# Total no of 1’s in numbers

Given an integer n, count the total number of digit 1 appearing in all positive integers less than or equal to n.

Examples:

```Input : n = 13
Output : 6
Explanation:
Here, no <= 13 containing 1 are 1, 10, 11,
12, 13. So, total 1s are 6.

Input : n = 5
Output : 1
Here, no <= 13 containing 1 are 1.
So, total 1s are 1.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1:

1. Iterate over i from 1 to n.
2. Convert i to string and count ’1’ in each integer string.
3. Add count of ’1’ in each string to the sum.

Below is the code for the above discussed approach.

## C++

 `// c++ code to count the frequency of 1 ` `// in numbers less than or equal to ` `// the given number. ` `#include ` `using` `namespace` `std; ` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``string str = to_string(i); ` `        ``countr += count(str.begin(), str.end(), ``'1'``); ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// driver function ` `int` `main() ` `{ ` `    ``int` `n = 13; ` `    ``cout << countDigitOne(n) << endl; ` `    ``n = 131; ` `    ``cout << countDigitOne(n) << endl; ` `    ``n = 159; ` `    ``cout << countDigitOne(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java code to count the frequency of 1 ` `// in numbers less than or equal to ` `// the given number. ` `class` `GFG ` `{ ` `static` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = ``0``; ` `    ``for` `(``int` `i = ``1``; i <= n; i++) ` `    ``{ ` `        ``String str = String.valueOf(i); ` `        ``countr += str.split(``"1"``, -``1``) . length - ``1``; ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``13``; ` `    ``System.out.println(countDigitOne(n)); ` `    ``n = ``131``; ` `    ``System.out.println(countDigitOne(n)); ` `    ``n = ``159``; ` `    ``System.out.println(countDigitOne(n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## Python3

 `# Python3 code to count the frequency  ` `# of 1 in numbers less than or equal  ` `# to the given number. ` ` `  `def` `countDigitOne(n): ` `    ``countr ``=` `0``; ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `        ``str1 ``=` `str``(i); ` `        ``countr ``+``=` `str1.count(``"1"``); ` ` `  `    ``return` `countr; ` ` `  `# Driver Code ` `n ``=` `13``; ` `print``(countDigitOne(n)); ` ` `  `n ``=` `131``; ` `print``(countDigitOne(n)); ` ` `  `n ``=` `159``; ` `print``(countDigitOne(n)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# code to count the frequency of 1 ` `// in numbers less than or equal to ` `// the given number. ` `using` `System; ` `class` `GFG ` `{ ` `static` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `    ``{ ` `        ``string` `str = i.ToString(); ` `        ``countr += str.Split(``"1"``) . Length - 1; ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 13; ` `    ``Console.WriteLine(countDigitOne(n)); ` `    ``n = 131; ` `    ``Console.WriteLine(countDigitOne(n)); ` `    ``n = 159; ` `    ``Console.WriteLine(countDigitOne(n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

Output:

```6
66
96
```

Time complexity: O(nlog(n))

Approach 2:

1. Initialize countr as 0.
2. Iterate over i from 1 to n incrementing by 10 each time.
3. Add (n / (i * 10 ) ) * i to countr after each (i*10) interval.
4. Add min( max((n mod (i*10) – i + 1, 0), i) to countr.

Below is the code for the above discussed approach.

## C++

 `// c++ code to count the frequency of 1 ` `// in numbers less than or equal to ` `// the given number. ` `#include ` `using` `namespace` `std; ` ` `  `// function to count the frequency of 1. ` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = 0; ` `    ``for` `(``int` `i = 1; i <= n; i *= 10) { ` `        ``int` `divider = i * 10; ` `        ``countr += (n / divider) * i +  ` `               ``min(max(n % divider - i + 1, 0), i); ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// driver function ` `int` `main() ` `{ ` `    ``int` `n = 13; ` `    ``cout << countDigitOne(n) << endl; ` `    ``n = 113; ` `    ``cout << countDigitOne(n) << endl; ` `    ``n = 205; ` `    ``cout << countDigitOne(n) << endl; ` `} `

## Java

 `/// Java code to count the  ` `// frequency of 1 in numbers ` `// less than or equal to ` `// the given number. ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` `     `  `// function to count  ` `// the frequency of 1. ` `static` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = ``0``; ` `    ``for` `(``int` `i = ``1``; ` `             ``i <= n; i *= ``10``)  ` `    ``{ ` `        ``int` `divider = i * ``10``; ` `        ``countr += (n / divider) * i +  ` `                ``Math.min(Math.max(n %  ` `                         ``divider - i +  ` `                            ``1``, ``0``), i); ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``13``; ` `    ``System.out.println(countDigitOne(n)); ` `     `  `    ``n = ``113``; ` `    ``System.out.println(countDigitOne(n)); ` `     `  `    ``n = ``205``; ` `    ``System.out.println(countDigitOne(n)); ` `} ` `} ` ` `  `// This code is contributed by akt_mit `

## Python3

 `# Python3 code to count the  ` `# frequency of 1 in numbers  ` `# less than or equal to ` `# the given number. ` ` `  `# function to count the  ` `# frequency of 1. ` `def` `countDigitOne(n): ` `    ``countr ``=` `0``; ` `    ``i ``=` `1``; ` `    ``while``(i <``=` `n): ` `        ``divider ``=` `i ``*` `10``; ` `        ``countr ``+``=` `(``int``(n ``/` `divider) ``*` `i ``+` `                 ``min``(``max``(n ``%` `divider ``-``i ``+`  `                              ``1``, ``0``), i)); ` `        ``i ``*``=` `10``; ` `     `  `    ``return` `countr; ` ` `  `# Driver Code ` `n ``=` `13``; ` `print``(countDigitOne(n)); ` `n ``=` `113``; ` `print``(countDigitOne(n)); ` `n ``=` `205``; ` `print``(countDigitOne(n)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# code to count the  ` `// frequency of 1 in numbers ` `// less than or equal to ` `// the given number. ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// function to count  ` `// the frequency of 1. ` `static` `int` `countDigitOne(``int` `n) ` `{ ` `    ``int` `countr = 0; ` `    ``for` `(``int` `i = 1;  ` `             ``i <= n; i *= 10)  ` `    ``{ ` `        ``int` `divider = i * 10; ` `        ``countr += (n / divider) * i +  ` `                   ``Math.Min(Math.Max(n % divider -  ` `                                     ``i + 1, 0), i); ` `    ``} ` `    ``return` `countr; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `n = 13; ` `    ``Console.WriteLine(countDigitOne(n)); ` `     `  `    ``n = 113; ` `    ``Console.WriteLine(countDigitOne(n)); ` `     `  `    ``n = 205; ` `    ``Console.WriteLine(countDigitOne(n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

Output:

```6
40
141
```

Time complexity: O(log(n))

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