# Different substrings in a string that start and end with given strings

Given a string s and two other strings begin and end, find the number of different substrings in the string which begin and end with the given begin and end strings.

Examples:

```Input : s = "geeksforgeeks"
begin = "geeks"
end = "for"
Output : 1

Input : s = "vishakha"
begin = "h"
end = "a"
Output : 2
Two different sub-strings are "ha" and "hakha".```

Approach: Find all occurrences of string begin and string end. Store the index of each string in two different arrays. After that traverse through the whole string and add one symbol per iteration to already seen sub-strings and map new strings to some non-negative integers. As the ends and beginnings of strings and different strings of equal length are mapped to different numbers (and equal strings are mapped equally), simply count the number of necessary sub-strings of a certain length.

Implementation:

## C++

 `// Cpp program to find number of` `// different sub strings` `#include ` `using` `namespace` `std;`   `// function to return number of different ` `// sub-strings` `int` `numberOfDifferentSubstrings(string s, string a, ` `                                          ``string b)` `{` `    ``// initially our answer is zero.` `    ``int` `ans = 0;`   `    ``// find the length of given strings` `    ``int` `ls = s.size(), la = a.size(), lb = b.size();`   `    ``// currently make array and initially put zero.` `    ``int` `x[ls] = { 0 }, y[ls] = { 0 };`   `    ``// find occurrence of "a" and "b" in string "s"` `    ``for` `(``int` `i = 0; i < ls; i++) {` `        ``if` `(s.substr(i, la) == a)` `            ``x[i] = 1;` `        ``if` `(s.substr(i, lb) == b)` `            ``y[i] = 1;` `    ``}`   `    ``// We use a hash to make sure that same ` `    ``// substring is not counted twice.` `    ``unordered_set hash;    `   `    ``// go through all the positions to find ` `    ``// occurrence of "a" first.` `    ``string curr_substr = ``""``;` `    ``for` `(``int` `i = 0; i < ls; i++) {` `    `  `        ``// if we found occurrence of "a".` `        ``if` `(x[i]) {` `        `  `            ``// then go through all the positions` `            ``// to find occurrence of "b".` `            ``for` `(``int` `j = i; j < ls; j++) {` `            `  `                ``// if we do found "b" at index` `                ``// j then add it to already` `                ``// existed substring.` `                ``if` `(!y[j])` `                    ``curr_substr += s[j];`   `                ``// if we found occurrence of "b".` `                ``if` `(y[j]) {` `                `  `                    ``// now add string "b" to ` `                    ``// already existed substring.` `                    ``curr_substr += s.substr(j, lb);` `                    `  `                    ``// If current substring is not` `                    ``// included already.` `                    ``if` `(hash.find(curr_substr) == hash.end())` `                        ``ans++;`   `                    ``// put any non negative ` `                    ``// integer to make this` `                    ``// string as already ` `                    ``// existed.` `                    ``hash.insert(curr_substr);` `                ``}` `            ``}`   `            ``// make substring null.` `            ``curr_substr = ``""``;` `        ``}` `    ``}`   `    ``// return answer.` `    ``return` `ans;` `}`   `// Driver program for above function.` `int` `main()` `{` `    ``string s = ``"codecppforfood"``;` `    ``string begin = ``"c"``;` `    ``string end = ``"d"``;` `    ``cout << numberOfDifferentSubstrings(s, begin, end) ` `        ``<< endl;` `    ``return` `0;` `}`

## Java

 `// Java program to find number of ` `// different sub strings` `import` `java.util.HashSet;`   `class` `GFG` `{`   `// function to return number of ` `// different sub-strings` `static` `int` `numberOfDifferentSubstrings(String s, ` `                                       ``char` `a, ``char` `b) ` `{`   `    ``// initially our answer is zero.` `    ``int` `ans = ``0``;`   `    ``// find the length of given strings` `    ``int` `ls = s.length();`   `    ``// currently make array and ` `    ``// initially put zero.` `    ``int``[] x = ``new` `int``[ls];` `    ``int``[] y = ``new` `int``[ls];`   `    ``// find occurrence of "a" and "b" ` `    ``// in string "s"` `    ``for` `(``int` `i = ``0``; i < ls; i++)` `    ``{` `        ``if` `(s.charAt(i) == a)` `            ``x[i] = ``1``;` `        ``if` `(s.charAt(i) == b)` `            ``y[i] = ``1``;` `    ``}`   `    ``// We use a hash to make sure that same` `    ``// substring is not counted twice.` `    ``HashSet hash = ``new` `HashSet<>();`   `    ``// go through all the positions to find` `    ``// occurrence of "a" first.` `    ``String curr_substr = ``""``;` `    ``for` `(``int` `i = ``0``; i < ls; i++) ` `    ``{`   `        ``// if we found occurrence of "a".` `        ``if` `(x[i] != ``0``) ` `        ``{`   `            ``// then go through all the positions` `            ``// to find occurrence of "b".` `            ``for` `(``int` `j = i; j < ls; j++) ` `            ``{`   `                ``// if we do found "b" at index` `                ``// j then add it to already` `                ``// existed substring.` `                ``if` `(y[j] == ``0``)` `                    ``curr_substr += s.charAt(i);`   `                ``// if we found occurrence of "b".` `                ``if` `(y[j] != ``0``) ` `                ``{`   `                    ``// now add string "b" to` `                    ``// already existed substring.` `                    ``curr_substr += s.charAt(j);`   `                    ``// If current substring is not` `                    ``// included already.` `                    ``if` `(!hash.contains(curr_substr))` `                        ``ans++;`   `                    ``// put any non negative` `                    ``// integer to make this` `                    ``// string as already` `                    ``// existed.` `                    ``hash.add(curr_substr);` `                ``}` `            ``}`   `            ``// make substring null.` `            ``curr_substr = ``""``;` `        ``}` `    ``}`   `    ``// return answer.` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``String s = ``"codecppforfood"``;` `    ``char` `begin = ``'c'``;` `    ``char` `end = ``'d'``;` `    ``System.out.println(` `           ``numberOfDifferentSubstrings(s, begin, end));` `}` `}`   `// This code is contributed by` `// sanjeev2552`

## Python3

 `# Python 3 program to find number of` `# different sub strings`   `# function to return number of different ` `# sub-strings` `def` `numberOfDifferentSubstrings(s, a, b):`   `    ``# initially our answer is zero.` `    ``ans ``=` `0`   `    ``# find the length of given strings` `    ``ls ``=` `len``(s)` `    ``la ``=` `len``(a)` `    ``lb ``=` `len``(b)`   `    ``# currently make array and initially` `    ``# put zero.` `    ``x ``=` `[``0``] ``*` `ls` `    ``y ``=` `[``0``] ``*` `ls`   `    ``# find occurrence of "a" and "b" in string "s"` `    ``for` `i ``in` `range``(ls):` `        `  `        ``if` `(s[i: la ``+` `i] ``=``=` `a):` `            ``x[i] ``=` `1` `        ``if` `(s[i: lb ``+` `i] ``=``=` `b):` `            ``y[i] ``=` `1`   `    ``# We use a hash to make sure that same ` `    ``# substring is not counted twice.` `    ``hash` `=` `[] `   `    ``# go through all the positions to find ` `    ``# occurrence of "a" first.` `    ``curr_substr ``=` `""` `    ``for` `i ``in` `range``(ls):` `    `  `        ``# if we found occurrence of "a".` `        ``if` `(x[i]):` `        `  `            ``# then go through all the positions` `            ``# to find occurrence of "b".` `            ``for` `j ``in` `range``( i, ls):` `            `  `                ``# if we do found "b" at index` `                ``# j then add it to already` `                ``# existed substring.` `                ``if` `(``not` `y[j]):` `                    ``curr_substr ``+``=` `s[j]`   `                ``# if we found occurrence of "b".` `                ``if` `(y[j]):` `                `  `                    ``# now add string "b" to ` `                    ``# already existed substring.` `                `  `                    ``curr_substr ``+``=` `s[j: lb ``+` `j]` `                    `  `                    ``# If current substring is not` `                    ``# included already.` `                    ``if` `curr_substr ``not` `in` `hash``:` `                        ``ans ``+``=` `1`   `                    ``# put any non negative integer ` `                    ``# to make this string as already ` `                    ``# existed.` `                    ``hash``.append(curr_substr)`   `            ``# make substring null.` `            ``curr_substr ``=` `""`   `    ``# return answer.` `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``s ``=` `"codecppforfood"` `    ``begin ``=` `"c"` `    ``end ``=` `"d"` `    ``print``(numberOfDifferentSubstrings(s, begin, end))`   `# This code is contributed by ita_c`

## C#

 `// C# program to find number of ` `// different sub strings` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `// function to return number of ` `// different sub-strings` `static` `int` `numberOfDifferentSubstrings(String s, ` `                                    ``char` `a, ``char` `b) ` `{`   `    ``// initially our answer is zero.` `    ``int` `ans = 0;`   `    ``// find the length of given strings` `    ``int` `ls = s.Length;`   `    ``// currently make array and ` `    ``// initially put zero.` `    ``int``[] x = ``new` `int``[ls];` `    ``int``[] y = ``new` `int``[ls];`   `    ``// find occurrence of "a" and "b" ` `    ``// in string "s"` `    ``for` `(``int` `i = 0; i < ls; i++)` `    ``{` `        ``if` `(s[i] == a)` `            ``x[i] = 1;` `        ``if` `(s[i] == b)` `            ``y[i] = 1;` `    ``}`   `    ``// We use a hash to make sure that same` `    ``// substring is not counted twice.` `    ``HashSet hash = ``new` `HashSet();`   `    ``// go through all the positions to find` `    ``// occurrence of "a" first.` `    ``String curr_substr = ``""``;` `    ``for` `(``int` `i = 0; i < ls; i++) ` `    ``{`   `        ``// if we found occurrence of "a".` `        ``if` `(x[i] != 0) ` `        ``{`   `            ``// then go through all the positions` `            ``// to find occurrence of "b".` `            ``for` `(``int` `j = i; j < ls; j++) ` `            ``{`   `                ``// if we do found "b" at index` `                ``// j then add it to already` `                ``// existed substring.` `                ``if` `(y[j] == 0)` `                    ``curr_substr += s[i];`   `                ``// if we found occurrence of "b".` `                ``if` `(y[j] != 0) ` `                ``{`   `                    ``// now add string "b" to` `                    ``// already existed substring.` `                    ``curr_substr += s[j];`   `                    ``// If current substring is not` `                    ``// included already.` `                    ``if` `(!hash.Contains(curr_substr))` `                        ``ans++;`   `                    ``// put any non negative` `                    ``// integer to make this` `                    ``// string as already` `                    ``// existed.` `                    ``hash.Add(curr_substr);` `                ``}` `            ``}`   `            ``// make substring null.` `            ``curr_substr = ``""``;` `        ``}` `    ``}`   `    ``// return answer.` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``String s = ``"codecppforfood"``;` `    ``char` `begin = ``'c'``;` `    ``char` `end = ``'d'``;` `    ``Console.WriteLine(` `        ``numberOfDifferentSubstrings(s, begin, end));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`3`

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