feat(cses): more problems
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Everyone recommends CSES so I started with it, doing the first 8
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Everyone recommends CSES so I started with it, doing the first 8
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problems.
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problems.
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</p>
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</p>
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<h3>
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<ul>
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<li>
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<a href="https://cses.fi/problemset/task/1068" target="_blank"
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<a href="https://cses.fi/problemset/task/1068" target="_blank"
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>weird algorithm</a
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>weird algorithm</a
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>
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>: Trivial, but I forgot to print 1 at the end.
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</h3>
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<p>Trivial, but I forgot to print 1 at the end.</p>
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<p>
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<b>Return the exactly correct answer.</b>
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<b>Return the exactly correct answer.</b>
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</p>
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</li>
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<h3>
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<li>
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<a href="https://cses.fi/problemset/task/1083" target="_blank">
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<a href="https://cses.fi/problemset/task/1083" target="_blank">
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missing number
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missing number </a
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</a>
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>: N/A
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</h3>
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</li>
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<p>N/A</p>
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<li>
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<h3>
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<a href="https://cses.fi/problemset/task/1069" target="_blank">
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<a href="https://cses.fi/problemset/task/1069" target="_blank">
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repetitions
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repetitions </a
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</a>
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>: Use invariants.
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</h3>
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</li>
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<p>Use invariants.</p>
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<li>
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<h3>
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<a href="https://cses.fi/problemset/task/1094" target="_blank">
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<a href="https://cses.fi/problemset/task/1094" target="_blank">
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increasing array
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increasing array </a
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</a>
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>: Run through one iteration of the algorithm. Here, I erroneously
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</h3>
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<p>
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Run through one iteration of the algorithm. Here, I erroneously
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added <code>x - last</code> to a quantity,
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added <code>x - last</code> to a quantity,
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<i>after manipulating <code>x</code></i
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<i>after manipulating <code>x</code></i
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>.
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>.
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</p>
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</li>
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<h3>
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<li>
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<a href="https://cses.fi/problemset/task/1070/" target="_blank"
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<a href="https://cses.fi/problemset/task/1070/" target="_blank"
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>permutations</a
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>permutations</a
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>
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>: I'd seen this problem before yet struggled.
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</h3>
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<b>Fully understand the problem constraints</b>. In this case,
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<p>
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While I understood the definition of a permissible permutation, I
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I'd seen this problem before yet struggled.
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didn't fully internalize that you could place number
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<b>Fully understand the problem constraints</b>. In this case, While
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<i>wherever</i> you want. Instead, I was locked in on placing some
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I understood the definition of a permissible permutation, I didn't
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<code>x</code> at <code>i, i + 2, i + 4, ...</code>. Further, the
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fully internalize that you could place number <i>wherever</i> you
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fact that I didn't immediately recognize this solution means I
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want. Instead, I was locked in on placing some <code>x</code> at
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need to improve at <b>upsolving and reviewing problems</b>.
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<code>i, i + 2, i + 4, ...</code>. Further, the fact that I didn't
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</li>
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immediately recognize this solution means I need to improve at
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<li>
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<b>upsolving and reviewing problems</b>.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1071" target="_blank"
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<a href="https://cses.fi/problemset/task/1071" target="_blank"
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>permutations</a
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>permutations</a
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>
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>: Absolutely disastrous. I continually just f*dged with the
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</h3>
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offsets I was adding to my strategy until I happened to get the
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<p>
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answer right. <b>Don't guess</b>. Also,
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Absolutely disastrous. I continually just f*dged with the offsets I
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was adding to my strategy until I happened to get the answer right.
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<b>Don't guess</b>. Also,
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<b
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<b
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>don't be lazy—if an algorithm works, focus, write it out,
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>don't be lazy—if an algorithm works, focus, write it out,
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and enjoy being correct</b
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and enjoy being correct</b
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>.
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>.
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</p>
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</li>
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<h3>
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<li>
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<a href="https://cses.fi/problemset/task/1072" target="_blank"
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<a href="https://cses.fi/problemset/task/1072" target="_blank"
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>two knights</a
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>two knights</a
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>
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>: Required 2 hints from Sam Altman.
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</h3>
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<b>git gud at combinatorics</b>. Use the paradigm "count good,
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<p>
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remove bad." Lock in less on counting specifics—instead,
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Required 2 hints from Sam Altman. <b>git gud at combinatorics</b>.
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consider what objects <i>mean in aggregate</i>. In this case, a
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Use the paradigm "count good, remove bad." Lock in less on counting
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\(2\times3\) grid represents an "area" of attack, contributing 2
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specifics—instead, consider what objects
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bad knight pairs. This is much easier to digest then attempting to
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<i>mean in aggregate</i>. In this case, a \(2\times3\) grid
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remove overcounting per-knight. Fundamentally, the problem
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represents an "area" of attack, contributing 2 bad knight pairs.
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involves placing 2 knights, so breaking it down 2 knights at a
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This is much easier to digest then attempting to remove overcounting
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time is the most intuitive take.
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per-knight. Fundamentally, the problem involves placing 2 knights,
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</li>
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so breaking it down 2 knights at a time is the most intuitive take.
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<li>
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1092" target="_blank"
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<a href="https://cses.fi/problemset/task/1092" target="_blank"
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>two sets</a
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>two sets</a
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>
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>: <b>Don't lock in on one approach</b>. Here, this is dp. The
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</h3>
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fact that I knew the idea of partitioning the first \(n\) numbers
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<p>
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into two groups of size \(\frac{n(n+1)}{4}\) but failed to
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<b>Don't lock in on one approach</b>. Here, this is dp. The fact
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recognize the greedy approach means I didn't grasp the fundamental
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that I knew the idea of partitioning the first \(n\) numbers into
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arithmetic of the problem, nor the greedy idea: every number must
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two groups of size \(\frac{n(n+1)}{4}\) but failed to recognize the
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go into a set. If you add the largest number possible to set 1 to
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greedy approach means I didn't grasp the fundamental arithmetic of
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not exceed the target, this number can always be formed in the
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the problem, nor the greedy idea: every number must go into a set.
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other set by choosing \(1\) and \(x-1\). <b>git gud at greedy</b>.
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If you add the largest number possible to set 1 to not exceed the
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</li>
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target, this number can always be formed in the other set by
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</ul>
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choosing \(1\) and \(x-1\). <b>git gud at greedy</b>.
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<h2>more cses—22/2/2025</h2>
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</p>
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<ul>
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<li>
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<a href="https://cses.fi/problemset/task/2205" target="_blank"
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>gray code</a
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>: Missed the pattern + <b>gave up too <i>late</i></b>
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</li>
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<li>
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<a href="https://cses.fi/problemset/task/2165" target="_blank"
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>towers of hanoi</a
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>: <b>Recursive grasp is limp</b>—missed the idea.
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<b>Math/proof grasp too</b>—still don't understand how its
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\(2^n\).
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</li>
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<li>
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<a href="https://cses.fi/problemset/task/1623" target="_blank"
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>apple division</a
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>: I got distracted by the idea that it was NP-hard. Even when Sam
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Altman told me it was DP, I failed to simplify it to "add every
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element either to one or the other set".
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</li>
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<li>
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<a href="https://cses.fi/problemset/task/2431">digit queries</a>:
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got the idea + time complexity quickly, but the
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<b>math-based implementation is weak</b>. Jumped into the code
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<i>before</i> outlining a strict plan.
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</li>
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</ul>
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</article>
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</article>
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</div>
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</div>
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</main>
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</main>
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@ -15,6 +15,7 @@ header {
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ul {
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ul {
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list-style: unset;
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list-style: unset;
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list-style-type: "- ";
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}
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}
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li {
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li {
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