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 = 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. |
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.