Count of integers up to N which represent a Binary number
Last Updated :
12 Jul, 2025
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: 4
Explanation: Valid integers are 1, 10, 11, 100
Input: N = 20
Output: 3
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 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 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# 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
<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>
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 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 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# 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
<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>
Time Complexity: O(M)
Auxiliary Space: O(M)
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