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Maximum XOR-value of at-most k-elements from 1 to n
  • Difficulty Level : Basic
  • Last Updated : 07 Apr, 2021

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++




// Program to obtain maximum XOR value sub-array
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate maximum XOR value
int maxXOR(int n, int k) {
  int c = log2(n) + 1;
 
  // Return (2^c - 1)
  return ((1 << c) - 1);
}
 
// driver program
int main() {
  int n = 12;
  int k = 3;
  cout << maxXOR(n, k);
  return 0;
}

Java




// Program to obtain maximum
// XOR value sub-array
import java.lang.*;
 
class GFG
{
// function to calculate
// maximum XOR value
static int maxXOR(int n, int k)
{
int c = (int) (Math.log(n) /
               Math.log(2)) + 1;
 
// Return (2^c - 1)
return ((1 << c) - 1);
}
 
// Driver Code
public static void main(String[] args)
{
int n = 12;
int k = 3;
System.out.println(maxXOR(n, k));
}
}
 
// This code is contributed by Smitha

Python3




# Python3 program to obtain maximum
# XOR value sub-array
import math
 
# Function to calculate maximum XOR value
def maxXOR(n, k):
    c = int(math.log(n, 2)) + 1
 
    # Return (2^c - 1)
    return ((1 << c) - 1)
 
# Driver Code
n = 12; k = 3
print (maxXOR(n, k))
 
# This code is contributed by shreyanshi_arun.

C#




// Program to obtain maximum
// XOR value sub-array
using System;
 
class GFG
{
// function to calculate
// maximum XOR value
static int maxXOR(int n, int k)
{
int c = (int) (Math.Log(n) /
               Math.Log(2)) + 1;
 
// Return (2^c - 1)
return ((1 << c) - 1);
}
 
// Driver Code
public static void Main(String[] args)
{
int n = 12;
int k = 3;
Console.Write(maxXOR(n, k)) ;
}
}
 
// This code is contributed by Smitha

PHP




<?php
// Program to obtain maximum
// XOR value sub-array
 
// function to calculate
// maximum XOR value
function maxXOR($n, $k)
{
    $c = log($n, 2) + 1;
     
    // Return (2^c - 1)
    return ((1 << $c) - 1);
}
 
// Driver Code
$n = 12;
$k = 3;
echo maxXOR($n, $k);
 
// This code is contributed by aj_36
?>

Javascript




<script>
 
// JavaScript program to obtain maximum
// XOR value sub-array
 
// function to calculate
// maximum XOR value
function maxXOR(n, k)
{
let c = (Math.log(n) /
               Math.log(2)) + 1;
   
// Return (2^c - 1)
return ((1 << c) - 1);
}   
  
// Driver code
 
        let n = 12;
        let k = 3;
        document.write(maxXOR(n, k));
 
</script>
Output: 
15

 




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