Count numbers that don’t contain 3
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 Output: 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 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++
#include <bits/stdc++.h> using namespace std; /* 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 code int main() { cout << count(578) << " "; return 0; } // This code is contributed by rathbhupendra |
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 |
Python3
# 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 + 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 n = 578print (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. |
PHP
<?php /* returns count of numbers which are in range from 1 to n and don't contain 3 as a digit */function count1($n) { // Base cases (Assuming n is not negative) if ($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 $po = 1; for($x = intval($n/$po); $x > 9; $x = intval($n/$po)) $po = $po*10; // find the most significant digit (msd is 5 for 578) $msd = intval($n / $po); if ($msd != 3) // For 578, total will be 4*count(10^2 - 1) // + 4 + count(78) return count1($msd) * count1($po - 1) + count1($msd) + count1($n%$po); else // For 35, total will be equal to count(29) return count1($msd*$po - 1); } // Driver program to test above function echo count1(578); // This code is contributed by mits. ?> |
Javascript
<script> // javascript program to count numbers that not contain 3 // Function that returns count of numbers which // are in range from 1 to n // and not contain 3 as a digit function count(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 var po = 1; while (parseInt(n / po) > 9) po = po * 10; // find the most significant digit (msd is 5 for 578) var msd = parseInt (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 var n = 578; document.write(count(n)); // This code is contributed by gauravrajput1 </script> |
Output:
385
Time Complexity: O(log10n)
Auxiliary Space: O(1), since no extra space has been taken.
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