 Open in App
Not now

# Find the largest number with n set and m unset bits

• Difficulty Level : Easy
• Last Updated : 31 May, 2022

Given two non-negative numbers n and m. The problem is to find the largest number having n number of set bits and m number of unset bits in its binary representation.
Note : 0 bits before leading 1 (or leftmost 1) in binary representation are counted
Constraints: 1 <= n, 0 <= m, (m+n) <= 31
Examples :

```Input : n = 2, m = 2
Output : 12
(12)10 = (1100)2
We can see that in the binary representation of 12
there are 2 set and 2 unsets bits and it is the largest number.

Input : n = 4, m = 1
Output : 30```

Following are the steps:

1. Calculate num = (1 << (n + m)) – 1. This will produce a number num having (n + m) number of bits and all are set.
2. Now, toggle the last m bits of num and then return the toggled number. Refer this post.

## C++

 `// C++ implementation to find the largest number` `// with n set and m unset bits` `#include `   `using` `namespace` `std;`   `// function to toggle the last m bits` `unsigned ``int` `toggleLastMBits(unsigned ``int` `n,` `                            ``unsigned ``int` `m)` `{` `    ``// if no bits are required to be toggled` `    ``if` `(m == 0)` `        ``return` `n;`   `    ``// calculating a number 'num' having 'm' bits` `    ``// and all are set` `    ``unsigned ``int` `num = (1 << m) - 1;`   `    ``// toggle the last m bits and return the number` `    ``return` `(n ^ num);` `}`   `// function to find the largest number` `// with n set and m unset bits` `unsigned ``int` `largeNumWithNSetAndMUnsetBits(unsigned ``int` `n,` `                                        ``unsigned ``int` `m)` `{` `    ``// calculating a number 'num' having '(n+m)' bits` `    ``// and all are set` `    ``unsigned ``int` `num = (1 << (n + m)) - 1;`   `    ``// required largest number` `    ``return` `toggleLastMBits(num, m);` `}`   `// Driver program to test above` `int` `main()` `{` `    ``unsigned ``int` `n = 2, m = 2;` `    ``cout << largeNumWithNSetAndMUnsetBits(n, m);` `    ``return` `0;` `}`

## Java

 `// Java implementation to find the largest number` `// with n set and m unset bits` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// Function to toggle the last m bits` `    ``static` `int` `toggleLastMBits(``int` `n, ``int` `m)` `    ``{` `        ``// if no bits are required to be toggled` `        ``if` `(m == ``0``)` `            ``return` `n;` ` `  `        ``// calculating a number 'num' having 'm' bits` `        ``// and all are set` `        ``int` `num = (``1` `<< m) - ``1``;` ` `  `        ``// toggle the last m bits and return the number` `        ``return` `(n ^ num);` `    ``}` ` `  `    ``// Function to find the largest number` `    ``// with n set and m unset bits` `    ``static` `int` `largeNumWithNSetAndMUnsetBits(``int` `n, ``int` `m)` `    ``{` `        ``// calculating a number 'num' having '(n+m)' bits` `        ``// and all are set` `        ``int` `num = (``1` `<< (n + m)) - ``1``;` ` `  `        ``// required largest number` `        ``return` `toggleLastMBits(num, m);` `    ``}` `    `  `    ``// driver program` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `n = ``2``, m = ``2``;` `        ``System.out.println(largeNumWithNSetAndMUnsetBits(n, m));` `    ``}` `}`   `// Contributed by Pramod Kumar`

## Python3

 `# Python implementation to` `# find the largest number` `# with n set and m unset bits`   `# function to toggle` `# the last m bits` `def` `toggleLastMBits(n,m):`   `    ``# if no bits are required` `    ``# to be toggled` `    ``if` `(m ``=``=` `0``):` `        ``return` `n`   `    ``# calculating a number` `    ``# 'num' having 'm' bits` `    ``# and all are set` `    ``num ``=` `(``1` `<< m) ``-` `1`   `    ``# toggle the last m bits` `    ``# and return the number` `    ``return` `(n ^ num)`     `# function to find` `# the largest number` `# with n set and m unset bits` `def` `largeNumWithNSetAndMUnsetBits(n,m):`   `    ``# calculating a number` `    ``# 'num' having '(n+m)' bits` `    ``# and all are set` `    ``num ``=` `(``1` `<< (n ``+` `m)) ``-` `1`   `    ``# required largest number` `    ``return` `toggleLastMBits(num, m)`   `# Driver code`   `n ``=` `2` `m ``=` `2`   `print``(largeNumWithNSetAndMUnsetBits(n, m))`   `# This code is contributed` `# by Anant Agarwal.`

## C#

 `// C# implementation to find the largest number` `// with n set and m unset bits` `using` `System;`   `class` `GFG` `{ ` `    ``// Function to toggle the last m bits` `    ``static` `int` `toggleLastMBits(``int` `n, ``int` `m)` `    ``{` `        ``// if no bits are required to be toggled` `        ``if` `(m == 0)` `            ``return` `n;`   `        ``// calculating a number 'num' having 'm' bits` `        ``// and all are set` `        ``int` `num = (1 << m) - 1;`   `        ``// toggle the last m bits and return the number` `        ``return` `(n ^ num);` `    ``}`   `    ``// Function to find the largest number` `    ``// with n set and m unset bits` `    ``static` `int` `largeNumWithNSetAndMUnsetBits(``int` `n, ``int` `m)` `    ``{` `        ``// calculating a number 'num' having '(n+m)' bits` `        ``// and all are set` `        ``int` `num = (1 << (n + m)) - 1;`   `        ``// required largest number` `        ``return` `toggleLastMBits(num, m);` `    ``}` `    `  `    ``// Driver program` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `n = 2, m = 2;` `        ``Console.Write(largeNumWithNSetAndMUnsetBits(n, m));` `    ``}`   `}`   `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output :

`12`

Time Complexity : O(1)

Auxiliary Space: O(1)

For greater values of n and m, you can use long int and long long int datatypes to generate the required number.
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.