Скачать или смотреть Maximum Average Subarray I - Leetcode 643 Python

LeetCode 643 asked for an average. My first solution was... just plain 'mean'. 😠

(This is a long one, grab your ☕)

You see the "Easy" tag on LeetCode and you think, "Ah, a nice 10-minute warmup."

Famous last words.

The problem: "Maximum Average Subarray I." Given an array of numbers, find the contiguous subarray of a fixed length k that has the maximum average.

My brain, 5 seconds in: "EASY. I'll just be a brute."

Start at index 0.

Loop from i to i+k to sum the first window.

Divide by k, get the average. Store it.

Move to index 1.

Loop from i to i+k to sum the second window.

Divide, get the average. Compare.

Move to index 2...

...go make dinner while the code runs.

I wrote the code. It passed.

...and it was O(N*K).

My code was technically correct. It was also dumb.

It was re-calculating sums for elements it just saw.

Let's say [1, 12, -5, -6, 50, 3] and k=4.

Window 1: (1 + 12 - 5 - 6) Window 2: (12 - 5 - 6 + 50)

See it? My code literally calculated 12 - 5 - 6... and then immediately, with no shame, did it again.

This is the programming equivalent of:

Counting every person in a room.

One person leaves, one person enters.

Forgetting the first count... and counting EVERYONE again from scratch.

It's inefficient. It's disrespectful to your CPU. It's the "Nested Loop of Shame."

And in a real interview, it's the difference between "When can you start?" and "We'll let you know."

🪟 Enter: The Sliding Window 🪟

This pattern isn't just a "trick." It's a fundamental shift in thinking.

Stop thinking of it as "recalculating a new window every time." Start thinking of it as "maintaining a conga line."

Your k-sized window is a conga line of numbers.

Get the sum of the first k numbers. Do this ONLY ONCE. This is your currentSum. This currentSum / k is your first maxAverage.

Now, the conga line needs to move one step forward. This is NOT a for-loop inside a for-loop. This is two tiny operations:

➡️ The person at the front of the line (nums[i-k]) leaves. SUBTRACT their value from currentSum.

⬅️ A new person (nums[i]) joins the back of the line. ADD their value to currentSum.

That's it. Two simple O(1) operations.

You now have the sum of the next window. Compare its average with maxAverage. Repeat until you hit the end of the array.

You just turned your O(N*K) "Nested Loop of Shame" into an O(N) "One Pass Wonder."

This is what interviewers are really looking for when they give you an "easy" array problem. Can you see the pattern? Can you spot the redundant work and eliminate it?

Once you master this pattern, dozens of "Medium" and "Hard" problems (like "Longest Substring with K Distinct Characters") suddenly become easy.

I created a full visual walkthrough of this exact problem. I show the brute force, why it stinks, and exactly how the window "slides" its way to O(N) glory.

It’ll save you hours of confusion.

What's a LeetCode pattern that you struggled with until it finally "clicked" and changed the game for you?

(For me, it was this and anything related to Dynamic Programming...)

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