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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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
?>

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Output:

5


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