Maximum value of XOR among all triplets of an array

Given an array of integers ‘arr’, the task is to find the maximum XOR value of any triplet pair among all the possible triplet pairs.
Note: An array element can be used more than once.

Examples:

Input: arr[] = {3, 4, 5, 6}
Output: 7
The triplet with maximum XOR value is {4, 5, 6}.

Input: arr[] = {1, 3, 8, 15}
Output: 15

Approach:

  • Store all possible values of XOR between all possible two-element pairs from the array in a set.
  • Set data structure is used to avoid the repetitions of XOR values.
  • Now, XOR between every set element and array element to get the maximum value for any triplet pair.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// function to count maximum
// XOR value for a triplet
void Maximum_xor_Triplet(int n, int a[])
{
    // set is used to avoid repetitions
    set<int> s;
  
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
  
            // store all possible unique
            // XOR value of pairs
            s.insert(a[i] ^ a[j]);
        }
    }
  
    int ans = 0;
  
    for (auto i : s) {
        for (int j = 0; j < n; j++) {
  
            // store maximum value
            ans = max(ans, i ^ a[j]);
        }
    }
  
    cout << ans << "\n";
}
  
// Driver code
int main()
{
    int a[] = { 1, 3, 8, 15 };
    int n = sizeof(a) / sizeof(a[0]);
    Maximum_xor_Triplet(n, a);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
  
import java.util.HashSet;
  
class GFG 
{
  
    // function to count maximum
    // XOR value for a triplet
    static void Maximum_xor_Triplet(int n, int a[])
    {
        // set is used to avoid repetitions
        HashSet<Integer> s = new HashSet<Integer>();
  
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            {
  
                // store all possible unique
                // XOR value of pairs
                s.add(a[i] ^ a[j]);
            }
        }
  
        int ans = 0;
        for (Integer i : s) 
        {
            for (int j = 0; j < n; j++)
            {
  
                // store maximum value
                ans = Math.max(ans, i ^ a[j]);
            }
        }
        System.out.println(ans);
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int a[] = {1, 3, 8, 15};
        int n = a.length;
        Maximum_xor_Triplet(n, a);
    }
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# function to count maximum 
# XOR value for a triplet 
def Maximum_xor_Triplet(n, a): 
  
    # set is used to avoid repetitions 
    s = set() 
  
    for i in range(0, n): 
        for j in range(i, n): 
  
            # store all possible unique 
            # XOR value of pairs 
            s.add(a[i] ^ a[j]) 
  
    ans = 0
    for i in s: 
        for j in range(0, n): 
  
            # store maximum value 
            ans = max(ans, i ^ a[j]) 
  
    print(ans) 
  
# Driver code 
if __name__ == "__main__":
  
    a = [1, 3, 8, 15
    n = len(a) 
    Maximum_xor_Triplet(n, a) 
  
# This code is contributed 
# by Rituraj Jain

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // function to count maximum
    // XOR value for a triplet
    static void Maximum_xor_Triplet(int n, int []a)
    {
        // set is used to avoid repetitions
        HashSet<int> s = new HashSet<int>();
  
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            {
  
                // store all possible unique
                // XOR value of pairs
                s.Add(a[i] ^ a[j]);
            }
        }
  
        int ans = 0;
        foreach (int i in s) 
        {
            for (int j = 0; j < n; j++)
            {
  
                // store maximum value
                ans = Math.Max(ans, i ^ a[j]);
            }
        }
        Console.WriteLine(ans);
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int []a = {1, 3, 8, 15};
        int n = a.Length;
        Maximum_xor_Triplet(n, a);
    }
}
  
/* This code has been contributed 
by PrinciRaj1992*/

chevron_right


Output:

15


My Personal Notes arrow_drop_up