833. Find And Replace in String

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.

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Solution #1

因为参数里的index是乱序的，所以排序之后从前往后替换

O(Q*log Q + N)

class Solution {
    public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        int n = indexes.length;
        StringBuilder sb = new StringBuilder();
        
        List<int[]> bucket = new ArrayList<>();
        for (int i = 0; i < n; i++) bucket.add(new int[]{ indexes[i], i });
        Collections.sort(bucket, (a, b) -> a[0] - b[0]);
        int lastIndex = 0;
        
        for (int[] pair : bucket) {
            int index = pair[0];
            int i = pair[1];
            String sub = S.substring(index, index + sources[i].length());
            if (sub.equals(sources[i])) {
                sb.append(S.substring(lastIndex, index)).append(targets[i]);
                lastIndex = index + sources[i].length();
            }
        }
        
        sb.append(S.substring(lastIndex));
        return sb.toString();
    }
}

Solution #2, 用bucket sort
O(N) 时间空间

class Solution {
    public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        int n = indexes.length;
        StringBuilder sb = new StringBuilder();
        
        int[] bucket = new int[S.length()];
        for (int i = 0; i < S.length(); i++) bucket[i] = -1;
        for (int i = 0; i < n; i++) {
            bucket[indexes[i]] = i;
        }
        
        int lastIndex = 0;        
        for (int index = 0; index < S.length(); index++) {
            if (bucket[index] == -1) continue;
            int i = bucket[index];
            String sub = S.substring(index, index + sources[i].length());
            if (sub.equals(sources[i])) {
                sb.append(S.substring(lastIndex, index)).append(targets[i]);
                lastIndex = index + sources[i].length();
            }
        }
        
        sb.append(S.substring(lastIndex));
        return sb.toString();
    }
}