# 833. Find And Replace in String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions > 🚀 **Level Up Your Developer Identity** > > While mastering algorithms is key, showcasing your talent is what gets you hired. > > We recommend [**leader.me**](https://www.leader.me) — the ultimate all-in-one personal branding platform for programmers. > > **The All-In-One Career Powerhouse:** > - 📄 **Resume, Portfolio & Blog:** Integrate your skills, GitHub projects, and writing into one stunning site. > - 🌐 **Free Custom Domain:** Bind your own personal domain for free—forever. > - ✨ **Premium Subdomains:** Stand out with elite tech handle like `name.leader.me`. > > [**Build Your Programmer Brand at leader.me →**](https://www.leader.me) --- Visit original link: [833. Find And Replace in String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.blog/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?
Click to view the answer

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?
Click to view the answer

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: # Each item of this array is a string, not a char! We won't change the size of this array. 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) ``` ## Ruby ```ruby # @param {String} s # @param {Integer[]} indices # @param {String[]} sources # @param {String[]} targets # @return {String} def find_replace_string(s, indices, sources, targets) # Each item of this array is a string, not a char! We won't change the size of this array. result = s.clone.chars indices.each_with_index do |index, i| if s[index...index + sources[i].size] == sources[i] result[index] = targets[i] # The first one keep the whole targets[i] (1...sources[i].size).each do |j| result[index + j] = "" end end end result.join("") end ``` ## Other languages ```java // Welcome to create a PR to complete the code of this language, thanks! ``` > 🚀 **Level Up Your Developer Identity** > > While mastering algorithms is key, showcasing your talent is what gets you hired. > > We recommend [**leader.me**](https://www.leader.me) — the ultimate all-in-one personal branding platform for programmers. > > **The All-In-One Career Powerhouse:** > - 📄 **Resume, Portfolio & Blog:** Integrate your skills, GitHub projects, and writing into one stunning site. > - 🌐 **Free Custom Domain:** Bind your own personal domain for free—forever. > - ✨ **Premium Subdomains:** Stand out with elite tech handle like `name.leader.me`. > > [**Build Your Programmer Brand at leader.me →**](https://www.leader.me) --- Visit original link: [833. Find And Replace in String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.blog/en/leetcode/833-find-and-replace-in-string) for a better experience! GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).