# 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 codeint 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 codeif __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 approachusing 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



## Javascript


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

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