public class Solution extends Object
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:
sources[i] occurs at index indices[i] in the original string s.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.
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 <= 1000k == indices.length == sources.length == targets.length1 <= k <= 1000 <= indexes[i] < s.length1 <= sources[i].length, targets[i].length <= 50s consists of only lowercase English letters.sources[i] and targets[i] consist of only lowercase English letters.| Constructor and Description |
|---|
Solution() |
| Modifier and Type | Method and Description |
|---|---|
String |
findReplaceString(String s,
int[] indices,
String[] sources,
String[] targets) |
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