Dada una array arr[] , la tarea es contar el número de pares formados por elementos consecutivos en los que ambos elementos de un par son iguales.
Ejemplos:
Entrada: arr[] = {1, 2, 2, 3, 4, 4, 5, 5, 5, 5}
Salida: 5
(1, 2), (4, 5), (6, 7), (7 , 8) y (8, 9) son los pares de índices válidos
donde los elementos consecutivos son iguales.
Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Salida: 0
No hay dos elementos consecutivos iguales en la array dada.
Enfoque: inicialice count = 0 y recorra la array desde arr[0] hasta arr[n – 2] . Si el elemento actual es igual al siguiente elemento en la array, incremente el conteo . Imprime el conteo al final.
A continuación se muestra la implementación del enfoque anterior:
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>
5
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Premdeep Toppo y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA