833. Find And Replace in String - Medium

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();
    }
}