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 tripletvoid 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 codeint 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 approachimport 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 approachusing 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 tripletfunction 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 codelet 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
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