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Maximum value of XOR among all triplets of an array

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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:
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++

// 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;
}

                    

Java

// 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

                    

Python3

# 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

                    

C#

// 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*/

                    

Javascript

<script>
 
// JavaScript implementation of the approach
// function to count maximum
// XOR value for a triplet
function Maximum_xor_Triplet(n, a)
{
    // set is used to avoid repetitions
    let s = new Set();
 
    for (let i = 0; i < n; i++) {
        for (let j = i; j < n; j++) {
 
            // store all possible unique
            // XOR value of pairs
            s.add(a[i] ^ a[j]);
        }
    }
 
    let ans = 0;
 
    for (let i of s.values()) {
        for (let j = 0; j < n; j++) {
 
            // store maximum value
            ans = Math.max(ans, i ^ a[j]);
        }
    }
 
    document.write( ans, "<br>");
}
 
// Driver code
let a = [ 1, 3, 8, 15 ];
let n = a.length;
    Maximum_xor_Triplet(n, a);
 
</script>

                    

Output
15

Complexity Analysis:

  • Time Complexity: O(n*n*logn), as nested loops are used
  • Auxiliary Space: O(n), as extra space of size n is used to create a set


Last Updated : 07 Sep, 2022
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