<p>The <strong>count-and-say</strong> sequence is a sequence of digit strings defined by the recursive formula:</p>

<ul>
	<li><code>countAndSay(1) = &quot;1&quot;</code></li>
	<li><code>countAndSay(n)</code> is the run-length encoding of <code>countAndSay(n - 1)</code>.</li>
</ul>

<p><a href="http://en.wikipedia.org/wiki/Run-length_encoding" target="_blank">Run-length encoding</a> (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string <code>&quot;3322251&quot;</code> we replace <code>&quot;33&quot;</code> with <code>&quot;23&quot;</code>, replace <code>&quot;222&quot;</code> with <code>&quot;32&quot;</code>, replace <code>&quot;5&quot;</code> with <code>&quot;15&quot;</code> and replace <code>&quot;1&quot;</code> with <code>&quot;11&quot;</code>. Thus the compressed string becomes <code>&quot;23321511&quot;</code>.</p>

<p>Given a positive integer <code>n</code>, return <em>the </em><code>n<sup>th</sup></code><em> element of the <strong>count-and-say</strong> sequence</em>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">&quot;1211&quot;</span></p>

<p><strong>Explanation:</strong></p>

<pre>
countAndSay(1) = &quot;1&quot;
countAndSay(2) = RLE of &quot;1&quot; = &quot;11&quot;
countAndSay(3) = RLE of &quot;11&quot; = &quot;21&quot;
countAndSay(4) = RLE of &quot;21&quot; = &quot;1211&quot;
</pre>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">&quot;1&quot;</span></p>

<p><strong>Explanation:</strong></p>

<p>This is the base case.</p>
</div>

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

<ul>
	<li><code>1 &lt;= n &lt;= 30</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you solve it iteratively?
class Solution:
    def countAndSay(self, n: int) -> str:

        # idea: use stack to record char and its count
        # error1: initialize stac
        # T: O(n*m)
        # S: O(m)

        stack = []
        s = "1"

        for _ in range(n-1):
            for char in s:
                if stack and stack[-1][0] == char:
                    stack[-1][1] += 1
                else:
                    stack.append([char, 1])
            s = ""
            for char, count in stack:
                s += str(count)
                s += char
            stack = []
        return s


            





                






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