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++ 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 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 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# 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 |
<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> |
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