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.

Approach #1: String. [Java]

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));
                i++;
            }
        }
        return sb.toString();
    }
}

　　

Analysis:

Idea: Use a StringBuilder to build up the result.

1. Iterate through the indexes array and find out all indices that support replacement. Then, store mapping from those index values to their indices in the indexes array into a map named map.

2. Iterate through str, at each iteration i, check whether we can perform replacement, i.e., map.get(i) != null, if yes, append targets[i] to the StringBuilder and increase i by sources[i].length()-1. if no, append str.charAt(i) and increment i.

Reference:

https://leetcode.com/problems/find-and-replace-in-string/discuss/134758/Java-O(n)-solution