# 833. Find And Replace in String - LeetCode Best Practices Visit original link: [833. Find And Replace in String - LeetCode Best Practices](https://leetcoder.net/en/leetcode/833-find-and-replace-in-string) for a better experience! LeetCode link: [833. Find And Replace in String](https://leetcode.com/problems/find-and-replace-in-string), difficulty: **Medium**. ## LeetCode description of "833. Find And Replace in String" 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] ![](../../images/examples/833_1.png) **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] ![](../../images/examples/833_2.png) **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. ## Intuition This question looks simple, but it takes a lot of time to do it. - Question 1: For the target string `result`, you can clone it based on the original string or build it from an empty string. Which one is better?
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Cloning based on the original string is better. Because you save a lot of substring assignment operations.

- Question 2: After replacing the substring of `result` with `targets[i]`, the length of `result` may change, which makes subsequent replacement difficult. How to solve it?
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Use technical means to keep the length of `result` unchanged after string replacement.

## Complexity > N = s.length - Time complexity: `O(N)`. - Space complexity: `O(N)`. ## Python ```python class Solution: def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str: result = list(s) for i in range(len(indices)): index = indices[i] if s[index:index + len(sources[i])] == sources[i]: for j in range(index, index + len(sources[i])): if j == index: result[j] = targets[i] else: result[j] = '' return ''.join(result) ``` ## Other languages ```java // Welcome to create a PR to complete the code of this language, thanks! ``` Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website [leetcoder.net](https://leetcoder.net): Dare to claim the best practices of LeetCode solutions! Will save you a lot of time! Original link: [833. Find And Replace in String - LeetCode Best Practices](https://leetcoder.net/en/leetcode/833-find-and-replace-in-string). GitHub repository: [f*ck-leetcode](https://github.com/fuck-leetcode/fuck-leetcode).