---
id: sliding
title: "Sliding Window"
author: Darren Yao?, Benjamin Qi
prerequisites:
- Silver - Stacks & Queues
- Silver - Introduction to Ordered Maps & Sets
description: "?"
frequency: 2
---
import { Problem } from "../models";
export const metadata = {
problems: {
sample: [
new Problem("CSES", "Sum of Two Values", "1141", "Easy", false, []),
],
slide: [
new Problem("CSES", "Playlist", "1141", "Easy", false, [],"classic example of 2P"),
],
general: [
new Problem("CSES", "Subarray Sums I", "1660", "Easy", false, [], ""),
new Problem("CSES", "Sliding Median", "1076", "Easy", false, [], ""),
new Problem("CSES", "Sliding Cost", "1077", "Hard", false, [], ""),
new Problem("CSES", "Max Subarray Sum II", "1644", "Normal", false, [], ""),
new Problem("Silver", "Diamond Collector", "643", "Easy", false, ["2P", "Sorting"], ""),
new Problem("Silver", "Paired Up", "738", "Normal", false, ["2P", "Sorting"]),
new Problem("Gold", "Haybale Feast", "767", "Normal", false, ["Set", "Sliding Window"]),
new Problem("CF", "Books", "problemset/problem/279/B", "Normal", false, []),
new Problem("CF", "Cellular Network", "problemset/problem/702/C", "Normal", false, []),
new Problem("CF", "USB vs. PS/2", "problemset/problem/762/B", "Normal", false, []),
new Problem("CF", "K-Good Segment", "problemset/problem/616/D", "Normal", false, []),
new Problem("CF", "Garland", "problemset/problem/814/C", "Normal", false, []),
new Problem("CF", "Jury Meeting", "problemset/problem/853/B", "Normal", false, []),
new Problem("Plat", "Fort Moo", "600", "Very Hard", false, ["Sliding Window"]),
],
qs: [
new Problem("YS","Queue Composite","queue_operate_all_composite","Hard",true,[],""),
],
}
};
## Two Pointers
Two pointers refers to iterating two monotonic pointers across an array to search for a pair of indices satisfying some condition in linear time.
## Sliding Window
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. For example, we could 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 $O(\log N)$.
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 subtracting $a_i$ from $s$ and adding $a_{j+1}$ to $s$.
### Problems
## Sliding Window Minimum in $O(N)$
Mentions two ways to solve this (both are important)!
In particular, the second method allows us to solve the following generalization in linear time as well: