<p>You are stacking blocks to form a pyramid. Each block has a color, which is represented by a single letter. Each row of blocks contains <strong>one less block</strong> than the row beneath it and is centered on top.</p>

<p>To make the pyramid aesthetically pleasing, there are only specific <strong>triangular patterns</strong> that are allowed. A triangular pattern consists of a <strong>single block</strong> stacked on top of <strong>two blocks</strong>. The patterns are given&nbsp;as a list of&nbsp;three-letter strings <code>allowed</code>, where the first two characters of a pattern represent the left and right bottom blocks respectively, and the third character is the top block.</p>

<ul>
	<li>For example, <code>&quot;ABC&quot;</code> represents a triangular pattern with a <code>&#39;C&#39;</code> block stacked on top of an <code>&#39;A&#39;</code> (left) and <code>&#39;B&#39;</code> (right) block. Note that this is different from <code>&quot;BAC&quot;</code> where <code>&#39;B&#39;</code> is on the left bottom and <code>&#39;A&#39;</code> is on the right bottom.</li>
</ul>

<p>You start with a bottom row of blocks <code>bottom</code>, given as a single string, that you <strong>must</strong> use as the base of the pyramid.</p>

<p>Given <code>bottom</code> and <code>allowed</code>, return <code>true</code><em> if you can build the pyramid all the way to the top such that <strong>every triangular pattern</strong> in the pyramid is in </em><code>allowed</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/26/pyramid1-grid.jpg" style="width: 600px; height: 232px;" />
<pre>
<strong>Input:</strong> bottom = &quot;BCD&quot;, allowed = [&quot;BCC&quot;,&quot;CDE&quot;,&quot;CEA&quot;,&quot;FFF&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> The allowed triangular patterns are shown on the right.
Starting from the bottom (level 3), we can build &quot;CE&quot; on level 2 and then build &quot;A&quot; on level 1.
There are three triangular patterns in the pyramid, which are &quot;BCC&quot;, &quot;CDE&quot;, and &quot;CEA&quot;. All are allowed.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/26/pyramid2-grid.jpg" style="width: 600px; height: 359px;" />
<pre>
<strong>Input:</strong> bottom = &quot;AAAA&quot;, allowed = [&quot;AAB&quot;,&quot;AAC&quot;,&quot;BCD&quot;,&quot;BBE&quot;,&quot;DEF&quot;]
<strong>Output:</strong> false
<strong>Explanation:</strong> The allowed triangular patterns are shown on the right.
Starting from the bottom (level 4), there are multiple ways to build level 3, but trying all the possibilites, you will get always stuck before building level 1.
</pre>

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

<ul>
	<li><code>2 &lt;= bottom.length &lt;= 6</code></li>
	<li><code>0 &lt;= allowed.length &lt;= 216</code></li>
	<li><code>allowed[i].length == 3</code></li>
	<li>The letters in all input strings are from the set <code>{&#39;A&#39;, &#39;B&#39;, &#39;C&#39;, &#39;D&#39;, &#39;E&#39;, &#39;F&#39;}</code>.</li>
	<li>All the values of <code>allowed</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def pyramidTransition(self, bottom: str, allowed: List[str]) -> bool:
        # we're building a bottom-up triangle -> dynamic programming
        # how long can bottom be: 6
        # how long is allowed: 216
        # also are only 6 letters

        # backtracking
        # let's say we're at any row, of length n
        # then we have n blocks to place, and if we're going to place a block if the config is allowed
        # build out all entire row combinations first, then send each of these new row combinations to the next level, and the base case is when we have a 2 level pyramid, so when n == 2, we just look at the base and see if any of them are in allowed

        # O(B)
        # O(B - 1) * O(6) * O(1)
        allowed = set(allowed)
        def recur(row):
            if len(row) == 2:
                for letter in "ABCDEF":
                    if row + letter in allowed:
                        return True
                return False
            new_rows = defaultdict(list)
            # we need to place new blocks
            for i in range(len(row)-1):
                bottom_blocks = row[i] + row[i+1]
                for letter in "ABCDEF":
                    if bottom_blocks + letter in allowed:
                        if i > 0:
                            for prev_row in new_rows[i-1]:
                                new_rows[i].append(prev_row + letter)
                        else:
                            new_rows[i].append(letter)
            # new_last will be the last index of new_rows
            new_row_length = len(row) - 1
            ret = False
            # print(new_rows[new_row_length-1])
            for next_row in new_rows[new_row_length-1]:
                ret = ret or recur(next_row)
            return ret
        return recur(bottom)
17400