1910. Remove All Occurrences of a Substring

Problem Link

Description

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Given two strings s and part, perform the following operation on s until all occurrences of the substring part are removed:

• Find the leftmost occurrence of the substring part and remove it from s.

Return s after removing all occurrences of part.

A substring is a contiguous sequence of characters in a string.

 

Example 1:

Input: s = "daabcbaabcbc", part = "abc"
Output: "dab"
Explanation: The following operations are done:
- s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc".
- s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc".
- s = "dababc", remove "abc" starting at index 3, so s = "dab".
Now s has no occurrences of "abc".

Example 2:

Input: s = "axxxxyyyyb", part = "xy"
Output: "ab"
Explanation: The following operations are done:
- s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb".
- s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb".
- s = "axxyyb", remove "xy" starting at index 2 so s = "axyb".
- s = "axyb", remove "xy" starting at index 1 so s = "ab".
Now s has no occurrences of "xy".

 

Constraints:

• 1 <= s.length <= 1000
• 1 <= part.length <= 1000
• s​​​​​​ and part consists of lowercase English letters.

Solution

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Python3

class Solution:
    def removeOccurrences(self, s: str, part: str) -> str:
        queue = collections.deque()
        pn = len(part)
        
        def check():
            i = -1
            for _ in range(pn):
                if queue[i] != part[i]: return False
                i -= 1
            
            return True
        
        for x in s:
            queue.append(x)
            
            while len(queue) >= pn and check():
                for _ in range(pn):
                    queue.pop()
            
        return "".join(queue)