LeetCode 833 Find And Replace in String

LeetCode

You are given a 0-indexed string s that you must perform k replacement operations on. The replacement operations are given as three 0-indexed parallel arrays, indices, sources, and targets, all of length k.

To complete the ith replacement operation:

1. Check if the substring sources[i] occurs at index indices[i] in the original string s.
2. If it does not occur, do nothing.
3. Otherwise if it does occur, replace that substring with targets[i].

For example, if s = "abcd", indices[i] = 0, sources[i] = "ab", and targets[i] = "eee", then the result of this replacement will be "eeecd".

All replacement operations must occur simultaneously, meaning the replacement operations should not affect the indexing of each other. The testcases will be generated such that the replacements will not overlap.

For example, a testcase with s = "abc", indices = [0, 1], and sources = ["ab","bc"] will not be generated because the "ab" and "bc" replacements overlap.

Return the resulting string after performing all replacement operations on s.

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

Example 1:

Input: s = “abcd”, indices = [0, 2], sources = [“a”, “cd”], targets = [“eee”, “ffff”]
Output: “eeebffff”
Explanation:
“a” occurs at index 0 in s, so we replace it with “eee”.
“cd” occurs at index 2 in s, so we replace it with “ffff”.

Example 2:

Input: s = “abcd”, indices = [0, 2], sources = [“ab”,”ec”], targets = [“eee”,”ffff”]
Output: “eeecd”
Explanation:
“ab” occurs at index 0 in s, so we replace it with “eee”.
“ec” does not occur at index 2 in s, so we do nothing.

Constraints: 1 <= s.length <= 1000
k == indices.length == sources.length == targets.length 1 <= k <= 100
0 <= indexes[i] < s.length 1 <= sources[i].length, targets[i].length <= 50
s consists of only lowercase English letters. sources[i] and targets[i] consist of only lowercase English letters.

题目大意：

整洁题。找到位置，然后验证，最后替换

解题思路：

N/A

解题步骤：

N/A

注意事项：

i是循环外的变量，所以poplate index_dict注意不能重名

Python代码：

def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
	res = ''
	index_dict = {}
	for _i, _n in enumerate(indices):
		index_dict[_n] = _i # 0 -> 0, 2 -> 1
	i = 0
	while i < len(s):
		if i in index_dict and s[i:i + len(sources[index_dict[i]])] == sources[index_dict[i]]:
			res += targets[index_dict[i]]
			i += len(sources[index_dict[i]])
		else:
			res += s[i]
			i += 1
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)