Find the smallest number with n set and m unset bits

• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

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

Examples:

Input : n = 2, m = 2
Output : 9
(9)10 = (1001)2
We can see that in the binary representation of 9
there are 2 set and 2 unsets bits and it is the
smallest number.

Input : n = 4, m = 1
Output : 23

Approach: 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 bits in the range from n to (n+m-1) in num, i.e, to toggle bits from the rightmost nth bit to the rightmost (n+m-1)th bit and then return the toggled number. Refer this post.

C++

 // C++ implementation to find the smallest number // with n set and m unset bits #include   using namespace std;   // function to toggle bits in the given range unsigned int toggleBitsFromLToR(unsigned int n,                                 unsigned int l,                                 unsigned int r) {     // for invalid range     if (r < l)         return n;       // calculating a number 'num' having 'r'     // number of bits and bits in the range l     // to r are the only set bits     int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);       // toggle bits in the range l to r in 'n'     // and return the number     return (n ^ num); }   // function to find the smallest number // with n set and m unset bits unsigned int smallNumWithNSetAndMUnsetBits(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 smallest number     return toggleBitsFromLToR(num, n, n + m - 1); }   // Driver program to test above int main() {     unsigned int n = 2, m = 2;     cout << smallNumWithNSetAndMUnsetBits(n, m);     return 0; }

Java

 // Java implementation to find the smallest number // with n set and m unset bits   class GFG {     // Function to toggle bits in the given range     static int toggleBitsFromLToR(int n, int l, int r)     {         // for invalid range         if (r < l)             return n;            // calculating a number 'num' having 'r'         // number of bits and bits in the range l         // to r are the only set bits         int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);            // toggle bits in the range l to r in 'n'         // and return the number         return (n ^ num);     }           // Function to find the smallest number     // with n set and m unset bits     static int smallNumWithNSetAndMUnsetBits(int n, int m)     {         // calculating a number 'num' having '(n+m)' bits         // and all are set         int num = (1 << (n + m)) - 1;            // required smallest number         return toggleBitsFromLToR(num, n, n + m - 1);     }           // driver program     public static void main (String[] args)     {         int n = 2, m = 2;         System.out.println(smallNumWithNSetAndMUnsetBits(n, m));     } }   // Contributed by Pramod Kumar

Python3

 # Python3 implementation to find # the smallest number with n set # and m unset bits   # function to toggle bits in the # given range def toggleBitsFromLToR(n, l, r):       # for invalid range     if (r < l):         return n        # calculating a number 'num'     # having 'r' number of bits     # and bits in the range l     # to r are the only set bits     num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)        # toggle bits in the range     # l to r in 'n' and return the number     return (n ^ num)   # function to find the smallest number # with n set and m unset bits def smallNumWithNSetAndMUnsetBits(n, m):       # calculating a number 'num' having     # '(n+m)' bits and all are set     num = (1 << (n + m)) - 1        # required smallest number     return toggleBitsFromLToR(num, n, n + m - 1);      # Driver program to test above n = 2 m = 2   ans = smallNumWithNSetAndMUnsetBits(n, m) print (ans)   # This code is contributed by Saloni Gupta

C#

 // C# implementation to find the smallest number // with n set and m unset bits using System;   class GFG {     // Function to toggle bits in the given range     static int toggleBitsFromLToR(int n, int l, int r)     {         // for invalid range         if (r < l)             return n;           // calculating a number 'num' having 'r'         // number of bits and bits in the range l         // to r are the only set bits         int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);           // toggle bits in the range l to r in 'n'         // and return the number         return (n ^ num);     }           // Function to find the smallest number     // with n set and m unset bits     static int smallNumWithNSetAndMUnsetBits(int n, int m)     {         // calculating a number 'num' having '(n+m)' bits         // and all are set         int num = (1 << (n + m)) - 1;           // required smallest number         return toggleBitsFromLToR(num, n, n + m - 1);     }           // Driver program     public static void Main ()     {         int n = 2, m = 2;         Console.Write(smallNumWithNSetAndMUnsetBits(n, m));     } }   // This code is contributed by Sam007



Javascript



Output:

9

For greater values of n and m, you can use long int and long long int datatypes to generate the required number.
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