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Count of integers up to N which represent a Binary number
  • Difficulty Level : Medium
  • Last Updated : 15 Apr, 2021
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Given an integer N, the task is to count every number i from 1 to N (both inclusive) such that i is a binary representation of some integer where N can be any value within the range[1, 109]

Examples: 

Input: N = 100 
Output:
Explanation: Valid integers are 1, 10, 11, 100

Input: N = 20 
Output:
Explanation: Valid integers are 1, 10, 11 

Naive approach: Since maximum number of digits in N can be 10 so store every binary combination of 10 digits and then use Binary search or upper_bound to check the largest integer in the given range of N.
Time Complexity: O(MAX + log(MAX)) where MAX = 1024 (210)

Efficient approach: We can observe that for any value of N, the maximum number of such possible representations is 2count of digits of N – 1. Hence, we need to follow the following steps: 



  • Extract digits of N from right to left and store the position of the current digit in a variable ctr.
  • If the current digit exceeds 1, it means that maximum possible representations using ctr digits can be obtained. Thus, set answer equal to 2ctr – 1.
  • Otherwise, if the current digit is 1, then add 2ctr – 1 to the answer obtained so far.
  • The final value obtained after traversing all the digits gives the answer.

Below is the implementation of the above approach:

C++




// C++ Program to count the
// number of integers upto N
// which are of the form of
// binary representations
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
int countBinaries(int N)
{
 
    int ctr = 1;
    int ans = 0;
    while (N > 0) {
 
        // If the current last
        // digit is 1
        if (N % 10 == 1) {
 
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += pow(2, ctr - 1);
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1) {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = pow(2, ctr) - 1;
        }
 
        ctr++;
        N /= 10;
    }
 
    return ans;
}
// Driver Code
int main()
{
 
    int N = 20;
    cout << countBinaries(N);
 
    return 0;
}

Java




// Java program to count the number
// of integers upto N which are of
// the form of binary representations
import java.util.*;
class GFG{
 
// Function to return the count
static int countBinaries(int N)
{
    int ctr = 1;
    int ans = 0;
    while (N > 0)
    {
         
        // If the current last
        // digit is 1
        if (N % 10 == 1)
        {
 
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += Math.pow(2, ctr - 1);
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1)
        {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = (int) (Math.pow(2, ctr) - 1);
        }
        ctr++;
        N /= 10;
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 20;
    System.out.print(countBinaries(N));
}
}
 
// This code is contributed by amal kumar choubey

Python3




# Python3 program to count the
# number of integers upto N
# which are of the form of
# binary representations
from math import *
 
# Function to return the count
def countBinaries(N):
     
    ctr = 1
    ans = 0
     
    while (N > 0):
         
        # If the current last
        # digit is 1
        if (N % 10 == 1):
             
            # Add 2^(ctr - 1) possible
            # integers to the answer
            ans += pow(2, ctr - 1)
 
        # If the current digit exceeds 1
        elif (N % 10 > 1):
             
            # Set answer as 2^ctr - 1
            # as all possible binary
            # integers with ctr number
            # of digits can be obtained
            ans = pow(2, ctr) - 1
 
        ctr += 1
        N //= 10
 
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    N = 20
     
    print(int(countBinaries(N)))
 
# This code is contributed by Bhupendra_Singh

C#




// C# program to count the number
// of integers upto N which are of
// the form of binary representations
using System;
 
class GFG{
 
// Function to return the count
static int countBinaries(int N)
{
    int ctr = 1;
    int ans = 0;
    while (N > 0)
    {
         
        // If the current last
        // digit is 1
        if (N % 10 == 1)
        {
             
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += (int)Math.Pow(2, ctr - 1);
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1)
        {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = (int)(Math.Pow(2, ctr) - 1);
        }
        ctr++;
        N /= 10;
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 20;
    Console.Write(countBinaries(N));
}
}
 
// This code is contributed by amal kumar choubey

Javascript




<script>
 
    // Javascript Program to count the
    // number of integers upto N
    // which are of the form of
    // binary representations
 
    // Function to return the count
    function countBinaries(N)
    {
 
        let ctr = 1;
        let ans = 0;
        while (N > 0) {
 
            // If the current last
            // digit is 1
            if (N % 10 == 1) {
 
                // Add 2^(ctr - 1) possible
                // integers to the answer
                ans += Math.pow(2, ctr - 1);
            }
 
            // If the current digit exceeds 1
            else if (N % 10 > 1) {
 
                // Set answer as 2^ctr - 1
                // as all possible binary
                // integers with ctr number
                // of digits can be obtained
                ans = Math.pow(2, ctr) - 1;
            }
 
            ctr++;
            N /= 10;
        }
 
        return ans;
    }
     
    let N = 20;
    document.write(countBinaries(N));
     
    // This code is contributed by divyesh072019.
     
</script>
Output: 
3

 

Time Complexity: O(M2) where M is the count of digits in N 
Auxiliary Space: O(1)

Optimization: The above approach can be optimized by pre-computing the powers of 2 up to M (count of digits up to M of N) by the help of a prefix product array.

Below is the implementation of the optimized solution:

C++




// C++ Program to count the
// number of integers upto N
// which are of the form of
// binary representations
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
int countBinaries(int N)
{
    // PreCompute and store
    // the powers of 2
    vector<int> powersOfTwo(11);
 
    powersOfTwo[0] = 1;
    for (int i = 1; i < 11; i++) {
        powersOfTwo[i]
= powersOfTwo[i - 1]
* 2;
    }
 
    int ctr = 1;
    int ans = 0;
    while (N > 0) {
 
        // If the current last
        // digit is 1
        if (N % 10 == 1) {
 
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += powersOfTwo[ctr - 1];
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1) {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = powersOfTwo[ctr] - 1;
        }
 
        ctr++;
        N /= 10;
    }
 
    return ans;
}
// Driver Code
int main()
{
 
    int N = 20;
    cout << countBinaries(N);
 
    return 0;
}

Java




// Java program to count the number of
// integers upto N which are of the
// form of binary representations
import java.util.*;
 
class GFG{
 
// Function to return the count
static int countBinaries(int N)
{
     
    // PreCompute and store
    // the powers of 2
    Vector<Integer> powersOfTwo = new Vector<Integer>(11);
    powersOfTwo.add(1);
     
    for(int i = 1; i < 11; i++)
    {
       powersOfTwo.add(powersOfTwo.get(i - 1) * 2);
    }
 
    int ctr = 1;
    int ans = 0;
    while (N > 0)
    {
 
        // If the current last
        // digit is 1
        if (N % 10 == 1)
        {
 
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += powersOfTwo.get(ctr - 1);
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1)
        {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = powersOfTwo.get(ctr) - 1;
        }
        ctr++;
        N /= 10;
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 20;
    System.out.print(countBinaries(N));
}
}
 
// This code is contributed by amal kumar choubey

Python3




# Python3 program to count the
# number of integers upto N
# which are of the form of
# binary representations
 
# Function to return the count
def countBinaries(N):
 
    # PreCompute and store
    # the powers of 2
    powersOfTwo = [0] * 11
 
    powersOfTwo[0] = 1
     
    for i in range(1, 11):
        powersOfTwo[i] = powersOfTwo[i - 1] * 2
 
    ctr = 1
    ans = 0
     
    while (N > 0):
 
        # If the current last
        # digit is 1
        if (N % 10 == 1):
 
            # Add 2^(ctr - 1) possible
            # integers to the answer
            ans += powersOfTwo[ctr - 1]
 
        # If the current digit exceeds 1
        elif (N % 10 > 1):
 
            # Set answer as 2^ctr - 1
            # as all possible binary
            # integers with ctr number
            # of digits can be obtained
            ans = powersOfTwo[ctr] - 1
 
        ctr += 1
        N = N // 10
 
    return ans
 
# Driver code
N = 20
 
print(countBinaries(N))
 
# This code is contributed by divyeshrabadiya07

C#




// C# program to count the number of
// integers upto N which are of the
// form of binary representations
using System;
using System.Collections.Generic;
class GFG{
 
// Function to return the count
static int countBinaries(int N)
{
     
    // PreCompute and store
    // the powers of 2
    List<int> powersOfTwo = new List<int>();
    powersOfTwo.Add(1);
     
    for(int i = 1; i < 11; i++)
    {
        powersOfTwo.Add(powersOfTwo[i - 1] * 2);
    }
 
    int ctr = 1;
    int ans = 0;
    while (N > 0)
    {
 
        // If the current last
        // digit is 1
        if (N % 10 == 1)
        {
 
            // Add 2^(ctr - 1) possible
            // integers to the answer
            ans += powersOfTwo[ctr - 1];
        }
 
        // If the current digit exceeds 1
        else if (N % 10 > 1)
        {
 
            // Set answer as 2^ctr - 1
            // as all possible binary
            // integers with ctr number
            // of digits can be obtained
            ans = powersOfTwo[ctr] - 1;
        }
        ctr++;
        N /= 10;
    }
    return ans;
}
 
// Driver Code
static public void Main ()
{
    int N = 20;
    Console.Write(countBinaries(N));
}
}
 
// This code is contributed by ShubhamCoder

Javascript




<script>
    // Javascript program to count the number of
    // integers upto N which are of the
    // form of binary representations
     
    // Function to return the count
    function countBinaries(N)
    {
 
        // PreCompute and store
        // the powers of 2
        let powersOfTwo = [];
        powersOfTwo.push(1);
 
        for(let i = 1; i < 11; i++)
        {
            powersOfTwo.push(powersOfTwo[i - 1] * 2);
        }
 
        let ctr = 1;
        let ans = 0;
        while (N > 0)
        {
 
            // If the current last
            // digit is 1
            if (N % 10 == 1)
            {
 
                // Add 2^(ctr - 1) possible
                // integers to the answer
                ans += powersOfTwo[ctr - 1];
            }
 
            // If the current digit exceeds 1
            else if (N % 10 > 1)
            {
 
                // Set answer as 2^ctr - 1
                // as all possible binary
                // integers with ctr number
                // of digits can be obtained
                ans = powersOfTwo[ctr] - 1;
            }
            ctr++;
            N /= 10;
        }
        return ans;
    }
     
    let N = 20;
    document.write(countBinaries(N));
</script>
Output: 
3

 

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

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