feat(cses): more problems

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