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Maximum value of XOR among all triplets of an array
  • Difficulty Level : Easy
  • Last Updated : 28 Jan, 2019

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++




// 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*/
Output:
15



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