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

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.
```

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

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 `

Output:

`Yes`

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Improved By : 29AjayKumar, sanjeev2552

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