Cuente el número de grupos en la array dada

Dada una array arr[] de N enteros, la tarea es contar el número de grupos en la array dada.
 

Agrupación se define como una serie de 2 o más elementos adyacentes del mismo valor.

Ejemplos: 
 

Entrada: arr[] = { 13, 15, 66, 66, 37, 8, 8, 11, 52 }; 
Salida: 2 
Explicación: 
Hay dos grupos en la array dada {66, 66} y {8, 8}.
Entrada: arr[] = {1, 2, 1, 4, 3, 2} 
Salida: 0 
Explicación: 
No hay grupos en la array dada. 
 

Enfoque: Para resolver el problema, debemos seguir los siguientes pasos: 
 

1. Recorra la array y compruebe si aparece el mismo elemento en dos índices consecutivos.
2. Para cualquier ocurrencia de este tipo, repite hasta que ocurra un número diferente.
3. Aumente el conteo de grupos en 1 solo después de la ejecución del paso 2. Si aún no se recorre toda la array, repita los pasos anteriores para los siguientes elementos.
4. Imprima el conteo final de grupos después de recorrer toda la array.

A continuación se muestra la implementación del enfoque anterior:
 

// C++ program to calculate
// the number of clumps in
// an array
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// clumps in the given array arr[]
int countClumps(int arr[], int N)
{
 
    // Initialise count of clumps as 0
    int clumps = 0;
 
    // Traverse the arr[]
    for (int i = 0; i < N - 1; i++) {
 
        int flag = 0;
        // Whenever a sequence of same
        // value is encountered
        while (arr[i] == arr[i + 1]) {
            flag = 1;
            i++;
        }
 
        if (flag)
            clumps++;
    }
 
    // Return the count of clumps
    return clumps;
}
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 13, 15, 66, 66, 66, 37, 37,
                  8, 8, 11, 11 };
 
    // length of the given array arr[]
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << countClumps(arr, N) << '\n';
    return 0;
}

// Java program for the above approach
class Test {
 
    // Given array arr[]
    static int arr[] = { 13, 15, 66, 66, 66,
                         37, 37, 8, 8, 11, 11 };
 
    // Function to count the number of
    // clumps in the given array arr[]
    static int countClumps()
    {
        int l = arr.length;
 
        // Initialise count of clumps as 0
        int clumps = 0;
 
        // Traverse the arr[]
        for (int i = 0; i < l - 1; i++) {
 
            int flag = 0;
            // Whenever a sequence of same
            // value is encountered
            while (i < l - 1
                   && arr[i] == arr[i + 1]) {
                flag = 1;
                i++;
            }
 
            if (flag == 1)
                clumps++;
        }
 
        // Return the count of clumps
        return clumps;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Function Call
        System.out.println(countClumps());
    }
}

# Python3 program to calculate
# the number of clumps in
# an array
 
# Function to count the number of
# clumps in the given array arr[]
def countClumps(arr, N):
     
    # Initialise count of clumps as 0
    clumps = 0
 
    # Traverse the arr[]
    i = 0
    while(i < N - 1):
         
        flag = 0
         
        # Whenever a sequence of same
        # value is encountered
        while (i + 1 < N and
               arr[i] == arr[i + 1]):
            flag = 1
            i += 1
 
        if (flag):
            clumps += 1
             
        i += 1
 
    # Return the count of clumps
    return clumps
     
# Driver Code
 
# Given array
arr = [ 13, 15, 66, 66, 66,
        37, 37, 8, 8, 11, 11 ]
 
# length of the given array arr[]
N = len(arr)
 
# Function Call
print(countClumps(arr, N))
 
# This code is contributed by yatin

// C# program for the above approach
using System;
class GFG{
 
// Given array arr[]
static int []arr = { 13, 15, 66, 66, 66,
                     37, 37, 8, 8, 11, 11 };
 
// Function to count the number of
// clumps in the given array arr[]
static int countClumps()
{
    int l = arr.Length;
 
    // Initialise count of clumps as 0
    int clumps = 0;
 
    // Traverse the arr[]
    for (int i = 0; i < l - 1; i++)
    {
        int flag = 0;
         
        // Whenever a sequence of same
        // value is encountered
        while (i < l - 1 && arr[i] == arr[i + 1])
        {
            flag = 1;
            i++;
        }
 
        if (flag == 1)
            clumps++;
    }
 
    // Return the count of clumps
    return clumps;
}
 
// Driver Code
public static void Main()
{
 
    // Function Call
    Console.WriteLine(countClumps());
}
}
 
// This code is contributed by shivanisinghss2110

<script>
 
    // Javascript program to calculate
    // the number of clumps in
    // an array 
   
    // Function to count the number of
    // clumps in the given array arr[]
    function countClumps(arr, N)
    {
 
        // Initialise count of clumps as 0
        let clumps = 0;
 
        // Traverse the arr[]
        for (let i = 0; i < N - 1; i++) {
 
            let flag = 0;
            // Whenever a sequence of same
            // value is encountered
            while (arr[i] == arr[i + 1]) {
                flag = 1;
                i++;
            }
 
            if (flag)
                clumps++;
        }
 
        // Return the count of clumps
        return clumps;
    }
     
    // Given array
    let arr = [ 13, 15, 66, 66, 66, 37, 37, 8, 8, 11, 11 ];
   
    // length of the given array arr[]
    let N = arr.length;
   
    // Function Call
    document.write(countClumps(arr, N));
 
</script>

4

Complejidad de tiempo: O(N) , donde N es el número de elementos en la array dada. 
Espacio Auxiliar: O(1)