Given a number N. The task is to find the number of integers from 1 to n which contains digits 0’s and 1’s only.
Examples:
Input : N = 15 Output : 3 Explanation : 1, 10, 11 are such integers. Input : N = 120 Output : 7 Explanation : 1, 10, 11, 100, 101, 110, 111 are such integers.
Approach: An efficient approach is to build integers which contain 1’s and 0’s only using a recursive function starting from the number 1. For each number check whether it is less than n or not.
Below is the implementation of the above approach:
C++
// C++ program to find the number of integers// from 1 to n which contains digits 0's and 1's only#include <bits/stdc++.h>using namespace std;// Function to find the number of integers// from 1 to n which contains 0's and 1's onlyint countNumbers(int x, int n){ // If number is greater than n if (x > n) return 0; // otherwise add count this number and // call two functions return 1 + countNumbers(x * 10, n) + countNumbers(x * 10 + 1, n);}// Driver codeint main(){ int n = 120; cout << countNumbers(1, n); return 0;} |
Java
// Java program to find the number of integers// from 1 to n which contains digits 0's and 1's onlyclass GFG{ // Function to find the number of integers// from 1 to n which contains 0's and 1's onlystatic int countNumbers(int x, int n){ // If number is greater than n if (x > n) return 0; // otherwise add count this number and // call two functions return 1 + countNumbers(x * 10, n) + countNumbers(x * 10 + 1, n);}// Driver codepublic static void main (String[] args) { int n = 120; System.out.println(countNumbers(1, n));}}// This code is contributed by chandan_jnu |
Python3
# Python3 program to find the number of # integers from 1 to n which contains # digits 0's and 1's only # Function to find the number of integers # from 1 to n which contains 0's and 1's only def countNumbers(x, n): # If number is greater than n if x > n : return 0 # otherwise add count this number and # call two functions return (1 + countNumbers(x * 10, n) + countNumbers(x * 10 + 1, n)) # Driver code if __name__ == '__main__': n = 120; print(countNumbers(1, n)); # This code is contributed by Arnab Kundu |
C#
// C# program to find the number of integers // from 1 to n which contains digits 0's and 1's only using System;class GFG { // Function to find the number of integers // from 1 to n which contains 0's and 1's only static int countNumbers(int x, int n) { // If number is greater than n if (x > n) return 0; // otherwise add count this number and // call two functions return 1 + countNumbers(x * 10, n) + countNumbers(x * 10 + 1, n); } // Driver code public static void Main() { int n = 120; Console.WriteLine(countNumbers(1, n)); } } // This code is contributed by Ryuga |
PHP
<?php// PHP program to find the number of // integers from 1 to n which contains// digits 0's and 1's only// Function to find the number of integers// from 1 to n which contains 0's and 1's onlyfunction countNumbers($x, $n){ // If number is greater than n if ($x > $n) return 0; // otherwise add count this number and // call two functions return 1 + countNumbers($x * 10, $n) + countNumbers($x * 10 + 1, $n);}// Driver code$n = 120;echo(countNumbers(1, $n));// This code is contributed // by Code_Mech.?> |
Javascript
<script>// JavaScript program to find the// number of integers from 1 to n// which contains digits 0's and 1's only// Function to find the number of// integers from 1 to n which // contains 0's and 1's onlyfunction countNumbers(x, n){ // If number is greater than n if (x > n) return 0; // Otherwise add count this number and // call two functions return 1 + countNumbers(x * 10, n) + countNumbers(x * 10 + 1, n);}// Driver codelet n = 120;document.write(countNumbers(1, n));// This code is contributed by Manoj.</script> |
Output:
7
Time Complexity: O((log10n))
Auxiliary Space: O((log10n))
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