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
func movesToStamp(stamp string, target string) (ans []int) {
	m, n := len(stamp), len(target)
	indeg := make([]int, n-m+1)
	for i := range indeg {
		indeg[i] = m
	}
	g := make([][]int, n)
	q := []int{}
	for i := 0; i < n-m+1; i++ {
		for j := range stamp {
			if target[i+j] == stamp[j] {
				indeg[i]--
				if indeg[i] == 0 {
					q = append(q, i)
				}
			} else {
				g[i+j] = append(g[i+j], i)
			}
		}
	}
	vis := make([]bool, n)
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		ans = append(ans, i)
		for j := range stamp {
			if !vis[i+j] {
				vis[i+j] = true
				for _, k := range g[i+j] {
					indeg[k]--
					if indeg[k] == 0 {
						q = append(q, k)
					}
				}
			}
		}
	}
	for _, v := range vis {
		if !v {
			return []int{}
		}
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return
}