Count characters with same neighbors in C++

We are given a string with, let’s say, str and the task is to compute the count of characters in a string str having the same neighbors and that will include both left and right side of a character in a string. Also, in this scenario the first and last character in a string will always be considered as they have only one adjacent character.

For Example

Input − string str = “poiot”
Output − count is 3

Explanation − In the given string characters p, t and i are having the same neighbors so the count will be increased to 3.

Input − string str = “nitihig”
Output − count is 4

Explanation − In the given string characters n, t, h and g are having the same neighbors so the count will be increased to 4.

Approach used in the below program is as follows

Input the string in a variable let’s say str
Calculate the length of string str using length() function that will return an integer value as per the number of letters in the string including the spaces.
If the length of the string is less than or equals 2 then return the count as the length of the string because they all will be counted.
If the length of the string is greater than 2, initialise the count with 2.
Now, start the loop with i to 1 till i is less than (length-1)
Inside the loop, check if (str[i-1] = str[i+1]) then increase the count by 1
Now return the total value of count
Print the result.

Example

Live Demo

#include <iostream>
using namespace std;
// To count the characters
int countChar(string st){
   int size = st.length();
   if (size <= 2){
      return size;
   }
   int result = 2;
   // Traverse the string
   for (int i = 1; i < size - 1; i++){
      // Increment the count by 1 if the previous
      // and next character is same
      if (st[i - 1] == st[i + 1]){
         result++;
      }
   }
   // Return result
   return result;
}
int main(){
   string st = "poiot";
   cout <<"count is " <<countChar(st);
   return 0;
}