Given a number n, write a function that returns count of numbers from 1 to n that don’t contain digit 3 in their decimal representation.

Examples:

Input: n = 10 Output: 9 Input: n = 45 Output: 31 // Numbers 3, 13, 23, 30, 31, 32, 33, 34, // 35, 36, 37, 38, 39, 43 contain digit 3. Input: n = 578 Ouput: 385

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**Solution:**

We can solve it recursively. Let count(n) be the function that counts such numbers.

'msd' --> the most significant digit in n 'd' --> number of digits in n. count(n) = n if n < 3 count(n) = n - 1 if 3 <= n < 10 count(n) = count(msd) * count(10^(d-1) - 1) + count(msd) + count(n % (10^(d-1))) if n > 10 and msd is not 3 count(n) = count( msd * (10^(d-1)) - 1) if n > 10 and msd is 3

Let us understand the solution with n = 578. count(578) = 4*count(99) + 4 + count(78) The middle term 4 is added to include numbers 100, 200, 400 and 500. Let us take n = 35 as another example. count(35) = count (3*10 - 1) = count(29)

## C/C++

#include <stdio.h> /* returns count of numbers which are in range from 1 to n and don't contain 3 as a digit */ int count(int n) { // Base cases (Assuming n is not negative) if (n < 3) return n; if (n >= 3 && n < 10) return n-1; // Calculate 10^(d-1) (10 raise to the power d-1) where d is // number of digits in n. po will be 100 for n = 578 int po = 1; while (n/po > 9) po = po*10; // find the most significant digit (msd is 5 for 578) int msd = n/po; if (msd != 3) // For 578, total will be 4*count(10^2 - 1) + 4 + count(78) return count(msd)*count(po - 1) + count(msd) + count(n%po); else // For 35, total will be equal to count(29) return count(msd*po - 1); } // Driver program to test above function int main() { printf ("%d ", count(578)); return 0; }

## Java

// Java program to count numbers that not contain 3 import java.io.*; class GFG { // Function that returns count of numbers which // are in range from 1 to n // and not contain 3 as a digit static int count(int n) { // Base cases (Assuming n is not negative) if (n < 3) return n; if (n >= 3 && n < 10) return n-1; // Calculate 10^(d-1) (10 raise to the power d-1) where d is // number of digits in n. po will be 100 for n = 578 int po = 1; while (n/po > 9) po = po*10; // find the most significant digit (msd is 5 for 578) int msd = n/po; if (msd != 3) // For 578, total will be 4*count(10^2 - 1) + 4 + count(78) return count(msd)*count(po - 1) + count(msd) + count(n%po); else // For 35, total will be equal to count(29) return count(msd*po - 1); } // Driver program public static void main (String[] args) { int n = 578; System.out.println(count(n)); } } // Contributed by Pramod Kumar

## Python

# Python program to count numbers upto n that don't contain 3 # Returns count of numbers which are in range from 1 to n # and don't contain 3 as a digit def count(n): # Base Cases ( n is not negative) if n < 3: return n elif n >= 3 and n < 10: return n-1 # Calculate 10^(d-1) ( 10 raise to the power d-1 ) where d # is number of digits in n. po will be 100 for n = 578 po = 1 while n/po > 9: po = po * 10 # Find the MSD ( msd is 5 for 578 ) msd = n/po if msd != 3: # For 578, total will be 4*count(10^2 - 1) + 4 + ccount(78) return count(msd) * count(po-1) + count(msd) + count(n%po) else: # For 35 total will be equal to count(29) return count(msd * po - 1) # Driver Program n = 578 print count(n) # Contributed by Harshit Agrawal

## C#

// C# program to count numbers that not // contain 3 using System; class GFG { // Function that returns count of // numbers which are in range from // 1 to n and not contain 3 as a // digit static int count(int n) { // Base cases (Assuming n is // not negative) if (n < 3) return n; if (n >= 3 && n < 10) return n-1; // Calculate 10^(d-1) (10 raise // to the power d-1) where d is // number of digits in n. po will // be 100 for n = 578 int po = 1; while (n / po > 9) po = po * 10; // find the most significant // digit (msd is 5 for 578) int msd = n / po; if (msd != 3) // For 578, total will be // 4*count(10^2 - 1) + 4 + // count(78) return count(msd) * count(po - 1) + count(msd) + count(n % po); else // For 35, total will be equal // to count(29) return count(msd * po - 1); } // Driver program public static void Main () { int n = 578; Console.Write(count(n)); } } // This code is contributed by Sam007.

Output:

385

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