Stamping The Sequence
<p>You are given two strings <code>stamp</code> and <code>target</code>. Initially, there is a string <code>s</code> of length <code>target.length</code> with all <code>s[i] == &#39;?&#39;</code>.</p>

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

<ul>
	<li>For example, if <code>stamp = &quot;abc&quot;</code> and <code>target = &quot;abcba&quot;</code>, then <code>s</code> is <code>&quot;?????&quot;</code> initially. In one turn you can:

    <ul>
    	<li>place <code>stamp</code> at index <code>0</code> of <code>s</code> to obtain <code>&quot;abc??&quot;</code>,</li>
    	<li>place <code>stamp</code> at index <code>1</code> of <code>s</code> to obtain <code>&quot;?abc?&quot;</code>, or</li>
    	<li>place <code>stamp</code> at index <code>2</code> of <code>s</code> to obtain <code>&quot;??abc&quot;</code>.</li>
    </ul>
    Note that <code>stamp</code> must be fully contained in the boundaries of <code>s</code> in order to stamp (i.e., you cannot place <code>stamp</code> at index <code>3</code> of <code>s</code>).</li>

</ul>

<p>We want to convert <code>s</code> to <code>target</code> using <strong>at most</strong> <code>10 * target.length</code> turns.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stamp = &quot;abca&quot;, target = &quot;aabcaca&quot;
<strong>Output:</strong> [3,0,1]
<strong>Explanation:</strong> Initially s = &quot;???????&quot;.
- Place stamp at index 3 to get &quot;???abca&quot;.
- Place stamp at index 0 to get &quot;abcabca&quot;.
- Place stamp at index 1 to get &quot;aabcaca&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= stamp.length &lt;= target.length &lt;= 1000</code></li>
	<li><code>stamp</code> and <code>target</code> consist of lowercase English letters.</li>
</ul>
Stack; Greedy; Queue; String
use std::collections::VecDeque;

impl Solution {
    pub fn moves_to_stamp(stamp: String, target: String) -> Vec<i32> {
        let m = stamp.len();
        let n = target.len();

        let mut indeg: Vec<usize> = vec![m; n - m + 1];
        let mut g: Vec<Vec<usize>> = vec![Vec::new(); n];
        let mut q: VecDeque<usize> = VecDeque::new();

        for i in 0..n - m + 1 {
            for j in 0..m {
                if target.chars().nth(i + j).unwrap() == stamp.chars().nth(j).unwrap() {
                    indeg[i] -= 1;
                    if indeg[i] == 0 {
                        q.push_back(i);
                    }
                } else {
                    g[i + j].push(i);
                }
            }
        }

        let mut ans: Vec<i32> = Vec::new();
        let mut vis: Vec<bool> = vec![false; n];

        while let Some(i) = q.pop_front() {
            ans.push(i as i32);

            for j in 0..m {
                if !vis[i + j] {
                    vis[i + j] = true;

                    for &k in g[i + j].iter() {
                        indeg[k] -= 1;
                        if indeg[k] == 0 {
                            q.push_back(k);
                        }
                    }
                }
            }
        }

        if vis.iter().all(|&v| v) {
            ans.reverse();
            ans
        } else {
            Vec::new()
        }
    }
}
