# Count Odd and Even numbers in a range from L to R

• Difficulty Level : Easy
• Last Updated : 30 Apr, 2021

Given two numbers L and R, the task is to count the number of odd numbers in the range L to R.
Examples:

Input: l = 3, r = 7
Output: 3 2
Count of odd numbers is 3 i.e. 3, 5, 7
Count of even numbers is 2 i.e. 4, 6
Input: l = 4, r = 8
Output:

Approach: Total numbers in the range will be (R – L + 1) i.e. N.

1. If N is even then the count of both odd and even numbers will be N/2.
2. If N is odd,
• If L or R is odd, then the count of odd number will be N/2 + 1 and even numbers = N – countofOdd.
• Else, count of odd numbers will be N/2 and even numbers = N – countofOdd.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ` `using` `namespace` `std;` `// Return the number of odd numbers``// in the range [L, R]``int` `countOdd(``int` `L, ``int` `R){` `    ``int` `N = (R - L) / 2;` `    ``// if either R or L is odd``    ``if` `(R % 2 != 0 || L % 2 != 0)``        ``N += 1;` `    ``return` `N;``}` `// Driver code``int` `main()``{``    ``int` `L = 3, R = 7;``    ``int` `odds = countOdd(L, R);``    ``int` `evens = (R - L + 1) - odds;``    ` `    ``cout << ``"Count of odd numbers is "` `<< odds << endl;``    ``cout << ``"Count of even numbers is "` `<< evens << endl;``    ``return` `0;``}` `// This code is contributed by Rituraj Jain`

## Java

 `// Java implementation of the above approach` `class` `GFG {` `    ``// Return the number of odd numbers``    ``// in the range [L, R]``    ``static` `int` `countOdd(``int` `L, ``int` `R)``    ``{``        ``int` `N = (R - L) / ``2``;` `        ``// if either R or L is odd``        ``if` `(R % ``2` `!= ``0` `|| L % ``2` `!= ``0``)``            ``N++;` `        ``return` `N;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `L = ``3``, R = ``7``;` `        ``int` `odds = countOdd(L, R);``        ``int` `evens = (R - L + ``1``) - odds;``        ``System.out.println(``"Count of odd numbers is "` `+ odds);``        ``System.out.println(``"Count of even numbers is "` `+ evens);``    ``}``}`

## Python 3

 `# Python 3 implementation of the``# above approach` `# Return the number of odd numbers``# in the range [L, R]``def` `countOdd(L, R):` `    ``N ``=` `(R ``-` `L) ``/``/` `2` `    ``# if either R or L is odd``    ``if` `(R ``%` `2` `!``=` `0` `or` `L ``%` `2` `!``=` `0``):``        ``N ``+``=` `1` `    ``return` `N` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``L ``=` `3``    ``R ``=` `7` `    ``odds ``=` `countOdd(L, R)``    ``evens ``=` `(R ``-` `L ``+` `1``) ``-` `odds``    ``print``(``"Count of odd numbers is"``, odds)``    ``print``(``"Count of even numbers is"``, evens)` `# This code is contributed by ita_c`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{` `    ``// Return the number of odd numbers``    ``// in the range [L, R]``    ``static` `int` `countOdd(``int` `L, ``int` `R)``    ``{``        ``int` `N = (R - L) / 2;` `        ``// if either R or L is odd``        ``if` `(R % 2 != 0 || L % 2 != 0)``            ``N++;` `        ``return` `N;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `L = 3, R = 7;` `        ``int` `odds = countOdd(L, R);``        ``int` `evens = (R - L + 1) - odds;``        ``Console.WriteLine(``"Count of odd numbers is "` `+ odds);``        ``Console.WriteLine(``"Count of even numbers is "` `+ evens);``    ``}``}` `// This code is contributed by Ryuga`

## PHP

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## Javascript

 ``
Output:
```Count of odd numbers is 3
Count of even numbers is 2```

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