1910. Remove All Occurrences of a Substring

Problem

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.

https://leetcode.com/problems/remove-all-occurrences-of-a-substring/

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.

Test Cases

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class Solution:
    def removeOccurrences(self, s: str, part: str) -> str:

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import pytest

from solution import Solution


@pytest.mark.parametrize('s, part, expected', [
    ("daabcbaabcbc", "abc", "dab"),
    ("axxxxyyyyb", "xy", "ab"),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, s, part, expected):
    assert sol.removeOccurrences(s, part) == expected

Thoughts

可以看作是 3174. Clear Digits 的进阶版，待匹配的 pattern 从两个字符（一个字母+一个数字）变成任意长度的字符串。

处理方法也类似，用一个栈记录结果字符串的字符。遍历 s，对于每个字符，先入栈，然后对比栈顶 m 个字符是否与 part 一致（设 part 的长度为 m），一致就把这些字符都弹出。

时间复杂度 O(n * m)，空间复杂度 O(n)。

Code

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class Solution:
    def removeOccurrences(self, s: str, part: str) -> str:
        stack: list[str] = []
        m = len(part)
        for c in s:
            stack.append(c)
            if len(stack) < m:
                continue

            for i in range(-1, -m - 1, -1):
                if stack[i] != part[i]:
                    break
            else:
                for _ in range(m):
                    stack.pop()

        return ''.join(stack)

参考资料

• 本文引用

  • 3174. Clear Digits

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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