Count of numbers having only 1 set bit in the range [0, n]

Given an integer n, the task is to count the numbers having only 1 set bit in the range [0, n].

Examples:

Input: n = 7
Output: 3
000, 001, 010, 011, 100, 101, 110 and 111 are the binary representation of all the numbers upto 7.
And there are only 3 numbers ( 001, 010 and 100 ) having only 1 set bit.

Input: n = 3
Output: 2

Approach: If k bits are required to represent n then there are k numbers possible as 1 can be positioned at k different positions each time.

Below is the implementation of the above approach

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the required count
int count(int n)
{
  
    // To store the count of numbers
    int cnt = 0;
    int p = 1;
    while (p <= n) {
        cnt++;
  
        // Every power of 2 contains
        // only 1 set bit
        p *= 2;
    }
    return cnt;
}
  
// Driver code
int main()
{
    int n = 7;
    cout << count(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
  
    // Function to return the required count
    static int count(int n)
    {
  
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
  
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 7;
        System.out.print(count(n));
    }
}

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C#

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// C# implementation of the approach
using System;
class GFG {
  
    // Function to return the required count
    static int count(int n)
    {
  
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
  
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
  
    // Driver code
    public static void Main()
    {
        int n = 7;
        Console.Write(count(n));
    }
}

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Python3

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# Python3 implementation of the approach
  
# Function to return the required count
def count(n):
      
    # To store the count of numbers
    cnt = 0
    p = 1
    while (p <= n):
        cnt = cnt + 1
          
        # Every power of 2 contains 
        # only 1 set bit
        p *= 2
    return cnt
  
# Driver code
n = 7
print(count(n));

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PHP

Output:

3

Time Complexity: O(log n)



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