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feat(post): collapse entries

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Barrett Ruth 2024-06-17 19:47:59 -05:00
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posts/two-pointers.html
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@ -42,37 +42,41 @@
</header>
<article class="post-article">
<h2>technique overview</h2>
<h3>
<a
target="blank"
href="https://leetcode.com/problems/container-with-most-water/"
>container with most water</a
>
</h3>
<p>Sometimes, the mathematical solution is the simplest.</p>
<p>
The area of a container bounded by the ground and its columns at
positions \((l, r)\) is: \[ \text{area} = \text{width} \cdot
\text{length} = (r - l) \cdot \min\{height[l], height[r]\} \]
</p>
<p>
At its core, this is a maximization problem: maximize the contained
area. \[ \max\{(r - l) \cdot \min\{height[l], height[r]\}\} \]
</p>
<p>
Given a new column position \(l_0 < l\) or \(r_0 < r\), the
contained area can only increase if the height of the corresponding
column increases.
</p>
<!-- TODO: add footnote -->
<p>
The following correct solution surveys all containers, initialized
with the widest columns positions, that are valid candidates for a
potentially new largest area. A running maximizum, the answer, is
maintained.
</p>
<div class="post-code">
<pre><code class="language-python">def maxArea(height: list[int]) -> int:
<div class="problem-header">
<h3>
<a
target="blank"
href="https://leetcode.com/problems/container-with-most-water/"
>container with most water</a
>
</h3>
</div>
<div class="problem-content">
<p>Sometimes, the mathematical solution is the simplest.</p>
<p>
The area of a container bounded by the ground and its columns at
positions \((l, r)\) is: \[ \text{area} = \text{width} \cdot
\text{length} = (r - l) \cdot \min\{height[l], height[r]\} \]
</p>
<p>
At its core, this is a maximization problem: maximize the
contained area. \[ \max\{(r - l) \cdot \min\{height[l],
height[r]\}\} \]
</p>
<p>
Given a new column position \(l_0 < l\) or \(r_0 < r\), the
contained area can only increase if the height of the
corresponding column increases.
</p>
<!-- TODO: add footnote -->
<p>
The following correct solution surveys all containers, initialized
with the widest columns positions, that are valid candidates for a
potentially new largest area. A running maximizum, the answer, is
maintained.
</p>
<div class="post-code">
<pre><code class="language-python">`def maxArea(height: list[int]) -> int:
ans = 0
l, r = 0, len(height) - 1
@ -85,41 +89,47 @@
while l < r and height[r] <= min_height:
r -= 1
return ans</code></pre>
return ans`</code></pre>
</div>
</div>
<h3>
<a
target="blank"
href="https://leetcode.com/problems/boats-to-save-people/"
>boats to save people</a
>
</h3>
<p>
Usually, the metaphorical problem description is a distraction.
However, I find that thinking within the confines of "boats" and
"people" yields an intuitive solution in this case.
</p>
<p>
<!-- TODO: footnote -->
Since only two people can fit in a boat at a time, pairing up
lightest and heaviest individuals will result in the least amount of
boats being used.
</p>
<p>
However, the weights are given in random order. Efficiently pairing
up the lightest and heaviest individuals, then, requires the most
common two-pointers prepreocessing step: sorting.
</p>
<p>Finally, flesh out any remaining aspects of the implementation:</p>
<ol>
<li>If one person remains, give them a boat.</li>
<li>
If both people don&apos;t fit, use the heavier person&mdash;the
lighter could maybe fit with someone else.
</li>
</ol>
<div class="post-code">
<pre><code class="language-python">def minimum_rescue_boats(people: list[int], limit: int) -> int:
<div class="problem-header">
<h3>
<a
target="blank"
href="https://leetcode.com/problems/boats-to-save-people/"
>boats to save people</a
>
</h3>
</div>
<div class="problem-content">
<p>
Usually, the metaphorical problem description is a distraction.
However, I find that thinking within the confines of "boats" and
"people" yields an intuitive solution in this case.
</p>
<p>
<!-- TODO: footnote -->
Since only two people can fit in a boat at a time, pairing up
lightest and heaviest individuals will result in the least amount
of boats being used.
</p>
<p>
However, the weights are given in random order. Efficiently
pairing up the lightest and heaviest individuals, then, requires
the most common two-pointers prepreocessing step: sorting.
</p>
<p>
Finally, flesh out any remaining aspects of the implementation:
</p>
<ol>
<li>If one person remains, give them a boat.</li>
<li>
If both people don&apos;t fit, use the heavier person&mdash;the
lighter could maybe fit with someone else.
</li>
</ol>
<div class="post-code">
<pre><code class="language-python">def minimum_rescue_boats(people: list[int], limit: int) -> int:
ans = 0
l, r = 0, len(people) - 1
@ -138,6 +148,7 @@
r -= 1
return ans</code></pre>
</div>
</div>
</article>
</div>