<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>

class Solution {
 public:
  vector<int> movesToStamp(string stamp, string target) {
    vector<int> ans;
    // stamped[i] := true if we already stamped target by stamping on index i
    vector<bool> stamped(target.length());
    int stampedCount = 0;  // Our goal is to make stampedCount = |target|.

    while (stampedCount < target.length()) {
      bool isStamped = false;
      // Try to stamp target[i..i + |stamp|) for each index.
      for (int i = 0; i <= target.length() - stamp.length(); ++i) {
        if (stamped[i])
          continue;
        const int stampified = stampify(stamp, target, i);
        if (stampified == 0)
          continue;
        stampedCount += stampified;
        isStamped = true;
        stamped[i] = true;
        ans.push_back(i);
      }
      // After trying to stamp on each index, we can't find a valid stamp.
      if (!isStamped)
        return {};
    }

    ranges::reverse(ans);
    return ans;
  }

 private:
  // Stamps target[i..i + |stamp|) and returns the number of newly stamped
  // characters.
  // e.g. stampify("abc", "ababc", 2) returns 3 because target becomes "ab***".
  int stampify(const string& stamp, string& target, int s) {
    int stampified = stamp.length();

    for (int i = 0; i < stamp.length(); ++i)
      if (target[s + i] == '*')  // It's already been stamped.
        --stampified;
      else if (target[s + i] != stamp[i])
        return 0;  // We can't stamp on the index i.

    if (stampified > 0)
      fill(target.begin() + s, target.begin() + s + stamp.length(), '*');

    return stampified;
  }
};
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