Sub-strings of a string that are prefix of the same string
Given a string str, the task is to count all possible sub-strings of the given string that are prefix of the same string.
Examples:
Input: str = “ababc”
Output: 7
All possible sub-string are “a”, “ab”, “aba”, “abab”, “ababc”, “a” and “ab”
Input: str = “abdabc”
Output: 8
Approach: Traverse the string character by character, if the current character is equal to the first character of the string then count all possible sub-strings starting from here that are also the prefixes of str and add it to count. After the complete string has been traversed, print the count.
Below is the implementation of the above approach:
C++14
// C++ implementation of the approach #include <iostream> #include <string> using namespace std; // Function to return the // count of sub-strings starting // from startIndex that are // also the prefixes of str int subStringsStartingHere(string str, int n, int startIndex) { int count = 0, i = 1; while (i <= n) { if (str.substr(0,i) == str.substr(startIndex, i)) { count++; } else break; i++; } return count; } // Function to return the // count of all possible sub-strings // of str that are also the prefixes of str int countSubStrings(string str, int n) { int count = 0; for (int i = 0; i < n; i++) { // If current character is equal to // the starting character of str if (str[i] == str[0]) count += subStringsStartingHere(str, n, i); } return count; } // Driver code int main() { string str = "abcda"; int n = str.length(); // Function Call cout << (countSubStrings(str, n)); } // This code is contributed by harshvijeta0 |
Java
// Java implementation of the approach public class GFG { // Function to return // the count of sub-strings starting // from startIndex that // are also the prefixes of str public static int subStringsStartingHere( String str, int n, int startIndex) { int count = 0, i = startIndex + 1; while (i <= n) { if (str.startsWith(str.substring( startIndex, i))) { count++; } else break; i++; } return count; } // Function to return the // count of all possible sub-strings // of str that are also the prefixes of str public static int countSubStrings(String str, int n) { int count = 0; for (int i = 0; i < n; i++) { // If current character is equal to // the starting character of str if (str.charAt(i) == str.charAt(0)) count += subStringsStartingHere(str, n, i); } return count; } // Driver code public static void main(String[] args) { String str = "ababc"; int n = str.length(); System.out.println(countSubStrings(str, n)); } } |
Python3
# Python3 implementation of the approach # Function to return the # count of sub-strings starting # from startIndex that are # also the prefixes of string def subStringsStartingHere(string, n, startIndex): count = 0 i = startIndex + 1 while(i <= n) : if string.startswith( string[startIndex : i]): count += 1 else : break i += 1 return count # Function to return the # count of all possible sub-strings # of string that are also # the prefixes of string def countSubStrings(string, n) : count = 0 for i in range(n) : # If current character is equal to # the starting character of str if string[i] == string[0] : count += subStringsStartingHere( string, n, i) return count # Driver Code if __name__ == "__main__" : string = "ababc" n = len(string) print(countSubStrings(string, n)) # this code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GFG { // Function to return the // count of sub-strings starting // from startIndex that // are also the prefixes of str static int subStringsStartingHere( String str, int n, int startIndex) { int count = 0, i = startIndex + 1; while (i <= n) { if (str.StartsWith(str.Substring( startIndex, i-startIndex))) { count++; } else break; i++; } return count; } // Function to return the // count of all possible sub-strings // of str that are also the prefixes of str static int countSubStrings(String str, int n) { int count = 0; for (int i = 0; i < n; i++) { // If current character is equal to // the starting character of str if (str[i] == str[0]) count += subStringsStartingHere( str, n, i); } return count; } // Driver code static public void Main(String []args) { String str = "ababc"; int n = str.Length; Console.WriteLine(countSubStrings(str, n)); } } //contributed by Arnab Kundu |
Output
6
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