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# Find the number of positive integers less than or equal to N that have an odd number of digits

• Difficulty Level : Basic
• Last Updated : 10 May, 2021

Given an integer N where 1 ≤ N ≤ 105, the task is to find the number of positive integers less than or equal to N that have an odd number of digits without leading zeros.
Examples:

Input: N = 11
Output:
1, 2, 3, …, 8 and 9 are the numbers ≤ 11
with odd number of digits.
Input: N = 893
Output: 803

Naive approach: Traverse from 1 to N and for each number check if it contains odd digits or not.
Efficient approach: For the values:

• When N < 10 then the count of valid numbers will be N.
• When N / 10 < 10 then 9.
• When N / 100 < 10 then 9 + N – 99.
• When N / 1000 < 10 then 9 + 900.
• When N / 10000 < 10 then 909 + N – 9999.
• Otherwise 90909.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the number of``// positive integers less than or equal``// to N that have odd number of digits``int` `odd_digits(``int` `n)``{``    ``if` `(n < 10)``        ``return` `n;``    ``else` `if` `(n / 10 < 10)``        ``return` `9;``    ``else` `if` `(n / 100 < 10)``        ``return` `9 + n - 99;``    ``else` `if` `(n / 1000 < 10)``        ``return` `9 + 900;``    ``else` `if` `(n / 10000 < 10)``        ``return` `909 + n - 9999;``    ``else``        ``return` `90909;``}` `// Driver code``int` `main()``{``    ``int` `n = 893;` `    ``cout << odd_digits(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the number of``// positive integers less than or equal``// to N that have odd number of digits``static` `int` `odd_digits(``int` `n)``{``    ``if` `(n < ``10``)``        ``return` `n;``    ``else` `if` `(n / ``10` `< ``10``)``        ``return` `9``;``    ``else` `if` `(n / ``100` `< ``10``)``        ``return` `9` `+ n - ``99``;``    ``else` `if` `(n / ``1000` `< ``10``)``        ``return` `9` `+ ``900``;``    ``else` `if` `(n / ``10000` `< ``10``)``        ``return` `909` `+ n - ``9999``;``    ``else``        ``return` `90909``;``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `n = ``893``;` `    ``System.out.println(odd_digits(n));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the number of``# positive integers less than or equal``# to N that have odd number of digits``def` `odd_digits(n) :` `    ``if` `(n < ``10``) :``        ``return` `n;``    ``elif` `(n ``/` `10` `< ``10``) :``        ``return` `9``;``    ``elif` `(n ``/` `100` `< ``10``) :``        ``return` `9` `+` `n ``-` `99``;``    ``elif` `(n ``/` `1000` `< ``10``) :``        ``return` `9` `+` `900``;``    ``elif` `(n ``/` `10000` `< ``10``) :``        ``return` `909` `+` `n ``-` `9999``;``    ``else` `:``        ``return` `90909``;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `893``;` `    ``print``(odd_digits(n));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``                    ` `class` `GFG``{` `// Function to return the number of``// positive integers less than or equal``// to N that have odd number of digits``static` `int` `odd_digits(``int` `n)``{``    ``if` `(n < 10)``        ``return` `n;``    ``else` `if` `(n / 10 < 10)``        ``return` `9;``    ``else` `if` `(n / 100 < 10)``        ``return` `9 + n - 99;``    ``else` `if` `(n / 1000 < 10)``        ``return` `9 + 900;``    ``else` `if` `(n / 10000 < 10)``        ``return` `909 + n - 9999;``    ``else``        ``return` `90909;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `n = 893;` `    ``Console.WriteLine(odd_digits(n));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`803`

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