# 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.

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

Approach: Initialize count = 0 and traverse the array from arr 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); ` `    ``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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