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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. 

Below is the implementation of the above approach:




// 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 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.


















                        






                        























                                    



                                    




                                    




                                    


                                



















# 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# 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.


















                        






                        























                                    



                                    




                                    




                                    


                                



















<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)
 

	

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