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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0901_1000.s0936_stamping_the_sequence;

// #Hard #String #Greedy #Stack #Queue #2022_03_30_Time_6_ms_(78.43%)_Space_45.6_MB_(27.45%)

import java.util.ArrayList;
import java.util.List;

/**
 * 936 - Stamping The Sequence\.
 *
 * Hard
 *
 * 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.
**/
public class Solution {
    public int[] movesToStamp(String stamp, String target) {
        List 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;
    }
}