Count numbers having 0 as a digit Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Count how many integers from 1 to N contain 0's as a digit.Examples: Input: n = 9 Output: 0 Input: n = 107 Output: 17 The numbers having 0 are 10, 20,..90, 100, 101..107 Input: n = 155 Output: 24 The numbers having 0 are 10, 20,..90, 100, 101..110, 120, ..150. The idea is to traverse all numbers from 1 to n. For every traversed number, traverse through its digits, if any digit is 0, increment count. Below is the implementation of the above idea : C++ // C++ program to count numbers from 1 to n with // 0 as a digit #include<bits/stdc++.h> using namespace std; // Returns 1 if x has 0, else 0 int has0(int x) { // Traverse through all digits of // x to check if it has 0. while (x) { // If current digit is 0, return true if (x % 10 == 0) return 1; x /= 10; } return 0; } // Returns count of numbers from 1 to n with 0 as digit int getCount(int n) { // Initialize count of numbers having 0 as digit int count = 0; // Traverse through all numbers and for every number // check if it has 0. for (int i=1; i<=n; i++) count += has0(i); return count; } // Driver program int main() { int n = 107; cout << "Count of numbers from 1" << " to " << n << " is " << getCount(n); } Java // Java program to count numbers // from 1 to n with 0 as a digit import java.io.*; class GFG { // Returns 1 if x has 0, else 0 static int has0(int x) { // Traverse through all digits // of x to check if it has 0. while (x != 0) { // If current digit is 0, // return true if (x % 10 == 0) return 1; x /= 10; } return 0; } // Returns count of numbers // from 1 to n with 0 as digit static int getCount(int n) { // Initialize count of // numbers having 0 as digit int count = 0; // Traverse through all numbers // and for every number // check if it has 0. for (int i = 1; i <= n; i++) count += has0(i); return count; } // Driver program public static void main(String args[]) { int n = 107; System.out.println("Count of numbers from 1" + " to " +n + " is " + getCount(n)); } } // This code is contributed by Nikita Tiwari. Python3 # Python3 program to count numbers # from 1 to n with 0 as a digit # Returns 1 if x has 0, else 0 def has0(x) : # Traverse through all digits # of x to check if it has 0. while (x != 0) : # If current digit is 0, # return true if (x % 10 == 0) : return 1 x = x // 10 return 0 # Returns count of numbers # from 1 to n with 0 as digit def getCount(n) : # Initialize count of numbers # having 0 as digit. count = 0 # Traverse through all numbers # and for every number check # if it has 0. for i in range(1, n + 1) : count = count + has0(i) return count # Driver program n = 107 print("Count of numbers from 1", " to ", n , " is " , getCount(n)) # This code is contributed by Nikita tiwari. C# // C# program to count numbers // from 1 to n with 0 as a digit using System; class GFG { // Returns 1 if x has 0, else 0 static int has0(int x) { // Traverse through all digits // of x to check if it has 0. while (x != 0) { // If current digit is 0, // return true if (x % 10 == 0) return 1; x /= 10; } return 0; } // Returns count of numbers // from 1 to n with 0 as digit static int getCount(int n) { // Initialize count of // numbers having 0 as digit int count = 0; // Traverse through all numbers // and for every number // check if it has 0. for (int i = 1; i <= n; i++) count += has0(i); return count; } // Driver Code public static void Main() { int n = 107; Console.WriteLine("Count of numbers from 1" + " to " +n + " is " + getCount(n)); } } // This code is contributed by Sam007 JavaScript <script> // JavaScript program to count numbers from 1 to n with // 0 as a digit // Returns 1 if x has 0, else 0 function has0(x) { // Traverse through all digits of // x to check if it has 0. while (x) { // If current digit is 0, return true if (x % 10 == 0) return 1; x = Math.floor(x / 10); } return 0; } // Returns count of numbers from 1 to n with 0 as digit function getCount(n) { // Initialize count of numbers having 0 as digit let count = 0; // Traverse through all numbers and for every number // check if it has 0. for (let i=1; i<=n; i++) count += has0(i); return count; } // Driver program let n = 107; document.write("Count of numbers from 1" + " to " + n + " is " + getCount(n)); // This code is contributed by Surbhi Tyagi. </script> OutputCount of numbers from 1 to 107 is 17 Time Complexity: O(n log n)Auxiliary Space: O(1)Refer below post for an optimized solution.Count numbers having 0 as a digit Comment More infoAdvertise with us Next Article Types of Asymptotic Notations in Complexity Analysis of Algorithms K kartik Improve Article Tags : Mathematical DSA Practice Tags : Mathematical Similar Reads Basics & PrerequisitesTime Complexity and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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