Summary of Sliding Window Pattern: 1/4 (Maximum Average Subarray I)

In this video, the speaker introduces the sliding window pattern, which includes fixed window size and dynamically sized window variants. The fixed window size is used to calculate the average of values within a specific subset of an array. Specifically, the speaker explains how to find the maximum average when considering all contiguous subarrays of size 4 within an array.

Key Points

Introduction to sliding window pattern
Fixed window size and dynamically sized window variants
Finding maximum average of subarrays of size 4

Naive Approach

Initially, the speaker discusses the naive approach to solving this problem, which involves iterating through k elements for each element of the input array, leading to inefficiency. They then provide a detailed step-by-step explanation of the naive solution by writing code in Go, highlighting the installation process, environment setup, and code implementation.

Sliding Window Solution

Subsequently, the speaker introduces the more efficient sliding window solution. This solution involves keeping track of start and end indexes of the window, updating the window sum, calculating the average, and moving the window to find the maximum average of subarrays of size 4 within the array. The implementation of the sliding window solution in Go is demonstrated using the same input array as the naive solution, showcasing its efficiency and discussing its time complexity.

Conclusion

The video concludes by comparing the time complexity of the naive and sliding window solutions, emphasizing the benefits of the latter. Additionally, the speaker briefly mentions the upcoming videos that will explain problems from the LeetCode patterns list, providing a comprehensive overview of the sliding window pattern and its practical applications.

Notable Quotes

— 48:02 — « code to calculate the average »

— 48:25 — « so our current window sum is 2. »

— 51:02 — « and then we'll go back to the top of our for loop »

— 51:33 — « calculate our max so our current window sum is 42 »

— 52:29 — « and our max is 12.75 which is what will get returned »