Number of strings which starts and ends with same character after rotations

Given a string str, the task is to find the number of strings that start and end with the same character after a rotation at every possible index of the given string.

Examples:

Input: str = “GeeksforGeeks”
Output: 2
Explanation:
All possible strings with rotations at every index are: “GeeksforGeeks”, “eeksforGeeksG”, “eksforGeeksGe”, “ksforGeeksGee”, “sforGeeksGeek”, “forGeeksGeeks”, “orGeeksGeeksf”, “rGeeksGeeksfo”, “GeeksGeeksfor”, “eeksGeeksforG”, “eksGeeksforGe”, “ksGeeksforGee”, “sGeeksforGeek”.
Out of the above strings formed only 2 string starts and ends with the same characters: “eksforGeeksGe” and “eksGeeksforGe”.

Input: str = “aaabcdd”
Output: 3
Explanation:
All possible strings with rotations at every index are: “aaabcdd”, “aabcdda”, “abcddaa”, “bcddaaa”, “cddaaab”, “ddaaabc”, “daaabcd”.
Out of the above strings formed only 3 string starts and ends with the same characters: “aabcdda”, “abcddaa” and “daaabcd”.

Naive Approach: The idea is to generate all the possible rotations of the given string and check whether the each string formed after rotation starts and ends with the same character or not. If Yes then include this string in the count. Print the final count.



Efficient Approach: The efficient approach to counting the possible string is to rotate the given string at those indexes which have continuous same characters. Therefore, the final count is the (number of continuos same characters – 1) for each continuous characters in the given string.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above appraoch
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the count of string
// with equal end after rotations
int countStrings(string s)
{
    // To store the final count
    int cnt = 0;
  
    // Traverse the string
    for (int i = 1; s[i]; i++) {
        // If current character is same
        // as the previous character then
        // increment the count
        if (s[i] == s[i + 1]) {
            cnt++;
        }
    }
  
    // Return the final count
    return cnt;
}
  
// Driver Code
int main()
{
    // Given string
    string str("aacbb");
  
    // Function Call
    cout << countStrings(str);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
class GFG{
  
// Function to find the count of string
// with equal end after rotations
static int countStrings(String s)
{
      
    // To store the final count
    int cnt = 0;
  
    // Traverse the string
    for(int i = 1; i < s.length() - 1; i++)
    {
          
       // If current character is same
       // as the previous character then
       // increment the count
       if (s.charAt(i) == s.charAt(i + 1))
       {
           cnt++;
       }
    }
  
    // Return the final count
    return cnt;
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given string
    String str = "aacbb";
      
    // Function call
    System.out.println(countStrings(str));
}
}
  
// This code is contributed by rutvik_56

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
  
# Function to find the count of string
# with equal end after rotations
def countStrings(s):
  
    # To store the final count
    cnt = 0;
  
    # Traverse the string
    for i in range(1, len(s) - 1):
  
        # If current character is same
        # as the previous character then
        # increment the count
        if (s[i] == s[i + 1]):
            cnt += 1;
              
    # Return the final count
    return cnt;
  
# Driver Code
if __name__ == '__main__':
  
    # Given string
    str = "aacbb";
  
    # Function call
    print(countStrings(str));
  
# This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the count of string
// with equal end after rotations
static int countStrings(String s)
{
      
    // To store the final count
    int cnt = 0;
  
    // Traverse the string
    for(int i = 1; i < s.Length - 1; i++)
    {
          
       // If current character is same
       // as the previous character then
       // increment the count
       if (s[i] == s[i + 1])
       {
           cnt++;
       }
    }
      
    // Return the final count
    return cnt;
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given string
    String str = "aacbb";
      
    // Function call
    Console.WriteLine(countStrings(str));
}
}
  
// This code is contributed by sapnasingh4991

chevron_right


Output:

1

Time Complexity: O(N), where N is the length of the given string.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.