Count of Binary Digit numbers smaller than N

Given a limit N, we need to find out the count of binary digit numbers which are smaller than N. Binary digit numbers are those numbers which contains only 0 and 1 as their digits as 1, 10, 101 etc are binary digit numbers.

Examples:

Input : N = 200
Output : 7
Count of binary digit number smaller than N is 7, 
enumerated below,
1, 10, 11, 110, 101, 100, 111



One simple way to solve this problem is to loop from 1 till N and check each number whether it is a binary digit number or not. If it is a binary digit number, increase the count of such numbers but this procedure will take O(N) time. We can do better, as we know that count of such numbers will be much smaller than N, we can iterate over binary digit numbers only and check whether generated numbers are smaller than N or not.
In below code, BFS like approach is implemented to iterate over only binary digit numbers. We start with 1 and each time we will push (t*10) and (t*10 + 1) into the queue where t is the popped element, if t is a binary digit number then (t*10) and (t*10 + 1) will also binary digit number, so we will iterate over these numbers only using queue. We will stop pushing elements in the queue when popped number crosses the N.

C++

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// C++ program to count all binary digit
// numbers smaller than N
#include <bits/stdc++.h>
using namespace std;
  
//  method returns count of binary digit
//  numbers smaller than N
int countOfBinaryNumberLessThanN(int N)
{
    //  queue to store all intermediate binary
    // digit numbers
    queue<int> q;
  
    //  binary digits start with 1
    q.push(1);
    int cnt = 0;
    int t;
  
    //  loop untill we have element in queue
    while (!q.empty())
    {
        t = q.front();
        q.pop();
  
        //  push next binary digit numbers only if
        // current popped element is N
        if (t <= N)
        {
            cnt++;
  
            // uncomment below line to print actual
            // number in sorted order
            // cout << t << " ";
  
            q.push(t * 10);
            q.push(t * 10 + 1);
        }
    }
  
    return cnt;
}
  
//    Driver code to test above methods
int main()
{
    int N = 200;
    cout << countOfBinaryNumberLessThanN(N);
    return 0;
}

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Java

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import java.util.LinkedList;
import java.util.Queue;
  
// java program to count all binary digit
// numbers smaller than N
public class GFG {
  
//  method returns count of binary digit
//  numbers smaller than N
    static int countOfBinaryNumberLessThanN(int N) {
        //  queue to store all intermediate binary
        // digit numbers
        Queue<Integer> q = new LinkedList<>();
  
        //  binary digits start with 1
        q.add(1);
        int cnt = 0;
        int t;
  
        //  loop untill we have element in queue
        while (q.size() > 0) {
            t = q.peek();
            q.remove();
  
            //  push next binary digit numbers only if
            // current popped element is N
            if (t <= N) {
                cnt++;
  
                // uncomment below line to print actual
                // number in sorted order
                // cout << t << " ";
                q.add(t * 10);
                q.add(t * 10 + 1);
            }
        }
  
        return cnt;
    }
  
//    Driver code to test above methods
    static public void main(String[] args) {
        int N = 200;
        System.out.println(countOfBinaryNumberLessThanN(N));
    }
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to count all binary digit
# numbers smaller than N
from collections import deque
  
# method returns count of binary digit
# numbers smaller than N
def countOfBinaryNumberLessThanN(N):
    # queue to store all intermediate binary
    # digit numbers
    q = deque()
  
    # binary digits start with 1
    q.append(1)
    cnt = 0
  
    # loop untill we have element in queue
    while (q):
        t = q.popleft()
          
        # push next binary digit numbers only if
        # current popped element is N
        if (t <= N):
            cnt = cnt + 1
            # uncomment below line to print actual
            # number in sorted order
            q.append(t * 10)
            q.append(t * 10 + 1)
  
    return cnt
  
# Driver code to test above methods
if __name__=='__main__':
    N = 200
    print(countOfBinaryNumberLessThanN(N))
  
# This code is contributed by
# Sanjit_Prasad

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C#

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// C# program to count all binary digit 
// numbers smaller than N 
using System;
using System.Collections.Generic;
  
class GFG 
  
    // method returns count of binary digit 
    // numbers smaller than N 
    static int countOfBinaryNumberLessThanN(int N) 
    {
          
        // queue to store all intermediate binary 
        // digit numbers 
        Queue<int> q = new Queue<int>(); 
  
        // binary digits start with 1 
        q.Enqueue(1); 
        int cnt = 0; 
        int t; 
  
        // loop untill we have element in queue 
        while (q.Count > 0)
        
            t = q.Peek(); 
            q.Dequeue(); 
  
            // push next binary digit numbers only if 
            // current popped element is N 
            if (t <= N) 
            
                cnt++; 
  
                // uncomment below line to print actual 
                // number in sorted order 
                q.Enqueue(t * 10); 
                q.Enqueue(t * 10 + 1); 
            
        
  
        return cnt; 
    
  
    // Driver code  
    static void Main() 
    
        int N = 200; 
        Console.WriteLine(countOfBinaryNumberLessThanN(N)); 
    
  
// This code is contributed by mits

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PHP

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<?php
// PHP program to count all binary digit
// numbers smaller than N
  
// method returns count of binary digit
// numbers smaller than N
function countOfBinaryNumberLessThanN($N)
{
    // queue to store all intermediate 
    // binary digit numbers
    $q = array();
  
    // binary digits start with 1
    array_push($q, 1);
    $cnt = 0;
    $t = 0;
  
    // loop untill we have element in queue
    while (!empty($q))
    {
        $t = array_pop($q);
  
        // push next binary digit numbers only 
        // if current popped element is N
        if ($t <= $N)
        {
            $cnt++;
  
            // uncomment below line to print 
            // actual number in sorted order
            // cout << t << " ";
  
            array_push($q, $t * 10);
            array_push($q, ($t * 10 + 1));
        }
    }
  
    return $cnt;
}
  
// Driver Code
$N = 200;
echo countOfBinaryNumberLessThanN($N);
  
// This code is contributed by mits
?>

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Output:

7

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