Contar pares consecutivos de los mismos elementos

Dada una array arr[] , la tarea es contar el número de pares formados por elementos consecutivos en los que ambos elementos de un par son iguales.
Ejemplos: 
 

Entrada: arr[] = {1, 2, 2, 3, 4, 4, 5, 5, 5, 5} 
Salida: 5 
(1, 2), (4, 5), (6, 7), (7 , 8) y (8, 9) son los pares de índices válidos 
donde los elementos consecutivos son iguales.
Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 
Salida: 0 
No hay dos elementos consecutivos iguales en la array dada. 
 

Enfoque: inicialice count = 0 y recorra la array desde arr[0] hasta arr[n – 2] . Si el elemento actual es igual al siguiente elemento en la array, incremente el conteo . Imprime el conteo al final.
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 consecutive
// elements in the array which are equal
int countCon(int ar[], int n)
{
    int cnt = 0;
 
    for (int i = 0; i < n - 1; i++) {
 
        // If consecutive elements are same
        if (ar[i] == ar[i + 1])
            cnt++;
    }
    return cnt;
}
 
// Driver code
int main()
{
    int ar[] = { 1, 2, 2, 3, 4, 4, 5, 5, 5, 5 };
    int n = sizeof(ar) / sizeof(ar[0]);
    cout << countCon(ar, n);
 
    return 0;
}

// Java implementation of the approach
public class GfG
{
 
    // Function to return the count of consecutive
    // elements in the array which are equal
    static int countCon(int ar[], int n)
    {
        int cnt = 0;
     
        for (int i = 0; i < n - 1; i++)
        {
     
            // If consecutive elements are same
            if (ar[i] == ar[i + 1])
                cnt++;
        }
        return cnt;
    }
     
    // Driver Code
    public static void main(String []args){
         
        int ar[] = { 1, 2, 2, 3, 4, 4, 5, 5, 5, 5 };
        int n = ar.length;
        System.out.println(countCon(ar, n));
    }
}
 
// This code is contributed by Rituraj Jain

# Python3 implementation of the approach
 
# Function to return the count of consecutive
# elements in the array which are equal
def countCon(ar, n):
    cnt = 0
 
    for i in range(n - 1):
 
        # If consecutive elements are same
        if (ar[i] == ar[i + 1]):
            cnt += 1
     
    return cnt
 
# Driver code
ar = [1, 2, 2, 3, 4,
      4, 5, 5, 5, 5]
n = len(ar)
print(countCon(ar, n))
 
# This code is contributed by mohit kumar

// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to return the count of consecutive
    // elements in the array which are equal
    static int countCon(int[] ar, int n)
    {
        int cnt = 0;
     
        for (int i = 0; i < n - 1; i++)
        {
     
            // If consecutive elements are same
            if (ar[i] == ar[i + 1])
                cnt++;
        }
        return cnt;
    }
     
    // Driver Code
    public static void Main()
    {
         
        int[] ar = { 1, 2, 2, 3, 4, 4, 5, 5, 5, 5 };
        int n = ar.Length;
        Console.WriteLine(countCon(ar, n));
    }
}
 
// This code is contributed by Code_Mech.

<?php
// PHP implementation of the approach
 
// Function to return the count of consecutive
// elements in the array which are equal
function countCon($ar, $n)
{
    $cnt = 0;
 
    for ($i = 0; $i < $n - 1; $i++)
    {
 
        // If consecutive elements are same
        if ($ar[$i] == $ar[$i + 1])
            $cnt++;
    }
    return $cnt;
}
 
// Driver code
$ar = array(1, 2, 2, 3, 4, 4, 5, 5, 5, 5);
$n = sizeof($ar);
echo countCon($ar, $n);
 
// This code is contributed
// by Akanksha Rai
?>

<script>
// Javascript implementation of the approach
 
    // Function to return the count of consecutive
    // elements in the array which are equal
    function countCon(ar,n)
    {
        let cnt = 0;
       
        for (let i = 0; i < n - 1; i++)
        {
       
            // If consecutive elements are same
            if (ar[i] == ar[i + 1])
                cnt++;
        }
        return cnt;
    }
     
    // Driver Code
    let ar = [1, 2, 2, 3, 4, 4, 5, 5, 5, 5 ];
    let n = ar.length;
    document.write(countCon(ar, n));
 
// This code is contributed by unknown2108
</script>

5

Complejidad temporal: O(n)
Espacio auxiliar: O(1)