# Maximum XOR using K numbers from 1 to n

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.

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

If we have k = 1 then the maximum possible xor sum is 1. Now for k > 1 we can always have an number with its all bits set to 1. So result will be maximum number greater than n with its all bits set to 1.

## C++

 `// CPP program to find max xor sum ` `// of 1 to n using atmost k numbers ` `#include ` `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

 ` `

Output :

```7
```

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