Stamping The Sequence - LeetCode solutions

Problem

You are given two strings stamp and target. Initially, there is a string s of length target.length with all s[i] == '?'.

In one turn, you can place stamp over s and replace every letter in the s with the corresponding letter from stamp.

For example, if stamp = "abc" and target = "abcba", then s is "?????" initially. In one turn you can:

place stamp at index 0 of s to obtain "abc??",
place stamp at index 1 of s to obtain "?abc?", or
place stamp at index 2 of s to obtain "??abc".

Note that stamp must be fully contained in the boundaries of s in order to stamp (i.e., you cannot place stamp at index 3 of s).

We want to convert s to target using at most 10 * target.length turns.

Return an array of the index of the left-most letter being stamped at each turn. If we cannot obtain target from s within 10 * target.length turns, return an empty array.

Example 1:

Input: stamp = "abc", target = "ababc"
Output: [0,2]
Explanation: Initially s = "?????".
- Place stamp at index 0 to get "abc??".
- Place stamp at index 2 to get "ababc".
[1,0,2] would also be accepted as an answer, as well as some other answers.

Example 2:

Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Explanation: Initially s = "???????".
- Place stamp at index 3 to get "???abca".
- Place stamp at index 0 to get "abcabca".
- Place stamp at index 1 to get "aabcaca".

Constraints:

1 <= stamp.length <= target.length <= 1000
stamp and target consist of lowercase English letters.

Solution

class Solution {
    public int[] movesToStamp(String stamp, String target) {
        List<Integer> moves = new ArrayList<>();
        char[] s = stamp.toCharArray();
        char[] t = target.toCharArray();
        int stars = 0;
        boolean[] visited = new boolean[target.length()];
        while (stars < target.length()) {
            boolean doneReplace = false;
            for (int i = 0; i <= target.length() - stamp.length(); i++) {
                if (!visited[i] && canReplace(t, i, s)) {
                    stars = doReplace(t, i, s, stars);
                    doneReplace = true;
                    visited[i] = true;
                    moves.add(i);
                    if (stars == t.length) {
                        break;
                    }
                }
            }
            if (!doneReplace) {
                return new int[0];
            }
        }

        int[] result = new int[moves.size()];
        for (int i = 0; i < moves.size(); i++) {
            result[i] = moves.get(moves.size() - i - 1);
        }
        return result;
    }

    private boolean canReplace(char[] t, int i, char[] s) {
        for (int j = 0; j < s.length; j++) {
            if (t[i + j] != '*' && t[i + j] != s[j]) {
                return false;
            }
        }
        return true;
    }

    private int doReplace(char[] t, int i, char[] s, int stars) {
        for (int j = 0; j < s.length; j++) {
            if (t[i + j] != '*') {
                t[i + j] = '*';
                stars++;
            }
        }
        return stars;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).