833. Find And Replace in String

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.

1. 首先遍历S，根据indexes，找S中对应的字符串是否为sources中对应字符串，如果是，则存入hash table中

2. 遍历hash table，用StringBuilder构建最后输出

time = O(N + M * L),　　N = S.length(), M = # of possible replaces, L = avg length of targets

space = O(N)

class Solution {
    public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < indexes.length; i++) {
            if(S.startsWith(sources[i], indexes[i])) {
                map.put(indexes[i], i);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < S.length(); ) {
            if(map.containsKey(i)) {
                sb.append(targets[map.get(i)]);
                i += sources[map.get(i)].length();
            } else {
                sb.append(S.charAt(i++));
            }
        }
        return sb.toString();
    }
}