feat(two-pointers): boats to save people

posts/two-pointers.html

@@ -84,13 +84,59 @@
             r -= 1
     return area</code></pre>
           </div>
-          <!-- <h3> -->
+          <h3>
-          <!--   <a -->
+            <a
-          <!--     target="blank" -->
+              target="blank"
-          <!--     href="https://leetcode.com/problems/boats-to-save-people/" -->
+              href="https://leetcode.com/problems/boats-to-save-people/"
-          <!--     >boats to save people</a -->
+              >boats to save people</a
-          <!--   > -->
+            >
-          <!-- </h3> -->
+          </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>