Count consecutive pairs of same elements

Given an array arr[], the task is to count the number of pairs formed by consecutive elements in which both of the elements in a pair are same.
Examples:

Input: arr[] = {1, 2, 2, 3, 4, 4, 5, 5, 5, 5} 
Output: 5 
(1, 2), (4, 5), (6, 7), (7, 8) and (8, 9) are the valid index pairs 
where consecutive elements are equal.
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 
Output: 0 
No two consecutive elements in the given array are equal. 

Approach: Initialize count = 0 and traverse the array from arr[0] to arr[n – 2]. If the current element is equal to the next element in the array then increment the count. Print the count in the end.
Below is the implementation of the above approach: 

C++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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>

Output:

5

Time Complexity: O(n)
Auxiliary Space: O(1)