<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 {
    int n, m;
private:
    int findInd(string stamp, string target){
        for(int i=0; i<=m-n; i++){
            int s=0, t=i;
            bool flag = false;
            while(s<n && t<m && (target[t]=='?' || target[t]==stamp[s])){
                if(target[t]!='?') flag=true;
                s++, t++;
            }
            if(s==n && flag) return i;
        }
        return -1;
    }
public:
    vector<int> movesToStamp(string stamp, string target) {
        n = stamp.size(), m = target.size();
        string temp = "";
        for(int i=0; i<m; i++) temp+='?';
        vector<int> ans;

        while(temp!=target){
            int ind = findInd(stamp,target);
            if(ind<0) return {};
            else{
                ans.push_back(ind);
                for(int i=ind; i<ind+n; i++)
                    target[i] = '?';
            }
        }

        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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