<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) {
		int n = target.size(), m = stamp.size();
		vector<int> ans;
		for(int i = 0; i < n; ++i) {
			int j = 0;
			int k = i;
			while(k < n && j < m && stamp[j] == target[k]) {
				k++;
				j++;
				}
				if(j == m) {
					ans.push_back(i);
					for(int itr = i; itr < i + m; ++itr) {
						target[itr] = '?';
						}
					}
			}
			//cout << target << endl;
			for(int i = 0; i < n; ++i) {
				int j = 0;
				int k = i;
				bool ok = true;
				while(k < n && j < m) {
					if(stamp[j] == target[k] || target[k] == '?') {
						if(target[k] != '?') ok = false;
						k++; j++;
						}
					else break;
					}
					if(j == m && ok == false) {
						ans.push_back(i);
						for(int itr = i; itr < i + m; ++itr) {
							target[itr] = '?';
							}
						}
				}
				for(int i = n - 1; i >= 0; --i) {
					int j = 0;
					int k = i;
					bool ok = true;
					while(k < n && j < m) {
						if(stamp[j] == target[k] || target[k] == '?') {
							if(target[k] != '?') ok = false;
							k++;
							j++;
							}
							else break;
						}
						if(j == m && ok == false) {
							ans.push_back(i);
							for(int itr = i; itr < i + m; ++itr) {
								target[itr] = '?';
								}
							}
					}
					for(auto it : target) {
						if(it != '?') {
							ans.clear();
							break;
							}
						}
						reverse(ans.begin(), ans.end());
						return ans;
    }
};
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