Contar números pares e impares en un rango de L a R

Dados dos números L y R, la tarea es contar el número de números impares en el rango L a R.
Ejemplos: 

Entrada: l = 3, r = 7 
Salida: 3 2  La
cuenta de números impares es 3, es decir, 3, 5, 7  La
cuenta de números pares es 2, es decir, 4, 6
Entrada: l = 4, r = 8 
Salida: 2 
 

Enfoque: los números totales en el rango serán (R – L + 1), es decir, N.  

1. Si N es par, la cuenta de los números pares e impares será N/2 .
2. Si N es impar, 
   • Si L o R son impares, entonces el conteo de los números impares será N/2 + 1, y los números pares = N – conteo de impares .
   • De lo contrario, el recuento de números impares será N/2 y los números pares = N – countofOdd .

A continuación se muestra la implementación del enfoque anterior:

// 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

// 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 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# 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
// 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 intval($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
?>

<script>
 
// Javascript implementation
// of the above approach
 
 
// Return the number of odd numbers
// in the range [L, R]
function countOdd( L, R){
 
    let N = Math.floor((R - L) / 2);
 
    // if either R or L is odd
    if (R % 2 != 0 || L % 2 != 0)
        N += 1;
 
    return N;
}
 
 
    // Driver Code
     
    let L = 3, R = 7;
    let odds = countOdd(L, R);
    let evens = (R - L + 1) - odds;
     
    document.write(
    "Count of odd numbers is " + odds + "</br>"
    );
    document.write(
    "Count of even numbers is " + evens + "</br>"
    );
     
     
</script>

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

Complejidad de Tiempo: O(1), ya que solo hay una operación aritmética básica que toma tiempo constante.
Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.