Maximum XOR-value of at-most k-elements from 1 to n
You are given two positive integer n and k. You have to calculate the maximum possible XOR value of at most k-elements from 1 to n.
Note:k > 1
Examples :
Input : n = 7, k = 3
Output : 7
Explanation : You can select 1, 2, 4 for maximum XOR-value
Input : n = 7, k = 2
Output : 7
Explanation : You can select 3 and 4 for maximum value.
For any value of k we can select atleast two numbers from 1 to n and for the required result we have to take a closer look on the bit-representation of n. So lets understand it through an example. Suppose n = 6 and k = 2:
Bit representation of 6 = 110
Bit representation of 5 = 101
Bit representation of 4 = 100
Bit representation of 3 = 011
Bit representation of 2 = 010
Bit representation of 1 = 001
Now, you can see that after selecting as much numbers you want and selecting any of them you can not obtain XOR value greater than 111 i.e 7. So, for a given n and k >1 the maximum possible XOR value is 2log2(n)+1-1 (that is the value when all bits of n are turned to 1).
C++
#include <bits/stdc++.h>
using namespace std;
int maxXOR( int n, int k) {
int c = log2(n) + 1;
return ((1 << c) - 1);
}
int main() {
int n = 12;
int k = 3;
cout << maxXOR(n, k);
return 0;
}
|
Java
import java.lang.*;
class GFG
{
static int maxXOR( int n, int k)
{
int c = ( int ) (Math.log(n) /
Math.log( 2 )) + 1 ;
return (( 1 << c) - 1 );
}
public static void main(String[] args)
{
int n = 12 ;
int k = 3 ;
System.out.println(maxXOR(n, k));
}
}
|
Python3
import math
def maxXOR(n, k):
c = int (math.log(n, 2 )) + 1
return (( 1 << c) - 1 )
n = 12 ; k = 3
print (maxXOR(n, k))
|
C#
using System;
class GFG
{
static int maxXOR( int n, int k)
{
int c = ( int ) (Math.Log(n) /
Math.Log(2)) + 1;
return ((1 << c) - 1);
}
public static void Main(String[] args)
{
int n = 12;
int k = 3;
Console.Write(maxXOR(n, k)) ;
}
}
|
PHP
<?php
function maxXOR( $n , $k )
{
$c = log( $n , 2) + 1;
return ((1 << $c ) - 1);
}
$n = 12;
$k = 3;
echo maxXOR( $n , $k );
?>
|
Javascript
<script>
function maxXOR(n, k)
{
let c = (Math.log(n) /
Math.log(2)) + 1;
return ((1 << c) - 1);
}
let n = 12;
let k = 3;
document.write(maxXOR(n, k));
</script>
|
Time Complexity : O(log(n))
Space Complexity : O(1)
Last Updated :
18 Apr, 2023
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