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.



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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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<bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

//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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right



Output:

Yes

This article is contributed by Himanshu Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : 29AjayKumar, sanjeev2552



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.