Count numbers 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  ; Base cases ( Assuming n is not negative ) ; 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 ; find the most significant digit ( msd is 5 for 578 ) ; For 578 , total will be 4 * count ( 10 ^ 2 - 1 ) + 4 + count ( 78 ) ; For 35 , total will be equal to count ( 29 ) ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; int count ( int n ) { if ( n < 3 ) return n ; if ( n >= 3 && n < 10 ) return n - 1 ; int po = 1 ; while ( n / po > 9 ) po = po * 10 ; int msd = n / po ; if ( msd != 3 ) return count ( msd ) * count ( po - 1 ) + count ( msd ) + count ( n % po ) ; else return count ( msd * po - 1 ) ; } int main ( ) { cout << count ( 578 ) << " ▁ " ; return 0 ; }