## Saturday, May 26, 2018

### LeetCode 833 - Find And Replace in String

https://leetcode.com/problems/find-and-replace-in-string/description/
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y.  The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y.  If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously.  It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".

Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.

Notes:
1. 0 <= indexes.length = sources.length = targets.length <= 100
2. 0 < indexes[i] < S.length <= 1000
3. All characters in given inputs are lowercase letters.

• Time Complexity: $O(NQ)$, where $N$ is the length of S, and we have $Q$ replacement operations. (Our complexity could be faster with a more accurate implementation, but it isn't necessary.)
• Space Complexity: $O(N)$, if we consider targets[i].length <= 100 as a constant bound.

public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
int N = S.length();
int[] match = new int[N];
Arrays.fill(match, -1);

for (int i = 0; i < indexes.length; ++i) {
int ix = indexes[i];
if (S.substring(ix, ix + sources[i].length()).equals(sources[i]))
match[ix] = i;
}

StringBuilder ans = new StringBuilder();
int ix = 0;
while (ix < N) {
if (match[ix] >= 0) {
ans.append(targets[match[ix]]);
ix += sources[match[ix]].length();
} else {
ans.append(S.charAt(ix++));
}
}
return ans.toString();
}