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feat(two-pointers): boats to save people

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Barrett Ruth 2024-06-17 13:51:30 -05:00
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@ -84,13 +84,59 @@
r -= 1
return area</code></pre>
</div>
<!-- <h3> -->
<!-- <a -->
<!-- target="blank" -->
<!-- href="https://leetcode.com/problems/boats-to-save-people/" -->
<!-- >boats to save people</a -->
<!-- > -->
<!-- </h3> -->
<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:
boats = 0
light = 0
heavy = len(people) - 1
people.sort()
while light <= heavy:
if light == heavy:
boats += 1
break
elif people[light] + people[heavy] <= limit:
boats += 1
light += 1
heavy -= 1
else:
boats += 1
heavy -= 1
return boats</code></pre>
</div>
</article>
</div>
</main>