<p>Given two strings <code>s</code> and <code>t</code>, transform string <code>s</code> into string <code>t</code> using the following operation any number of times:</p>

<ul>
	<li>Choose a <strong>non-empty</strong> substring in <code>s</code> and sort it in place so the characters are in <strong>ascending order</strong>.

	<ul>
		<li>For example, applying the operation on the underlined substring in <code>&quot;1<u>4234</u>&quot;</code> results in <code>&quot;1<u>2344</u>&quot;</code>.</li>
	</ul>
	</li>
</ul>

<p>Return <code>true</code> if <em>it is possible to transform <code>s</code> into <code>t</code></em>. Otherwise, return <code>false</code>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;84532&quot;, t = &quot;34852&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can transform s into t using the following sort operations:
&quot;84<u>53</u>2&quot; (from index 2 to 3) -&gt; &quot;84<u>35</u>2&quot;
&quot;<u>843</u>52&quot; (from index 0 to 2) -&gt; &quot;<u>348</u>52&quot;
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;34521&quot;, t = &quot;23415&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You can transform s into t using the following sort operations:
&quot;<u>3452</u>1&quot; -&gt; &quot;<u>2345</u>1&quot;
&quot;234<u>51</u>&quot; -&gt; &quot;234<u>15</u>&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;12345&quot;, t = &quot;12435&quot;
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>s.length == t.length</code></li>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> and <code>t</code> consist of only digits.</li>
</ul>

class Solution {
public:
    bool isTransformable(string s, string t) {
        int n = s.size();
        vector<vector<int>> idx(10);
        vector<int> pos(10);
        for(int i = 0; i < n; i++) {
            idx[s[i] - '0'].push_back(i);
        }

        for(char c: t) {
            int d = c - '0';
            if(pos[d] >= idx[d].size()) {
                return false;
            }
            for(int dd = 0; dd < d; dd++) {
                if(pos[dd] < idx[dd].size() && idx[dd][pos[dd]] < idx[d][pos[d]]) {
                    return false;
                }
            }
            pos[d] += 1;
        }
        return true;
    }
};
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