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

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: 2

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++

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// C++ implementation of the above approach 
#include <bits/stdc++.h>
  
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

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Java














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// 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); 


    } 


} 



















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Python 3














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# 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 



















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C#














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// 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 



















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PHP














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<?php 


// PHP implementation of the above approach 


  


// Return the number of odd numbers 


// in the range [L, R] 


function countOdd($L, $R) 


{ 


    $N = ($R - $L) / 2; 


  


    // if either R or L is odd 


    if ($R % 2 != 0 || $L % 2 != 0) 


        $N++; 


  


    return $N; 


} 


  


// Driver code 


$L = 3; $R = 7; 


  


$odds = countOdd($L, $R); 


$evens = ($R - $L + 1) - $odds; 


echo "Count of odd numbers is " . $odds . "\n"; 


echo "Count of even numbers is " . $evens; 


  


// This code is contributed 


// by Akanksha Rai 


?> 



















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Output:


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






          
          
          
          

            

    

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