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 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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. ` |

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## 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. ` `?> ` |

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Output:

385

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