What is the brute force approach and complexity for Max Consecutive Ones III?

For each start, extend right while the zero count stays ≤ k with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Max Consecutive Ones III?

Grow right; while zeros > k, shrink from the left; track the longest window with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Max Consecutive Ones III?

Look for key signals in the prompt: "consecutive 1s" indicates The answer is a contiguous block → think window.; "maximum number" indicates Longest window → shrink while invalid, record after fixing.; "flip at most k 0s" indicates A disguise. It just means: the window may contain at most k zeros..

Max Consecutive Ones III

Medium·LeetCode #1004 Pro OnlyMedium

LeetCode

Problem Statement

You are given an array nums that contains only 0s and 1s, and a number k.

You may change at most k of the 0s into 1s. After doing that, what is the longest run of consecutive 1s you can get?

Input

A binary array nums and an integer k.

Output

The length of the longest block of consecutive 1s possible after flipping at most k zeros.

Example 1

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2

Output: 6

Flip the zeros at positions 4 and 5 (0-based). Then positions 4 to 9 are all 1s: [0,0,1,1,1,1] becomes six 1s in a row.

Example 2

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3

Output: 10

Flip the zeros at positions 4, 5 and 9. Then positions 2 to 11 are all 1s — length 10.

Constraints

• 1 ≤ nums.length ≤ 10^5
• nums[i] is 0 or 1
• 0 ≤ k ≤ nums.length

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“consecutive 1s”

→ The answer is a contiguous block → think window.

“maximum number”

→ Longest window → shrink while invalid, record after fixing.

“flip at most k 0s”

→ A disguise. It just means: the window may contain at most k zeros.

Approach Complexity Ladder

Brute Force — try every startO(N²)O(1)

For each start, extend right while the zero count stays ≤ k

Optimal — variable sliding windowO(N)O(1)

Grow right; while zeros > k, shrink from the left; track the longest window

Brute Force Intuition

For every start, walk right and count the zeros. As long as the count is at most k, this block can be turned into all 1s, so its length is a candidate. Once the count goes over k, stop — longer blocks from this start only have more zeros.

Recognition rule for this sub-pattern:

Window grows and shrinks. Expand right; when the window breaks the rule, move left until it is valid again. Works when "bigger window = harder to be valid".

Remember this · Sliding Window

Trigger

Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".

Think

Can I keep a valid range and move it, instead of checking every subarray from scratch?

Key question

When is my current window valid?

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