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Count consecutive pairs of same elements
• Last Updated : 09 Apr, 2019

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ``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

 ``
Output:
```5
```

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