# Check if a string can become empty by recursively deleting a given sub-string

• Difficulty Level : Easy
• Last Updated : 19 Jun, 2022

Given a string “str” and another string “sub_str”. We are allowed to delete “sub_str” from “str” any number of times. It is also given that the “sub_str” appears only once at a time. The task is to find if “str” can become empty by removing “sub_str” again and again.
Examples:

```Input  : str = "GEEGEEKSKS", sub_str = "GEEKS"
Output : Yes
Explanation : In the string GEEGEEKSKS, we can first
delete the substring GEEKS from position 4.
The new string now becomes GEEKS. We can
again delete sub-string GEEKS from position 1.
Now the string becomes empty.

Input  : str = "GEEGEEKSSGEK", sub_str = "GEEKS"
Output : No
Explanation : In the string it is not possible to make the
string empty in any possible manner.```

Method#1: A simple solution to solve this problem is by using inbuilt string functions find() and erase(). First input the sub-string substr for searching purpose in the original string str, then iterate the original string to find the index of sub-string using find() which return starting index of the sub-string in the original string else -1 if not found and erase that sub-string using erase() until length of original string is greater than 0.
The above simple solutions works because the given substring appears only once at a time.

## C++

 `// C++ Program to check if a string can be``// converted to an empty string by deleting``// given sub-string from any position, any``// number of times.``#include``using` `namespace` `std;` `// Returns true if str can be made empty by``// recursively removing sub_str.``bool` `canBecomeEmpty(string str, string sub_str)``{``    ``while` `(str.size() > 0)``    ``{``        ``// idx: to store starting index of sub-``        ``//      string found in the original string``        ``int` `idx = str.find(sub_str);``        ``if` `(idx == -1)``            ``break``;` `        ``// Erasing the found sub-string from``        ``// the original string``        ``str.erase(idx, sub_str.size());``    ``}` `    ``return` `(str.size() == 0);``}` `// Driver code``int` `main()``{``    ``string str = ``"GEEGEEKSKS"``, sub_str = ``"GEEKS"``;``    ``if` `(canBecomeEmpty(str, sub_str))``        ``cout<<``"\nYes"``;``    ``else``        ``cout<<``"\nNo"``;``    ``return` `0;``}`

## Java

 `//Java program to check if a string can be``// converted to an empty string by deleting``// given sub-string from any position, any``// number of times.` `class` `GFG {` `// Returns true if str can be made empty by``// recursively removing sub_str.``    ``static` `boolean` `canBecomeEmpty(String str, String sub_str) {``        ``while` `(str.length() > ``0``) {``            ``// idx: to store starting index of sub-``            ``//      string found in the original string``            ``int` `idx = str.indexOf(sub_str);``            ``if` `(idx == -``1``) {``                ``break``;``            ``}` `            ``// Erasing the found sub-string from``            ``// the original string``            ``str = str.replaceFirst(sub_str,``""``);``        ``}` `        ``return` `(str.length() == ``0``);``    ``}` `// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``String str = ``"GEEGEEKSKS"``, sub_str = ``"GEEKS"``;``        ``if` `(canBecomeEmpty(str, sub_str)) {``            ``System.out.print(``"\nYes"``);``        ``} ``else` `{``            ``System.out.print(``"\nNo"``);``        ``}``    ``}``}``// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to check if a string can be``# converted to an empty string by deleting``# given sub-string from any position, any``# number of times.` `# Returns true if str can be made empty by``# recursively removing sub_str.``def` `canBecomeEmpty(string, sub_str):``    ``while` `len``(string) > ``0``:` `        ``# idx: to store starting index of sub-``        ``#     string found in the original string``        ``idx ``=` `string.find(sub_str)` `        ``if` `idx ``=``=` `-``1``:``            ``break` `        ``# Erasing the found sub-string from``        ``# the original string``        ``string ``=` `string.replace(sub_str, "", ``1``)` `    ``return` `(``len``(string) ``=``=` `0``)` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``string ``=` `"GEEGEEKSKS"``    ``sub_str ``=` `"GEEKS"``    ``if` `canBecomeEmpty(string, sub_str):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to check if a string can be``// converted to an empty string by deleting``// given sub-string from any position, any``// number of times.``using` `System;``    ` `class` `GFG``{` `    ``// Returns true if str can be made empty by``    ``// recursively removing sub_str.``    ``static` `Boolean canBecomeEmpty(String str, String sub_str)``    ``{``        ``while` `(str.Length > 0)``        ``{``            ``// idx: to store starting index of sub-``            ``//     string found in the original string``            ``int` `idx = str.IndexOf(sub_str);``            ``if` `(idx == -1)``            ``{``                ``break``;``            ``}` `            ``// Erasing the found sub-string from``            ``// the original string``            ``str = str.Replace(sub_str,``""``);``        ``}` `        ``return` `(str.Length == 0);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String str = ``"GEEGEEKSKS"``, sub_str = ``"GEEKS"``;``        ``if` `(canBecomeEmpty(str, sub_str))``        ``{``            ``Console.Write(``"\nYes"``);``        ``}``        ``else``        ``{``            ``Console.Write(``"\nNo"``);``        ``}``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

`Yes`

Method#2: One possible solution would be to use the regular expression. Regular expressions module are very helpful with string. Regular expression module has search function which is used to find the patterns in string. And replace function is used for replacing the characters of string. Follow are steps for our approach:

• Use the re.search function on string to find the pattern in string.
• Use the re.replace function to replace the sub-string from string.
• Repeat step one and two until there is sub-string in string after replacing it.
• At last if string is empty after all replacing then it can become empty else it cannot become empty.

## C++

 `// C++ Program to check if a string can be``// converted to an empty string by deleting``// given sub-string from any position, any``// number of times.``#include``using` `namespace` `std;`  `// Returns true if str can be made empty by``// recursively removing sub_str.``bool` `canBecomeEmpty(string str1 , string sub_str){``    ` `    ``regex r(sub_str);``    ``smatch m ;``    ``// matches words beginning by "Geek"``    ` `    ``while``(regex_search(str1, m, r)){``//     regex_replace() for replacing the match with 'geek'``   ` `    ``str1 = regex_replace(str1, r, ``""``);``    ``}` `// Returing result``    ``return` `true` `? str1==``""` `: ``false` `;``    ` `}` `// Driver code``int` `main()``{``    ``string s = ``"GeeksForGeeGeeksks"``;``    ``string k = ``"Geeks"``;``    ``if` `(canBecomeEmpty(s, k)){``        ``cout << ``"\nYes"` `;``    ``}``    ``else` `{``        ``cout << ``"\nNo"``;``        ``}``        ` `    ``return` `0;``}`

## Python3

 `# Python3 program to check if a string can be``# converted to an empty string by deleting``# given sub-string from any position, any``# number of times.`` ` `# Returns true if str can be made empty by``# recursively removing sub_str.``import` `re`  `def` `canBecomeEmpty(string, sub_str):``    ``# finding sub-string in string``    ``while` `sub_str ``in` `string :``        ``# Replacing sub-string from string``        ``string ``=` `re.sub(sub_str, "", string)``    ``# Returning result``    ``return` `True` `if` `string ``=``=` `"" ``else` `False``    ` `    ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``string ``=` `"GeeksforGeeGeeksks"``    ``sub_str ``=` `"Geeks"``    ``if` `canBecomeEmpty(string, sub_str):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)``        ` ` ` `# This code is contributed by``# sanjeev2552`

## Javascript

 `// Javascript program to check if a string can be``// converted to an empty string by deleting``// given sub-string from any position, any``// number of times.` `// Returns true if str can be made empty by``// recursively removing sub_str``function` `canBecomeEmpty(str,sub_str)``{  ``    ``// Regular expression for sub_string``    ``const regex = ``new` `RegExp(sub_str);``    ``// Iterating over string untill it matches sub_string``    ``while``(regex.test(str)){``        ``// Replacing sub-string from string``    ``str = str.replace(regex, ``''``)``        ` `    ``}``    ``// Returning result``        ``return` `false` `? str : ``true` `;``}` `// Driver code``let str = ``"GeeksForGeeGeeksks"``, sub_str = ``"Geeks"``;``if` `(canBecomeEmpty(str, sub_str)) {``    ``console.log(``"Yes"``);``} ``else` `{``    ``console.log(``"No"``);``}`  `// This code is contributed by sam snehil`

Output

`No`

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