Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Count numbers up to N having Kth bit set

  • Last Updated : 23 Apr, 2021

Given two integers N and K, the task is to find the count of numbers up to N having K-th bit set.

Examples:

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.

Input: N = 14, K = 2
Output: 7
Explanation: 
The numbers less than equal to 14, having 2nd bit set are 4, 5, 6, 7, 12, 13, and 14.

Input: N = 6, K = 1
Output: 3
Explanation
The numbers less than equal to 6 having 1st bit set are 1, 3, 5.

 

Naive Approach: The simplest approach is to traverse from 1 to N, and check for each number whether its K-th bit is set or not.



Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by dividing the task into two parts:

  1. First, right shift N, K+1 times followed by left shifting the result K times, which gives the count of numbers satisfying the given condition till the nearest power of 2 less than N.
  2. Now, check if the Kth bit is set in N or not.
  3. If the Kth bit is set in N, then add the count of numbers from the nearest power of 2 less than N to the answer.

Below is the implementation of the above approach:

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of number of 1's at ith bit
// in a range [1, n - 1]
long long getcount(long long n, int k)
{
    // Store count till nearest
    // power of 2 less than N
    long long res = (n >> (k + 1)) << k;
 
    // If K-th bit is set in N
    if ((n >> k) & 1)
 
        // Add to result the nearest
        // power of 2 less than N
        res += n & ((1ll << k) - 1);
 
    // Return result
    return res;
}
 
// Driver Code
int main()
{
 
    long long int N = 14;
    int K = 2;
 
    // Function Call
    cout << getcount(N + 1, K) << endl;
 
    return 0;
}

Java




// Java program for above approach
class GFG
{
 
  // Function to return the count
  // of number of 1's at ith bit
  // in a range [1, n - 1]
  static long getcount(long n, int k)
  {
 
    // Store count till nearest
    // power of 2 less than N
    long res = (n >> (k + 1)) << k;
 
    // If K-th bit is set in N
    if (((n >> k) & 1) != 0)
 
      // Add to result the nearest
      // power of 2 less than N
      res += n & ((1 << k) - 1);
 
    // Return result
    return res;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    long N = 14;
    int K = 2;
 
    // Function Call
    System.out.println(getcount(N + 1, K));
  }
}
 
// This code is contributed by divyesh072019

Python3




# Python3 program for above approach
 
# Function to return the count
# of number of 1's at ith bit
# in a range [1, n - 1]
def getcount(n, k):
     
    # Store count till nearest
    # power of 2 less than N
    res = (n >> (k + 1)) << k
 
    # If K-th bit is set in N
    if ((n >> k) & 1):
         
        # Add to result the nearest
        # power of 2 less than N
        res += n & ((1 << k) - 1)
 
    # Return result
    return res
 
# Driver Code
if __name__ == '__main__':
 
    N = 14
    K = 2
 
    # Function Call
    print (getcount(N + 1, K))
 
# This code is contributed by mohit kumar 29

C#




// C# program for above approach
using System;
 
class GFG{
 
// Function to return the count
// of number of 1's at ith bit
// in a range [1, n - 1]
static long getcount(long n, int k)
{
     
    // Store count till nearest
    // power of 2 less than N
    long res = (n >> (k + 1)) << k;
     
    // If K-th bit is set in N
    if (((n >> k) & 1) != 0)
     
        // Add to result the nearest
        // power of 2 less than N
        res += n & ((1 << k) - 1);
     
    // Return result
    return res;
}
 
// Driver Code 
static void Main()
{
    long N = 14;
    int K = 2;
     
    // Function Call
    Console.WriteLine(getcount(N + 1, K));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
 
    // Javascript program for above approach
     
    // Function to return the count
    // of number of 1's at ith bit
    // in a range [1, n - 1]
    function getcount(n, k)
    {
 
        // Store count till nearest
        // power of 2 less than N
        let res = (n >> (k + 1)) << k;
 
        // If K-th bit is set in N
        if (((n >> k) & 1) != 0)
 
            // Add to result the nearest
            // power of 2 less than N
            res += n & ((1 << k) - 1);
 
        // Return result
        return res;
    }
     
    let N = 14;
    let K = 2;
      
    // Function Call
    document.write(getcount(N + 1, K));
   
</script>
Output: 
7

 

Time Complexity: O(1)
Auxiliary Space: O(1)




My Personal Notes arrow_drop_up
Recommended Articles
Page :