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feat(cses): more problems

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Barrett Ruth 2025-02-23 18:41:56 -05:00
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@ -191,97 +191,108 @@
Everyone recommends CSES so I started with it, doing the first 8 Everyone recommends CSES so I started with it, doing the first 8
problems. problems.
</p> </p>
<h3> <ul>
<a href="https://cses.fi/problemset/task/1068" target="_blank" <li>
>weird algorithm</a <a href="https://cses.fi/problemset/task/1068" target="_blank"
> >weird algorithm</a
</h3> >: Trivial, but I forgot to print 1 at the end.
<p>Trivial, but I forgot to print 1 at the end.</p> <b>Return the exactly correct answer.</b>
<p> </li>
<b>Return the exactly correct answer.</b> <li>
</p> <a href="https://cses.fi/problemset/task/1083" target="_blank">
<h3> missing number </a
<a href="https://cses.fi/problemset/task/1083" target="_blank"> >: N/A
missing number </li>
</a> <li>
</h3> <a href="https://cses.fi/problemset/task/1069" target="_blank">
<p>N/A</p> repetitions </a
<h3> >: Use invariants.
<a href="https://cses.fi/problemset/task/1069" target="_blank"> </li>
repetitions <li>
</a> <a href="https://cses.fi/problemset/task/1094" target="_blank">
</h3> increasing array </a
<p>Use invariants.</p> >: Run through one iteration of the algorithm. Here, I erroneously
<h3> added <code>x - last</code> to a quantity,
<a href="https://cses.fi/problemset/task/1094" target="_blank"> <i>after manipulating <code>x</code></i
increasing array >.
</a> </li>
</h3> <li>
<p> <a href="https://cses.fi/problemset/task/1070/" target="_blank"
Run through one iteration of the algorithm. Here, I erroneously >permutations</a
added <code>x - last</code> to a quantity, >: I'd seen this problem before yet struggled.
<i>after manipulating <code>x</code></i <b>Fully understand the problem constraints</b>. In this case,
>. While I understood the definition of a permissible permutation, I
</p> didn't fully internalize that you could place number
<h3> <i>wherever</i> you want. Instead, I was locked in on placing some
<a href="https://cses.fi/problemset/task/1070/" target="_blank" <code>x</code> at <code>i, i + 2, i + 4, ...</code>. Further, the
>permutations</a fact that I didn't immediately recognize this solution means I
> need to improve at <b>upsolving and reviewing problems</b>.
</h3> </li>
<p> <li>
I'd seen this problem before yet struggled. <a href="https://cses.fi/problemset/task/1071" target="_blank"
<b>Fully understand the problem constraints</b>. In this case, While >permutations</a
I understood the definition of a permissible permutation, I didn't >: Absolutely disastrous. I continually just f*dged with the
fully internalize that you could place number <i>wherever</i> you offsets I was adding to my strategy until I happened to get the
want. Instead, I was locked in on placing some <code>x</code> at answer right. <b>Don't guess</b>. Also,
<code>i, i + 2, i + 4, ...</code>. Further, the fact that I didn't <b
immediately recognize this solution means I need to improve at >don't be lazy&mdash;if an algorithm works, focus, write it out,
<b>upsolving and reviewing problems</b>. and enjoy being correct</b
</p> >.
<h3> </li>
<a href="https://cses.fi/problemset/task/1071" target="_blank" <li>
>permutations</a <a href="https://cses.fi/problemset/task/1072" target="_blank"
> >two knights</a
</h3> >: Required 2 hints from Sam Altman.
<p> <b>git gud at combinatorics</b>. Use the paradigm "count good,
Absolutely disastrous. I continually just f*dged with the offsets I remove bad." Lock in less on counting specifics&mdash;instead,
was adding to my strategy until I happened to get the answer right. consider what objects <i>mean in aggregate</i>. In this case, a
<b>Don't guess</b>. Also, \(2\times3\) grid represents an "area" of attack, contributing 2
<b bad knight pairs. This is much easier to digest then attempting to
>don't be lazy&mdash;if an algorithm works, focus, write it out, remove overcounting per-knight. Fundamentally, the problem
and enjoy being correct</b involves placing 2 knights, so breaking it down 2 knights at a
>. time is the most intuitive take.
</p> </li>
<h3> <li>
<a href="https://cses.fi/problemset/task/1072" target="_blank" <a href="https://cses.fi/problemset/task/1092" target="_blank"
>two knights</a >two sets</a
> >: <b>Don't lock in on one approach</b>. Here, this is dp. The
</h3> fact that I knew the idea of partitioning the first \(n\) numbers
<p> into two groups of size \(\frac{n(n+1)}{4}\) but failed to
Required 2 hints from Sam Altman. <b>git gud at combinatorics</b>. recognize the greedy approach means I didn't grasp the fundamental
Use the paradigm "count good, remove bad." Lock in less on counting arithmetic of the problem, nor the greedy idea: every number must
specifics&mdash;instead, consider what objects go into a set. If you add the largest number possible to set 1 to
<i>mean in aggregate</i>. In this case, a \(2\times3\) grid not exceed the target, this number can always be formed in the
represents an "area" of attack, contributing 2 bad knight pairs. other set by choosing \(1\) and \(x-1\). <b>git gud at greedy</b>.
This is much easier to digest then attempting to remove overcounting </li>
per-knight. Fundamentally, the problem involves placing 2 knights, </ul>
so breaking it down 2 knights at a time is the most intuitive take. <h2>more cses&mdash;22/2/2025</h2>
</p> <ul>
<h3> <li>
<a href="https://cses.fi/problemset/task/1092" target="_blank" <a href="https://cses.fi/problemset/task/2205" target="_blank"
>two sets</a >gray code</a
> >: Missed the pattern + <b>gave up too <i>late</i></b>
</h3> </li>
<p> <li>
<b>Don't lock in on one approach</b>. Here, this is dp. The fact <a href="https://cses.fi/problemset/task/2165" target="_blank"
that I knew the idea of partitioning the first \(n\) numbers into >towers of hanoi</a
two groups of size \(\frac{n(n+1)}{4}\) but failed to recognize the >: <b>Recursive grasp is limp</b>&mdash;missed the idea.
greedy approach means I didn't grasp the fundamental arithmetic of <b>Math/proof grasp too</b>&mdash;still don't understand how its
the problem, nor the greedy idea: every number must go into a set. \(2^n\).
If you add the largest number possible to set 1 to not exceed the </li>
target, this number can always be formed in the other set by <li>
choosing \(1\) and \(x-1\). <b>git gud at greedy</b>. <a href="https://cses.fi/problemset/task/1623" target="_blank"
</p> >apple division</a
>: I got distracted by the idea that it was NP-hard. Even when Sam
Altman told me it was DP, I failed to simplify it to "add every
element either to one or the other set".
</li>
<li>
<a href="https://cses.fi/problemset/task/2431">digit queries</a>:
got the idea + time complexity quickly, but the
<b>math-based implementation is weak</b>. Jumped into the code
<i>before</i> outlining a strict plan.
</li>
</ul>
</article> </article>
</div> </div>
</main> </main>

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