Maximum value of XOR among all triplets of an array
Last Updated :
07 Sep, 2022
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++
#include <bits/stdc++.h>
using namespace std;
void Maximum_xor_Triplet( int n, int a[])
{
set< int > s;
for ( int i = 0; i < n; i++) {
for ( int j = i; j < n; j++) {
s.insert(a[i] ^ a[j]);
}
}
int ans = 0;
for ( auto i : s) {
for ( int j = 0; j < n; j++) {
ans = max(ans, i ^ a[j]);
}
}
cout << ans << "\n" ;
}
int main()
{
int a[] = { 1, 3, 8, 15 };
int n = sizeof (a) / sizeof (a[0]);
Maximum_xor_Triplet(n, a);
return 0;
}
|
Java
import java.util.HashSet;
class GFG
{
static void Maximum_xor_Triplet( int n, int a[])
{
HashSet<Integer> s = new HashSet<Integer>();
for ( int i = 0 ; i < n; i++)
{
for ( int j = i; j < n; j++)
{
s.add(a[i] ^ a[j]);
}
}
int ans = 0 ;
for (Integer i : s)
{
for ( int j = 0 ; j < n; j++)
{
ans = Math.max(ans, i ^ a[j]);
}
}
System.out.println(ans);
}
public static void main(String[] args)
{
int a[] = { 1 , 3 , 8 , 15 };
int n = a.length;
Maximum_xor_Triplet(n, a);
}
}
|
Python3
def Maximum_xor_Triplet(n, a):
s = set ()
for i in range ( 0 , n):
for j in range (i, n):
s.add(a[i] ^ a[j])
ans = 0
for i in s:
for j in range ( 0 , n):
ans = max (ans, i ^ a[j])
print (ans)
if __name__ = = "__main__" :
a = [ 1 , 3 , 8 , 15 ]
n = len (a)
Maximum_xor_Triplet(n, a)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void Maximum_xor_Triplet( int n, int []a)
{
HashSet< int > s = new HashSet< int >();
for ( int i = 0; i < n; i++)
{
for ( int j = i; j < n; j++)
{
s.Add(a[i] ^ a[j]);
}
}
int ans = 0;
foreach ( int i in s)
{
for ( int j = 0; j < n; j++)
{
ans = Math.Max(ans, i ^ a[j]);
}
}
Console.WriteLine(ans);
}
public static void Main(String[] args)
{
int []a = {1, 3, 8, 15};
int n = a.Length;
Maximum_xor_Triplet(n, a);
}
}
|
Javascript
<script>
function Maximum_xor_Triplet(n, a)
{
let s = new Set();
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
s.add(a[i] ^ a[j]);
}
}
let ans = 0;
for (let i of s.values()) {
for (let j = 0; j < n; j++) {
ans = Math.max(ans, i ^ a[j]);
}
}
document.write( ans, "<br>" );
}
let a = [ 1, 3, 8, 15 ];
let n = a.length;
Maximum_xor_Triplet(n, a);
</script>
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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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