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 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 continuous same characters – 1) for each continuous character in the given string.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach

#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 = 0; 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("aaa");

    // Function Call
    cout << countStrings(str);
    return 0;
}

Java

// 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 = 0; 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

Python3

# 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(0, 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

C#

// 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 = 0; 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

Javascript

<script>

// javascript program for the above approach


// Function to find the count of string
// with equal end after rotations
function countStrings( s)
{
    // To store the final count
    let cnt = 0;

    // Traverse the string
    for (let i = 0; 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

    // Given string
    let str = "aacbb";

    // Function Call
        document.write(countStrings(str));


// This code contributed by gauravrajput1

</script>

Output

2

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1) as constant space for variables is being used