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Count numbers from 1 to n that have 4 as a digit
• Difficulty Level : Hard
• Last Updated : 08 Apr, 2021

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``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;``}`

## 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.`

## Python3

 `# 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.`

## 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`

## PHP

 ``

## Javascript

 ``

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(10d) =   9 * count(10d - 1) + 10d - 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 10d - 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 * 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``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;``   ``// d=1 a = count of numbers from 0 to 9 = 1``   ``// d=2 a = count of numbers from 0 to 99 = a*9 + 10 = 19``   ``// d=3 a = count of numbers from 0 to 999 = a*19 + 100 = 171``   ``int` `*a = ``new` `int``[d+1];``   ``a = 0, a = 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 * 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;``}`

## 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;``    ``// d=1 a = count of numbers from``    ``// 0 to 9 = 1``    ``// d=2 a = count of numbers from``    ``// 0 to 99 = a*9 + 10 = 19``    ``// d=3 a = count of numbers from``    ``// 0 to 999 = a*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 * 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`

## 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``    ``# d=1 a = count of numbers from 0 to 9 = 1``    ``# d=2 a = count of numbers from``    ``#            0 to 99 = a*9 + 10 = 19``    ``# d=3 a = count of numbers from``    ``#            0 to 999 = a*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 * 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`

## 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;``    ``// d=1 a = count of numbers from``    ``// 0 to 9 = 1``    ``// d=2 a = count of numbers from``    ``// 0 to 99 = a*9 + 10 = 19``    ``// d=3 a = count of numbers from``    ``// 0 to 999 = a*19 + 100 = 171``    ``int``[] a = ``new` `int``[d+2];``    ``a = 0;``    ``a = 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 * 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`

## PHP

 ` 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 * 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``?>`

## Javascript

 ``

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

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