

    // return the vector.
    return sum;
  }

  // Driver code
  public static void main (String[] args) {
    int []arr = { 2, 4, 7, 1, 8, 7, 2 };

    int []sum = MaximumSumPartition(arr);
    ShowPartition(sum, arr);

  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python code to implement the approach
import sys

# Utility function to print the partitions
def ShowPartition(sum, arr)  :

    print("First partition: ", end = '')
    for i in range(sum[0]+1) :
        print(arr[i] , end = " ")

    print("\nSecond partition: ", end = '')
    for i in range(sum[0]+1, sum[1]+1) :
        print(arr[i] , end = " ")

    print("\nThird partition: ", end = '')
    for i in range(sum[1]+1, sum[2]+1) :
        print(arr[i] , end = " ")

    print("\nSum: ", end = '')
    print(sum[3])


  # Function to maximise the partitions sum
def MaximumSumPartition(arr) :

    i = 0
    j = 0
    k = 0
    n = len(arr)
    sum = [0] * 4
    for i in range(0, 4):
        sum[i] = 0


    # initialise the dummy sum values.
    s1 = 0
    s2 = 0
    s3 = 0
    s = -sys.maxsize -1
    x = 0
    y = 0
    z = 0

    # nested for  loop
    for i in range(0, n-2, 1):

        # XOR sum of first partition.
        s1 += 1 ^ arr[i]
        x = i

    for j in range(i + 1, n - 1, 1) :

        # XOR sum of second partition.
        s2 += 2 ^ arr[j]
        y = j

    for k in range(j + 1, n, 1) :

        # XOR sum of third partition.
        s3 += 3 ^ arr[k]
        z = k

        # XOR sum of all three partition.
        if (s1 + s2 + s3 > s) :

            s = s1 + s2 + s3
            sum[0] = x
            sum[1] = y
            sum[2] = z
            sum[3] = s

    # return the vector.
    return sum

# Driver code
arr = [ 2, 4, 7, 1, 8, 7, 2 ]

sum = MaximumSumPartition(arr);
ShowPartition(sum, arr);

# This code is contributed by code_hunt.

C#

// C# program for maximize the sum of
// bitwise XOR of each element of the array
// with it's partition number
using System;
class GFG
{

  // Utility function to print the partitions
  static void ShowPartition(int []sum, int []arr)
  {
    Console.Write("First partition: ");
    for (int i = 0; i <= sum[0]; i++)
      Console.Write(arr[i] + " ");

    Console.Write("\nSecond partition: ");
    for (int i = sum[0] + 1; i <= sum[1]; i++)
      Console.Write(arr[i] + " ");

    Console.Write("\nThird partition: ");
    for (int i = sum[1] + 1; i <= sum[2]; i++)
      Console.Write(arr[i] + " ");

    Console.Write("\nSum: ");
    Console.Write(sum[3]);
  }

  // Function to maximise the partitions sum
  static int[] MaximumSumPartition(int []arr)
  {
    int i = 0, j = 0, k = 0;
    int n = arr.Length;
    int []sum = new int[4];
    for(i = 0; i < 4; i++) {
      sum[i] = 0;
    }

    // initialise the dummy sum values.
    int s1 = 0, s2 = 0, s3 = 0, s = Int32.MinValue;
    int x = 0, y = 0, z = 0;

    // nested for  loop
    for (i = 0; i <= n - 3; i++) {

      // XOR sum of first partition.
      s1 += 1 ^ arr[i];
      x = i;

      for (j = i + 1; j <= n - 2; j++) {

        // XOR sum of second partition.
        s2 += 2 ^ arr[j];
        y = j;

        for (k = j + 1; k <= n - 1; k++) {

          // XOR sum of third partition.
          s3 += 3 ^ arr[k];
          z = k;

          // XOR sum of all three partition.
          if (s1 + s2 + s3 > s) {

            s = s1 + s2 + s3;
            sum[0] = x;
            sum[1] = y;
            sum[2] = z;
            sum[3] = s;
          }
        }
      }
    }

    // return the vector.
    return sum;
  }

  // Driver code
  public static void Main()
  {
    int []arr = { 2, 4, 7, 1, 8, 7, 2 };

    int []sum = MaximumSumPartition(arr);
    ShowPartition(sum, arr);
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript program for maximize the sum of
    // bitwise XOR of each element of the array
    // with it's partition number
    const INT_MIN = -2147483647 - 1;

    // Utility function to print the partitions
    const ShowPartition = (sum, arr) => {
        document.write("First partition: ");
        for (let i = 0; i <= sum[0]; i++)
            document.write(`${arr[i]} `);

        document.write("<br/>Second partition: ");
        for (let i = sum[0] + 1; i <= sum[1]; i++)
            document.write(`${arr[i]} `);

        document.write("<br/>Third partition: ");
        for (let i = sum[1] + 1; i <= sum[2]; i++)
            document.write(`${arr[i]} `);

        document.write("<br/>Sum: ");
        document.write(sum[3]);
    }

    // Function to maximise the partitions sum
    const MaximumSumPartition = (arr) => {
        let i, j, k;
        let n = arr.length;
        let sum = new Array(4).fill(0);

        // initialise the dummy sum values.
        let s1 = 0, s2 = 0, s3 = 0, s = INT_MIN;
        let x, y, z;

        // nested for loop
        for (i = 0; i <= n - 3; i++) {

            // XOR sum of first partition.
            s1 += 1 ^ arr[i];
            x = i;

            for (j = i + 1; j <= n - 2; j++) {

                // XOR sum of second partition.
                s2 += 2 ^ arr[j];
                y = j;

                for (k = j + 1; k <= n - 1; k++) {

                    // XOR sum of third partition.
                    s3 += 3 ^ arr[k];
                    z = k;

                    // XOR sum of all three partition.
                    if (s1 + s2 + s3 > s) {

                        s = s1 + s2 + s3;
                        sum[0] = x;
                        sum[1] = y;
                        sum[2] = z;
                        sum[3] = s;
                    }
                }
            }
        }
        // return the vector.
        return sum;
    }

    // Driver code
    let arr = [2, 4, 7, 1, 8, 7, 2];

    let sum = MaximumSumPartition(arr);
    ShowPartition(sum, arr);

    // This code is contributed by rakeshsahni

</script>

Output

First partition: 2 4 7 1 8 
Second partition: 7 
Third partition: 2 
Sum: 244

Time Complexity: O(N³)
Auxiliary Space: O(1)