# Count numbers from 1 to n that have 4 as a digit

Given a number n, find count of all numbers from 1 to n that have 4 as a digit.

**Examples : **

Input: n = 5 Output: 1 Only 4 has '4' as digit Input: n = 50 Output: 14 Input: n = 328 Output: 60

This problem is mainly a variation of previous article on Compute sum of digits in all numbers from 1 to n.

**Naive Solution:**

A naive solution is to go through every number x from 1 to n, and check if x has 4. To check if x has or not, we can traverse all digits of x. Below is the implementation of above idea :

## C++

`// A Simple C++ program to compute sum of digits in numbers from 1 to n ` `#include<iostream> ` `using` `namespace` `std; ` ` ` `bool` `has4(` `int` `x); ` ` ` `// Returns sum of all digits in numbers from 1 to n ` `int` `countNumbersWith4(` `int` `n) ` `{ ` ` ` `int` `result = 0; ` `// initialize result ` ` ` ` ` `// One by one compute sum of digits in every number from ` ` ` `// 1 to n ` ` ` `for` `(` `int` `x=1; x<=n; x++) ` ` ` `result += has4(x)? 1 : 0; ` ` ` ` ` `return` `result; ` `} ` ` ` `// A utility function to compute sum of digits in a ` `// given number x ` `bool` `has4(` `int` `x) ` `{ ` ` ` `while` `(x != 0) ` ` ` `{ ` ` ` `if` `(x%10 == 4) ` ` ` `return` `true` `; ` ` ` `x = x /10; ` ` ` `} ` ` ` `return` `false` `; ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ` `int` `n = 328; ` ` ` `cout << ` `"Count of numbers from 1 to "` `<< n ` ` ` `<< ` `" that have 4 as a a digit is "` ` ` `<< countNumbersWith4(n) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to compute sum of ` `// digits in numbers from 1 to n ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Returns sum of all digits ` ` ` `// in numbers from 1 to n ` ` ` `static` `int` `countNumbersWith4(` `int` `n) ` ` ` `{ ` ` ` `// initialize result ` ` ` `int` `result = ` `0` `; ` ` ` ` ` `// One by one compute sum of digits ` ` ` `// in every number from 1 to n ` ` ` `for` `(` `int` `x=` `1` `; x<=n; x++) ` ` ` `result += has4(x)? ` `1` `: ` `0` `; ` ` ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// A utility function to compute sum ` ` ` `// of digits in a given number x ` ` ` `static` `boolean` `has4(` `int` `x) ` ` ` `{ ` ` ` `while` `(x != ` `0` `) ` ` ` `{ ` ` ` `if` `(x%` `10` `== ` `4` `) ` ` ` `return` `true` `; ` ` ` `x = x /` `10` `; ` ` ` `} ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// Driver Program ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `n = ` `328` `; ` ` ` `System.out.println(` `"Count of numbers from 1 to "` ` ` `+ ` `" that have 4 as a a digit is "` ` ` `+ countNumbersWith4(n)) ; ` ` ` `} ` `} ` ` ` `// This code is contributed by Nikita Tiwari. ` |

*chevron_right*

*filter_none*

## Python

`# A Simple Python 3 program to compute ` `# sum of digits in numbers from 1 to n ` ` ` `# Returns sum of all digits in numbers from 1 to n ` `def` `countNumbersWith4(n) : ` ` ` `result ` `=` `0` `# initialize result ` ` ` ` ` `# One by one compute sum of digits ` ` ` `# in every number from 1 to n ` ` ` `for` `x ` `in` `range` `(` `1` `, n ` `+` `1` `) : ` ` ` `if` `(has4(x) ` `=` `=` `True` `) : ` ` ` `result ` `=` `result ` `+` `1` ` ` ` ` `return` `result ` ` ` `# A utility function to compute sum ` `# of digits in a given number x ` `def` `has4(x) : ` ` ` `while` `(x !` `=` `0` `) : ` ` ` `if` `(x` `%` `10` `=` `=` `4` `) : ` ` ` `return` `True` ` ` `x ` `=` `x ` `/` `/` `10` ` ` ` ` `return` `False` ` ` `# Driver Program ` `n ` `=` `328` `print` `(` `"Count of numbers from 1 to "` `, n, ` ` ` `" that have 4 as a a digit is "` `, ` ` ` `countNumbersWith4(n)) ` ` ` ` ` `# This code is contributed by Nikita Tiwari. ` |

*chevron_right*

*filter_none*

## C#

`// C# program to compute sum of ` `// digits in numbers from 1 to n ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ` `// Returns sum of all digits ` ` ` `// in numbers from 1 to n ` ` ` `static` `int` `countNumbersWith4(` `int` `n) ` ` ` `{ ` ` ` ` ` `// initialize result ` ` ` `int` `result = 0; ` ` ` ` ` `// One by one compute sum of digits ` ` ` `// in every number from 1 to n ` ` ` `for` `(` `int` `x = 1; x <= n; x++) ` ` ` `result += has4(x) ? 1 : 0; ` ` ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// A utility function to compute sum ` ` ` `// of digits in a given number x ` ` ` `static` `bool` `has4(` `int` `x) ` ` ` `{ ` ` ` `while` `(x != 0) ` ` ` `{ ` ` ` `if` `(x % 10 == 4) ` ` ` `return` `true` `; ` ` ` `x = x / 10; ` ` ` `} ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 328; ` ` ` `Console.WriteLine(` `"Count of numbers from 1 to "` ` ` `+ ` `" that have 4 as a a digit is "` ` ` `+ countNumbersWith4(n)) ; ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to compute sum of ` `// digits in numbers from 1 to n ` ` ` `// Returns sum of all digits ` `// in numbers from 1 to n ` `function` `countNumbersWith4(` `$n` `) ` `{ ` ` ` `$result` `= 0; ` `// initialize result ` ` ` ` ` `// One by one compute sum of ` ` ` `// digits in every number from 1 to n ` ` ` `for` `(` `$x` `= 1; ` `$x` `<= ` `$n` `; ` `$x` `++) ` ` ` `$result` `+= has4(` `$x` `) ? 1 : 0; ` ` ` ` ` `return` `$result` `; ` `} ` ` ` `// A utility function to compute ` `// sum of digits in a given number x ` `function` `has4(` `$x` `) ` `{ ` ` ` `while` `(` `$x` `!= 0) ` ` ` `{ ` ` ` `if` `(` `$x` `% 10 == 4) ` ` ` `return` `true; ` ` ` `$x` `= ` `intval` `(` `$x` `/ 10); ` ` ` `} ` ` ` `return` `false; ` `} ` ` ` `// Driver Code ` `$n` `= 328; ` `echo` `"Count of numbers from 1 to "` `. ` `$n` `. ` ` ` `" that have 4 as a a digit is "` `. ` ` ` `countNumbersWith4(` `$n` `); ` ` ` `// This code is contributed by Sam007 ` `?> ` |

*chevron_right*

*filter_none*

**Output :**

Count of numbers from 1 to 328 that have 4 as a a digit is 60

**Efficient Solution:**

Above is a naive solution. We can do it more efficiently by finding a pattern.

Let us take few examples.

Count of numbers from 0 to 9 = 1 Count of numbers from 0 to 99 = 1*9 + 10 = 19 Count of numbers from 0 to 999 = 19*9 + 100 = 271 In general, we can write count(10^{d}) = 9 * count(10^{d - 1}) + 10^{d - 1}

In below implementation, the above formula is implemented using dynamic programming as there are overlapping subproblems.

The above formula is one core step of the idea. Below is complete algorithm.

1) Find number of digits minus one in n. Let this value be 'd'. For 328, d is 2. 2) Compute some of digits in numbers from 1 to 10^{d}- 1. Let this sum be w. For 328, we compute sum of digits from 1 to 99 using above formula. 3) Find Most significant digit (msd) in n. For 328, msd is 3. 4.a) If MSD is 4. For example if n = 428, then count of numbers is sum of following. 1) Count of numbers from 1 to 399 2) Count of numbers from 400 to 428 which is 29. 4.b) IF MSD > 4. For example if n is 728, then count of numbers is sum of following. 1) Count of numbers from 1 to 399 and count of numbers from 500 to 699, i.e., "a[2] * 6" 2) Count of numbers from 400 to 499, i.e. 100 3) Count of numbers from 700 to 728, recur for 28 4.c) IF MSD < 4. For example if n is 328, then count of numbers is sum of following. 1) Count of numbers from 1 to 299 a 2) Count of numbers from 300 to 328, recur for 28

Below is implementation of above algorithm.

## C++

`// C++ program to count numbers having 4 as a digit ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to count numbers from 1 to n that have ` `// 4 as a digit ` `int` `countNumbersWith4(` `int` `n) ` `{ ` ` ` `// Base case ` ` ` `if` `(n < 4) ` ` ` `return` `0; ` ` ` ` ` `// d = number of digits minus one in n. For 328, d is 2 ` ` ` `int` `d = ` `log10` `(n); ` ` ` ` ` `// computing count of numbers from 1 to 10^d-1, ` ` ` `// d=0 a[0] = 0; ` ` ` `// d=1 a[1] = count of numbers from 0 to 9 = 1 ` ` ` `// d=2 a[2] = count of numbers from 0 to 99 = a[1]*9 + 10 = 19 ` ` ` `// d=3 a[3] = count of numbers from 0 to 999 = a[2]*19 + 100 = 171 ` ` ` `int` `*a = ` `new` `int` `[d+1]; ` ` ` `a[0] = 0, a[1] = 1; ` ` ` `for` `(` `int` `i=2; i<=d; i++) ` ` ` `a[i] = a[i-1]*9 + ` `ceil` `(` `pow` `(10,i-1)); ` ` ` ` ` `// Computing 10^d ` ` ` `int` `p = ` `ceil` `(` `pow` `(10, d)); ` ` ` ` ` `// Most significant digit (msd) of n, ` ` ` `// For 328, msd is 3 which can be obtained using 328/100 ` ` ` `int` `msd = n/p; ` ` ` ` ` `// If MSD is 4. For example if n = 428, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 ` ` ` `// 2) Count of numbers from 400 to 428 which is 29. ` ` ` `if` `(msd == 4) ` ` ` `return` `(msd)*a[d] + (n%p) + 1; ` ` ` ` ` `// IF MSD > 4. For example if n is 728, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 and count of numbers ` ` ` `// from 500 to 699, i.e., "a[2] * 6" ` ` ` `// 2) Count of numbers from 400 to 499, i.e. 100 ` ` ` `// 3) Count of numbers from 700 to 728, recur for 28 ` ` ` `if` `(msd > 4) ` ` ` `return` `(msd-1)*a[d] + p + countNumbersWith4(n%p); ` ` ` ` ` `// IF MSD < 4. For example if n is 328, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 299 a ` ` ` `// 2) Count of numbers from 300 to 328, recur for 28 ` ` ` `return` `(msd)*a[d] + countNumbersWith4(n%p); ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ` `int` `n = 328; ` ` ` `cout << ` `"Count of numbers from 1 to "` `<< n ` ` ` `<< ` `" that have 4 as a a digit is "` ` ` `<< countNumbersWith4(n) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to count numbers having 4 as a digit ` `class` `GFG ` `{ ` ` ` `// Function to count numbers from ` `// 1 to n that have 4 as a digit ` `static` `int` `countNumbersWith4(` `int` `n) ` `{ ` ` ` `// Base case ` ` ` `if` `(n < ` `4` `) ` ` ` `return` `0` `; ` ` ` ` ` `// d = number of digits minus ` ` ` `// one in n. For 328, d is 2 ` ` ` `int` `d = (` `int` `)Math.log10(n); ` ` ` ` ` `// computing count of numbers from 1 to 10^d-1, ` ` ` `// d=0 a[0] = 0; ` ` ` `// d=1 a[1] = count of numbers from ` ` ` `// 0 to 9 = 1 ` ` ` `// d=2 a[2] = count of numbers from ` ` ` `// 0 to 99 = a[1]*9 + 10 = 19 ` ` ` `// d=3 a[3] = count of numbers from ` ` ` `// 0 to 999 = a[2]*19 + 100 = 171 ` ` ` `int` `[] a = ` `new` `int` `[d + ` `2` `]; ` ` ` `a[` `0` `] = ` `0` `; ` ` ` `a[` `1` `] = ` `1` `; ` ` ` ` ` `for` `(` `int` `i = ` `2` `; i <= d; i++) ` ` ` `a[i] = a[i - ` `1` `] * ` `9` `+ (` `int` `)Math.ceil(Math.pow(` `10` `, i - ` `1` `)); ` ` ` ` ` `// Computing 10^d ` ` ` `int` `p = (` `int` `)Math.ceil(Math.pow(` `10` `, d)); ` ` ` ` ` `// Most significant digit (msd) of n, ` ` ` `// For 328, msd is 3 which can be obtained using 328/100 ` ` ` `int` `msd = n / p; ` ` ` ` ` `// If MSD is 4. For example if n = 428, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 ` ` ` `// 2) Count of numbers from 400 to 428 which is 29. ` ` ` `if` `(msd == ` `4` `) ` ` ` `return` `(msd) * a[d] + (n % p) + ` `1` `; ` ` ` ` ` `// IF MSD > 4. For example if n ` ` ` `// is 728, then count of numbers ` ` ` `// is sum of following. ` ` ` `// 1) Count of numbers from 1 to ` ` ` `// 399 and count of numbers from ` ` ` `// 500 to 699, i.e., "a[2] * 6" ` ` ` `// 2) Count of numbers from 400 ` ` ` `// to 499, i.e. 100 ` ` ` `// 3) Count of numbers from 700 to ` ` ` `// 728, recur for 28 ` ` ` `if` `(msd > ` `4` `) ` ` ` `return` `(msd - ` `1` `) * a[d] + p + ` ` ` `countNumbersWith4(n % p); ` ` ` ` ` `// IF MSD < 4. For example if n is 328, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 299 a ` ` ` `// 2) Count of numbers from 300 to 328, recur for 28 ` ` ` `return` `(msd) * a[d] + countNumbersWith4(n % p); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `n = ` `328` `; ` ` ` `System.out.println(` `"Count of numbers from 1 to "` `+ n + ` ` ` `" that have 4 as a digit is "` `+ countNumbersWith4(n)); ` `} ` `} ` ` ` `// This code is contributed by chandan_jnu ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to count numbers having 4 as a digit ` `import` `math as mt ` ` ` `# Function to count numbers from 1 to n ` `# that have 4 as a digit ` `def` `countNumbersWith4(n): ` ` ` ` ` `# Base case ` ` ` `if` `(n < ` `4` `): ` ` ` `return` `0` ` ` ` ` `# d = number of digits minus one in n. ` ` ` `# For 328, d is 2 ` ` ` `d ` `=` `int` `(mt.log10(n)) ` ` ` ` ` `# computing count of numbers from 1 to 10^d-1, ` ` ` `# d=0 a[0] = 0 ` ` ` `# d=1 a[1] = count of numbers from 0 to 9 = 1 ` ` ` `# d=2 a[2] = count of numbers from ` ` ` `# 0 to 99 = a[1]*9 + 10 = 19 ` ` ` `# d=3 a[3] = count of numbers from ` ` ` `# 0 to 999 = a[2]*19 + 100 = 171 ` ` ` `a ` `=` `[` `1` `for` `i ` `in` `range` `(d ` `+` `1` `)] ` ` ` `a[` `0` `] ` `=` `0` ` ` `if` `len` `(a) > ` `1` `: ` ` ` `a[` `1` `] ` `=` `1` ` ` `for` `i ` `in` `range` `(` `2` `, d ` `+` `1` `): ` ` ` `a[i] ` `=` `a[i ` `-` `1` `] ` `*` `9` `+` `mt.ceil(` `pow` `(` `10` `, i ` `-` `1` `)) ` ` ` ` ` `# Computing 10^d ` ` ` `p ` `=` `mt.ceil(` `pow` `(` `10` `, d)) ` ` ` ` ` `# Most significant digit (msd) of n, ` ` ` `# For 328, msd is 3 which can be ` ` ` `# obtained using 328/100 ` ` ` `msd ` `=` `n ` `/` `/` `p ` ` ` ` ` `# If MSD is 4. For example if n = 428, ` ` ` `# then count of numbers is sum of following. ` ` ` `# 1) Count of numbers from 1 to 399 ` ` ` `# 2) Count of numbers from 400 to 428 which is 29. ` ` ` `if` `(msd ` `=` `=` `4` `): ` ` ` `return` `(msd) ` `*` `a[d] ` `+` `(n ` `%` `p) ` `+` `1` ` ` ` ` `# IF MSD > 4. For example if n is 728, ` ` ` `# then count of numbers is sum of following. ` ` ` `# 1) Count of numbers from 1 to 399 and count ` ` ` `# of numbers from 500 to 699, i.e., "a[2] * 6" ` ` ` `# 2) Count of numbers from 400 to 499, i.e. 100 ` ` ` `# 3) Count of numbers from 700 to 728, recur for 28 ` ` ` `if` `(msd > ` `4` `): ` ` ` `return` `((msd ` `-` `1` `) ` `*` `a[d] ` `+` `p ` `+` ` ` `countNumbersWith4(n ` `%` `p)) ` ` ` ` ` `# IF MSD < 4. For example if n is 328, ` ` ` `# then count of numbers is sum of following. ` ` ` `# 1) Count of numbers from 1 to 299 a ` ` ` `# 2) Count of numbers from 300 to 328, recur for 28 ` ` ` `return` `(msd) ` `*` `a[d] ` `+` `countNumbersWith4(n ` `%` `p) ` ` ` `# Driver Code ` `n ` `=` `328` `print` `(` `"Count of numbers from 1 to"` `, n, ` ` ` `"that have 4 as a digit is"` `, countNumbersWith4(n)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

*chevron_right*

*filter_none*

## C#

`// C# program to count numbers having 4 as a digit ` `using` `System; ` ` ` `class` `GFG ` `{ ` `// Function to count numbers from ` `// 1 to n that have 4 as a digit ` `static` `int` `countNumbersWith4(` `int` `n) ` `{ ` ` ` `// Base case ` ` ` `if` `(n < 4) ` ` ` `return` `0; ` ` ` ` ` `// d = number of digits minus ` ` ` `// one in n. For 328, d is 2 ` ` ` `int` `d = (` `int` `)Math.Log10(n); ` ` ` ` ` `// computing count of numbers from 1 to 10^d-1, ` ` ` `// d=0 a[0] = 0; ` ` ` `// d=1 a[1] = count of numbers from ` ` ` `// 0 to 9 = 1 ` ` ` `// d=2 a[2] = count of numbers from ` ` ` `// 0 to 99 = a[1]*9 + 10 = 19 ` ` ` `// d=3 a[3] = count of numbers from ` ` ` `// 0 to 999 = a[2]*19 + 100 = 171 ` ` ` `int` `[] a = ` `new` `int` `[d+2]; ` ` ` `a[0] = 0; ` ` ` `a[1] = 1; ` ` ` ` ` `for` `(` `int` `i = 2; i <= d; i++) ` ` ` `a[i] = a[i - 1] * 9 + (` `int` `)Math.Ceiling(Math.Pow(10, i - 1)); ` ` ` ` ` `// Computing 10^d ` ` ` `int` `p = (` `int` `)Math.Ceiling(Math.Pow(10, d)); ` ` ` ` ` `// Most significant digit (msd) of n, ` ` ` `// For 328, msd is 3 which can be obtained using 328/100 ` ` ` `int` `msd = n / p; ` ` ` ` ` `// If MSD is 4. For example if n = 428, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 ` ` ` `// 2) Count of numbers from 400 to 428 which is 29. ` ` ` `if` `(msd == 4) ` ` ` `return` `(msd) * a[d] + (n % p) + 1; ` ` ` ` ` `// IF MSD > 4. For example if n is 728, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 and count of numbers ` ` ` `// from 500 to 699, i.e., "a[2] * 6" ` ` ` `// 2) Count of numbers from 400 to 499, i.e. 100 ` ` ` `// 3) Count of numbers from 700 to 728, recur for 28 ` ` ` `if` `(msd > 4) ` ` ` `return` `(msd - 1) * a[d] + p + countNumbersWith4(n % p); ` ` ` ` ` `// IF MSD < 4. For example if n is 328, then count of ` ` ` `// numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 299 a ` ` ` `// 2) Count of numbers from 300 to 328, recur for 28 ` ` ` `return` `(msd) * a[d] + countNumbersWith4(n % p); ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `int` `n = 328; ` ` ` `Console.WriteLine(` `"Count of numbers from 1 to "` `+ n + ` ` ` `" that have 4 as a digit is "` `+ countNumbersWith4(n)); ` `} ` `} ` ` ` `// This code is contributed by chandan_jnu ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to count numbers having ` `// 4 as a digit ` ` ` `// Function to count numbers from 1 to n ` `// that have 4 as a digit ` `function` `countNumbersWith4(` `$n` `) ` `{ ` ` ` `// Base case ` ` ` `if` `(` `$n` `< 4) ` ` ` `return` `0; ` ` ` ` ` `// d = number of digits minus one in n. ` ` ` `// For 328, d is 2 ` ` ` `$d` `= (int)log10(` `$n` `); ` ` ` ` ` `// computing count of numbers from 1 to 10^d-1, ` ` ` `// d=0 a[0] = 0; ` ` ` `// d=1 a[1] = count of numbers from 0 to 9 is 1 ` ` ` `// d=2 a[2] = count of numbers from 0 to 99 is ` ` ` `// a[1]*9 + 10 = 19 ` ` ` `// d=3 a[3] = count of numbers from 0 to 999 is ` ` ` `// a[2]*19 + 100 = 171 ` ` ` `$a` `= ` `array_fill` `(0, ` `$d` `+ 1, NULL); ` ` ` `$a` `[0] = 0; ` ` ` `$a` `[1] = 1; ` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$d` `; ` `$i` `++) ` ` ` `$a` `[` `$i` `] = ` `$a` `[` `$i` `- 1] * 9 + ` ` ` `ceil` `(pow(10, ` `$i` `- 1)); ` ` ` ` ` `// Computing 10^d ` ` ` `$p` `= ` `ceil` `(pow(10, ` `$d` `)); ` ` ` ` ` `// Most significant digit (msd) of n, ` ` ` `// For 328, msd is 3 which can be ` ` ` `// obtained using 328/100 ` ` ` `$msd` `= ` `intval` `(` `$n` `/ ` `$p` `); ` ` ` ` ` `// If MSD is 4. For example if n = 428, ` ` ` `// then count of numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 ` ` ` `// 2) Count of numbers from 400 to 428 which is 29. ` ` ` `if` `(` `$msd` `== 4) ` ` ` `return` `(` `$msd` `) * ` `$a` `[` `$d` `] + (` `$n` `% ` `$p` `) + 1; ` ` ` ` ` `// IF MSD > 4. For example if n is 728, ` ` ` `// then count of numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 399 and ` ` ` `// count of numbers from 500 to 699, i.e., "a[2] * 6" ` ` ` `// 2) Count of numbers from 400 to 499, i.e. 100 ` ` ` `// 3) Count of numbers from 700 to 728, recur for 28 ` ` ` `if` `(` `$msd` `> 4) ` ` ` `return` `(` `$msd` `- 1) * ` `$a` `[` `$d` `] + ` `$p` `+ ` ` ` `countNumbersWith4(` `$n` `% ` `$p` `); ` ` ` ` ` `// IF MSD < 4. For example if n is 328, then ` ` ` `// count of numbers is sum of following. ` ` ` `// 1) Count of numbers from 1 to 299 a ` ` ` `// 2) Count of numbers from 300 to 328, recur for 28 ` ` ` `return` `(` `$msd` `) * ` `$a` `[` `$d` `] + ` ` ` `countNumbersWith4(` `$n` `% ` `$p` `); ` `} ` ` ` `// Driver Code ` `$n` `= 328; ` `echo` `"Count of numbers from 1 to "` `. ` `$n` `. ` ` ` `" that have 4 as a digit is "` `. ` ` ` `countNumbersWith4(` `$n` `) . ` `"\n"` `; ` ` ` `// This code is contributed by ita_c ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

Count of numbers from 1 to 328 that have 4 as a a digit is 60

This article is contributed by **Shivam**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

## Recommended Posts:

- Count of Numbers in Range where first digit is equal to last digit of the number
- Count n digit numbers not having a particular digit
- Count of all N digit numbers such that num + Rev(num) = 10^N - 1
- Count numbers having 0 as a digit
- Count numbers having 0 as a digit
- Count of n digit numbers whose sum of digits equals to given sum
- Count n digit numbers divisible by given number
- Count numbers (smaller than or equal to N) with given digit sum
- Count of Binary Digit numbers smaller than N
- Count of N digit numbers possible which satisfy the given conditions
- Count numbers formed by given two digit with sum having given digits
- Count numbers with unit digit k in given range
- Count all possible N digit numbers that satisfy the given condition
- Count of Numbers in a Range where digit d occurs exactly K times
- Count of Numbers in a Range divisible by m and having digit d in even positions