Conteo de números del rango [L, R] que terminan con cualquiera de los dígitos dados

Dado un rango [L, R] , la tarea es encontrar el conteo de números del rango cuyo último dígito es 2 , 3 o 9 .
Ejemplos: 
 

Entrada: L = 1, R = 3 
Salida: 2 
2 y 3 son los únicos números válidos.
Entrada: L = 11, R = 33 
Salida: 8 
 

Enfoque: Inicialice un conteo de contador = 0 y ejecute un ciclo para cada elemento del rango, si el elemento actual termina con cualquiera de los dígitos dados, incremente el conteo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the count
// of the required numbers
int countNums(int l, int r)
{
    int cnt = 0;
 
    for (int i = l; i <= r; i++)
    {
 
        // Last digit of the current number
        int lastDigit = (i % 10);
 
        // If the last digit is equal to
        // any of the given digits
        if ((lastDigit % 10) == 2 || (lastDigit % 10) == 3
            || (lastDigit % 10) == 9)
        {
            cnt++;
        }
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int l = 11, r = 33;
    cout << countNums(l, r) ;
}
     
// This code is contributed by AnkitRai01

// Java implementation of the approach
class GFG {
 
    // Function to return the count
    // of the required numbers
    static int countNums(int l, int r)
    {
        int cnt = 0;
 
        for (int i = l; i <= r; i++) {
 
            // Last digit of the current number
            int lastDigit = (i % 10);
 
            // If the last digit is equal to
            // any of the given digits
            if ((lastDigit % 10) == 2 || (lastDigit % 10) == 3
                || (lastDigit % 10) == 9) {
                cnt++;
            }
        }
 
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int l = 11, r = 33;
        System.out.print(countNums(l, r));
    }
}

# Python3 implementation of the approach
 
# Function to return the count
# of the required numbers
def countNums(l, r) :
    cnt = 0;
 
    for i in range(l, r + 1) :
 
        # Last digit of the current number
        lastDigit = (i % 10);
 
        # If the last digit is equal to
        # any of the given digits
        if ((lastDigit % 10) == 2 or (lastDigit % 10) == 3
            or (lastDigit % 10) == 9) :
         
            cnt += 1;
 
    return cnt;
 
# Driver code
if __name__ == "__main__" :
     
    l = 11; r = 33;
     
    print(countNums(l, r)) ;
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the count
    // of the required numbers
    static int countNums(int l, int r)
    {
        int cnt = 0;
 
        for (int i = l; i <= r; i++)
        {
 
            // Last digit of the current number
            int lastDigit = (i % 10);
 
            // If the last digit is equal to
            // any of the given digits
            if ((lastDigit % 10) == 2 ||
                (lastDigit % 10) == 3 ||
                (lastDigit % 10) == 9)
            {
                cnt++;
            }
        }
        return cnt;
    }
 
    // Driver code
    public static void Main()
    {
        int l = 11, r = 33;
        Console.Write(countNums(l, r));
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
    // JavaScript implementation of the approach
     
    // Function to return the count
    // of the required numbers
    function countNums(l, r)
    {
        let cnt = 0;
   
        for (let i = l; i <= r; i++) {
   
            // Last digit of the current number
            let lastDigit = (i % 10);
   
            // If the last digit is equal to
            // any of the given digits
            if ((lastDigit % 10) == 2 || (lastDigit % 10) == 3
                || (lastDigit % 10) == 9) {
                cnt++;
            }
        }
   
        return cnt;
    }
     
    let l = 11, r = 33;
      document.write(countNums(l, r));
     
</script>

8

Complejidad de tiempo: O(r)

Espacio Auxiliar: O(1)