String slicing in Python to check if a string can become empty by recursive deletion
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.
We have existing solution for this problem please refer Check if a string can become empty by recursively deleting a given sub-string link. We will solve this problem in python using String Slicing. Approach is very simple,
- Use find() method of string to search given pattern sub-string.
- If sub-string lies in main string then find function will return index of it’s first occurrence.
- Now slice string in two parts, (i) from start of string till index-1 of founded sub-string, (ii) (start from first index of founded sub-string + length of sub-string) till end of string.
- Concatenate these two sliced part and repeat from step 1 until original string becomes empty or we don’t find sub-string anymore.
def checkEmpty(input, pattern): # If both are empty if len(input)== 0 and len(pattern)== 0: return 'true' # If only pattern is empty if len(pattern)== 0: return 'true' while (len(input) != 0): # find sub-string in main string index = input.find(pattern) # check if sub-string founded or not if (index ==(-1)): return 'false' # slice input string in two parts and concatenate input = input[0:index] + input[index + len(pattern):] return 'true' # Driver program if __name__ == "__main__": input ='GEEGEEKSKS' pattern ='GEEKS' print (checkEmpty(input, pattern)) |
Output:
true
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