Skip to content
Related Articles

Related Articles

Improve Article

Maximum XOR using K numbers from 1 to n

  • Difficulty Level : Medium
  • Last Updated : 08 Apr, 2021

Given an positive integer n and k. Find maximum xor of 1 to n using at most k numbers. Xor sum of 1 to n is defined as 1 ^ 2 ^ 3 ^ … ^ n.
Examples : 
 

Input :  n = 4, k = 3
Output : 7
Explanation
Maximum possible xor sum is 1 ^ 2 ^ 4 = 7.

Input : n = 11, k = 1
Output : 11
Explanation
Maximum Possible xor sum is 11.

 

If we have k = 1 then the maximum possible xor sum is ‘n’ itself. Now for k > 1 we can always have an number with its all bits set till the most significant set bit in ‘n’.
 

C++




// CPP program to find max xor sum
// of 1 to n using atmost k numbers
#include <bits/stdc++.h>
using namespace std;
 
// To return max xor sum of 1 to n
// using at most k numbers
int maxXorSum(int n, int k)
{
    // If k is 1 then maximum
    // possible sum is n
    if (k == 1)
        return n;
 
    // Finding number greater than
    // or equal to n with most significant
    // bit same as n. For example, if n is
    // 4, result is 7. If n is 5 or 6, result
    // is 7
    int res = 1;
    while (res <= n)
        res <<= 1;
 
    // Return res - 1 which denotes
    // a number with all bits set to 1
    return res - 1;
}
 
// Driver program
int main()
{
    int n = 4, k = 3;
    cout << maxXorSum(n, k);
    return 0;
}

Java




// Java program to find max xor sum
// of 1 to n using atmost k numbers
public class Main {
 
    // To return max xor sum of 1 to n
    // using at most k numbers
    static int maxXorSum(int n, int k)
    {
        // If k is 1 then maximum
        // possible sum is n
        if (k == 1)
            return n;
 
        // Finding number greater than
        // or equal to n with most significant
        // bit same as n. For example, if n is
        // 4, result is 7. If n is 5 or 6, result
        // is 7
        int res = 1;
        while (res <= n)
            res <<= 1;
 
        // Return res - 1 which denotes
        // a number with all bits set to 1
        return res - 1;
    }
 
    // Driver program to test maxXorSum()
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        System.out.print(maxXorSum(n, k));
    }
}

Python




# Python3 code to find max xor sum
# of 1 to n using atmost k numbers
 
# To return max xor sum of 1 to n
# using at most k numbers
def maxXorSum( n , k ):
    # If k is 1 then maximum
    # possible sum is n
    if k == 1:
        return n
     
    # Finding number greater than
    # or equal to n with most significant
    # bit same as n. For example, if n is
    # 4, result is 7. If n is 5 or 6, result
    # is 7
    res = 1
    while res <= n:
        res <<= 1
     
    # Return res - 1 which denotes
    # a number with all bits set to 1
    return res - 1
 
# Driver code
n = 4
k = 3
print( maxXorSum(n, k) )
 
# This code is contributed by Abhishek Sharma44.

C#




// C# program to find max xor sum
// of 1 to n using atmost k numbers
using System;
 
public class main {
 
    // To return max xor sum of 1 to n
    // using at most k numbers
    static int maxXorSum(int n, int k)
    {
        // If k is 1 then maximum
        // possible sum is n
        if (k == 1)
            return n;
 
        // Finding number greater than
        // or equal to n with most significant
        // bit same as n. For example, if n is
        // 4, result is 7. If n is 5 or 6, result
        // is 7
        int res = 1;
        while (res <= n)
            res <<= 1;
 
        // Return res - 1 which denotes
        // a number with all bits set to 1
        return res - 1;
    }
 
    // Driver program
    public static void Main()
    {
        int n = 4, k = 3;
        Console.WriteLine(maxXorSum(n, k));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find max xor sum
// of 1 to n using atmost k numbers
 
// To return max xor sum of 1 to n
// using at most k numbers
function maxXorSum($n, $k)
{
    // If k is 1 then maximum
    // possible sum is n
    if ($k == 1)
        return $n;
 
    // Finding number greater than
    // or equal to n with most
    // significant bit same as n.
    // For example, if n is 4, result
    // is 7. If n is 5 or 6, result is 7
    $res = 1;
    while ($res <= $n)
        $res <<= 1;
 
    // Return res - 1 which denotes
    // a number with all bits set to 1
    return $res - 1;
}
 
// Driver code
$n = 4;
$k = 3;
echo maxXorSum($n, $k);
 
// This code is contributed by Mithun Kumar
?>

Javascript




<script>
 
// JavaScript program to find max xor sum
// of 1 to n using atmost k numbers
 
    // To return max xor sum of 1 to n
    // using at most k numbers
    function maxXorSum(n, k)
    {
        // If k is 1 then maximum
        // possible sum is n
        if (k == 1)
            return n;
   
        // Finding number greater than
        // or equal to n with most significant
        // bit same as n. For example, if n is
        // 4, result is 7. If n is 5 or 6, result
        // is 7
        let res = 1;
        while (res <= n)
            res <<= 1;
   
        // Return res - 1 which denotes
        // a number with all bits set to 1
        return res - 1;
    }
       
 
// Driver code
         
        let n = 4, k = 3;
        document.write(maxXorSum(n, k));
 
</script>

Output : 

7

 



 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :