# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `int` `subStringsStartingHere(string str, ``int` `n, ``int` `index); ` ` `  `// 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; ` `} ` ` `  `// 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 = startIndex ; ` `    ``while` `(i <= n) ` `    ``{ ` `        ``if` `(str.substr(0,i) == str.substr(startIndex, i)) ` `        ``{ ` `            ``count++; ` `        ``} ` `        ``else` `            ``break``; ` `        ``i++; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  ` `  `// Driver code ` `int` `main()  ` `{ ` `    ``string str = ``"ababc"``; ` `    ``int` `n = str.length(); ` `    ``cout << (countSubStrings(str, n)); ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## 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:

```7
```

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