90 lines
No EOL
3.4 KiB
Text
90 lines
No EOL
3.4 KiB
Text
---
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id: amortized
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title: "Amortized Analysis"
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author: Darren Yao
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prerequisites:
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- Silver - Introduction to Sorting
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- Silver - Stacks & Queues
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description: "?"
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frequency: 2
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---
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import { Problem } from "../models";
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export const metadata = {
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problems: {
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sample: [
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new Problem("CSES", "Playlist", "1141", "Intro", false, []),
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new Problem("CSES", "Sum of Two Values", "1141", "Easy", false, []),
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],
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general: [
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new Problem("Silver", "Diamond Collector", "643", "Normal", false, ["2P", "Sorting"], ""),
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new Problem("Silver", "Paired Up", "738", "Normal", false, ["2P", "Sorting"]),
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new Problem("Gold", "Haybale Feast", "767", "Normal", false, ["Set", "Sliding Window"]),
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new Problem("Plat", "Fort Moo", "600", "Normal", false, ["Sliding Window"]),
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new Problem("CF", "Books", "problemset/problem/279/B", "Normal", false, []),
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new Problem("CF", "Cellular Network", "problemset/problem/702/C", "Normal", false, []),
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new Problem("CF", "USB vs. PS/2", "problemset/problem/762/B", "Normal", false, []),
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new Problem("CF", "K-Good Segment", "problemset/problem/616/D", "Normal", false, []),
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new Problem("CF", "Garland", "problemset/problem/814/C", "Normal", false, []),
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new Problem("CF", "Jury Meeting", "problemset/problem/853/B", "Normal", false, []),
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],
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qs: [
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new Problem("YS","Queue Composite","queue_operate_all_composite","Hard",false,[],""),
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],
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}
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};
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## Sample
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<problems-list problems={metadata.problems.sample} />
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## Two Pointers
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Two pointers refers to iterating two monotonic pointers across an array to search for a pair of indices satisfying some condition in linear time.
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- [CSES Playlist](https://cses.fi/problemset/task/1141)
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### Tutorials
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- Intro to USACO 14.1
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- CPH 8.1 (Amortized Analysis)
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## Sliding Window
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Let's envision a sliding window (or constant size subarray) of size $K$ moving left to right along an array, $a$. For each position of the window, we want to compute some information.
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Let's store a `std::set` of integers representing the integers inside the window. If the window currently spans the range $i \dots j$, we observe that moving the range forward to $i+1 \dots j+1$ only removes $a_i$ and adds $a_{j+1}$ to the window. We can support these two operations and query for the minimum/maximum in the set in $O(\log N)$.
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To compute the sum in the range, instead of using a set, we can store a variable $s$ representing the sum. As we move the window forward, we update $s$ by performing the operations $s -= a_i$ and $s += a_{j+1}$.
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### Min Queue
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- [cp-algorithms: Min Stack + Queue](https://cp-algorithms.com/data_structures/stack_queue_modification.html)
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- learn about the "min queue" that CPH describes.
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### Queue w/ Two Stacks
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<problems-list problems={metadata.problems.qs} />
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<optional-content title="Application">
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Used to remove a factor of $O(\log N)$ in [USACO Plat - Mowing Mischief](http://www.usaco.org/index.php?page=viewproblem2&cpid=926).
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</optional-content>
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### Further Reading
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- [Medium](https://levelup.gitconnected.com/an-introduction-to-sliding-window-algorithms-5533c4fe1cc7)
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- [G4G](https://www.geeksforgeeks.org/window-sliding-technique/)
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## Problems
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<problems-list problems={metadata.problems.general} />
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<optional-content title="Better Memory with Two Passes">
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- [Plat - Train Tracking](http://www.usaco.org/index.php?page=viewproblem2&cpid=841)
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- Extremely difficult.
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</optional-content> |