# Sub-strings of a string that are prefix of the same string

• Difficulty Level : Hard
• Last Updated : 22 Apr, 2022

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:
All possible sub-string are “a”, “ab”, “aba”, “abab”, “ababc”, “a” and “ab”

Input: str = “abdabc”
Output:

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 ``#include ``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)``            ``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)``                ``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`

## Javascript

 ``

Output

`6`

Time Complexity: O(N^2)
Auxiliary Space: O(1)

Efficient Approach:

Prerequisite: Z-Algorithm

Approach: Calculate z-array of the string such that  z[i] stores length of the longest substring starting from i which is also a prefix of string s. Then  to count all possible sub-strings of the string that are prefix of the same string, we just need to add all the values of the z-array since the total number of substrings matching would be equal to the length of longest substring.

Implementation of the above approach:-

## C++

 `#include ``using` `namespace` `std;` `// returns an array z such that  z[i]``// stores length of the longest substring starting``// from i which is also a prefix of string s``vector<``int``> z_function(string s)``{``    ``int` `n = (``int``)s.length();``    ``vector<``int``> z(n);``    ``// consider a window [l,r]``    ``// which matches with prefix of s``    ``int` `l = 0, r = 0;``    ``z = n;``    ``for` `(``int` `i = 1; i < n; ++i) {``        ``// when i<=r, we make use of already computed z``        ``// value for some smaller index``        ``if` `(i <= r)``            ``z[i] = min(r - i + 1, z[i - l]);` `        ``// if i>r nothing matches so we will calculate``        ``// z[i] using naive way.``        ``while` `(i + z[i] < n && s[z[i]] == s[i + z[i]])``            ``++z[i];``        ``// update window size``        ``if` `(i + z[i] - 1 > r)``            ``l = i, r = i + z[i] - 1;``    ``}``    ``return` `z;``}` `int` `main()``{``    ``string s = ``"abcda"``;` `    ``int` `n = s.length();` `    ``vector<``int``> z = z_function(s);` `    ``// stores the count of``    ``// Sub-strings of a string that``    ``// are prefix of the same string``    ``int` `count = 0;` `    ``for` `(``auto` `x : z)``        ``count += x;` `    ``cout << count << ``'\n'``;` `    ``return` `0;``}`

## Python3

 `# returns an array z such that  z[i]``# stores length of the longest substring starting``# from i which is also a prefix of s``def` `z_function(s):``    ``n ``=` `len``(s)``    ``z``=``[``0``]``*``n``    ``# consider a window [l,r]``    ``# which matches with prefix of s``    ``l ``=` `0``; r ``=` `0``    ``z[``0``] ``=` `n``    ``for` `i ``in` `range``(``1``, n) :``        ``# when i<=r, we make use of already computed z``        ``# value for some smaller index``        ``if` `(i <``=` `r):``            ``z[i] ``=` `min``(r ``-` `i ``+` `1``, z[i ``-` `l])` `        ``# if i>r nothing matches so we will calculate``        ``# z[i] using naive way.``        ``while` `(i ``+` `z[i] < n ``and` `s[z[i]] ``=``=` `s[i ``+` `z[i]]):``            ``z[i]``+``=``1``        ``# update window size``        ``if` `(i ``+` `z[i] ``-` `1` `> r):``            ``l ``=` `i; r ``=` `i ``+` `z[i] ``-` `1``    ` `    ``return` `z`  `if` `__name__ ``=``=` `'__main__'``:``    ``s ``=` `"abcda"` `    ``n ``=` `len``(s)` `    ``z ``=` `z_function(s)` `    ``# stores the count of``    ``# Sub-strings of a that``    ``# are prefix of the same string``    ``count ``=` `0` `    ``for` `x ``in` `z:``        ``count ``+``=` `x` `    ``print``(count)`

## Javascript

 ``

Output

`6`

Time Complexity: O(n)

Auxiliary Space: O(n)

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