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Java-based LeetCode algorithm problem solutions, regularly updated
833\. Find And Replace in String
Medium
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:**
![](https://assets.leetcode.com/uploads/2021/06/12/833-ex1.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:**
![](https://assets.leetcode.com/uploads/2021/06/12/833-ex2-1.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.