Why does flipping k zeros become a window problem?

After flipping, the answer is some contiguous block of all 1s. That block is exactly a window of the original array with at most k zeros in it. So the question becomes: what is the longest window containing at most k zeros?

Why shrink only until one zero leaves?

The window became invalid by gaining one zero too many. Moving left forward until a single zero drops out restores the limit, and any further shrinking would only make the window shorter.

No flips are allowed, so the window can never hold a zero, and the answer is the longest existing run of 1s, the same as Max Consecutive Ones.

Max Consecutive Ones III

Find the longest run of 1s after flipping at most k zeros, as the longest sliding window containing at most k zeros.

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Problem Understanding

Max Consecutive Ones III: given a binary array and an integer k, return the maximum number of consecutive 1s you can get if you may flip at most k zeros to 1.

Example: nums = [1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0], k = 2 returns 6: flip the two zeros at indices 4 and 5.

The Intuition: A Window With a Zero Budget

Don't actually flip anything. A run of 1s after flipping is just a window of the original array that holds at most k zeros; those zeros are the ones you would flip. So look for the longest such window with a variable-size sliding window:

Extend right, counting zeros as they enter.
If the count goes above k, move left forward until a zero leaves.
The window is valid again; record right − left + 1 if it is the longest.

Each index enters and leaves at most once, so this is O(n). The same "budget" shape appears in Longest Repeating Character Replacement, where the budget is replacements instead of zeros.

Interactive Walkthrough

The window L..R is shaded and the zeros inside it are dashed amber: those are the flips being spent. Watch the window slide forward whenever one zero too many comes in.

Dry Run Table

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

right	Window	Zeros	Length
4	0..4	2	5
5	4..5 after shrinking past index 3	2	2
9	4..9	2	6
10	5..10 after shrinking past index 4	2	6

Return 6.

The Solution Template

Variable-size window with at most k zeros.

Max Consecutive Ones III Code

1var longestOnes = function(nums, k) {
2    let left = 0, zeros = 0, best = 0;
3    for (let right = 0; right < nums.length; right++) {
4        if (nums[right] === 0) zeros++;
5        while (zeros > k) {
6            if (nums[left] === 0) zeros--;
7            left++;
8        }
9        best = Math.max(best, right - left + 1);
10    }
11    return best;
12};

Edge Cases & Common Mistakes

k = 0: the answer is the longest existing run of 1s.
k ≥ number of zeros: the whole array works.
All zeros: the answer is min(k, n).
Actually flipping the array: unnecessary; counting zeros in the window is enough.
Recording before shrinking: only record once zeros ≤ k.

Summary

Flipping at most k zeros means finding the longest window with at most k zeros. Grow the window on the right, shrink from the left whenever the zero count passes k, and record the longest valid window. O(n) time, O(1) space.

More practice: DSA interview questions with answers · patterns cheat sheet

Minimum Size Subarray Sum

Fruit Into Baskets

Max Consecutive Ones III

Problem Understanding

The Intuition: A Window With a Zero Budget

Interactive Walkthrough

Dry Run Table

The Solution Template

Max Consecutive Ones III Code

Edge Cases & Common Mistakes

Summary

The approach, step by step

Grow on the right

Shrink while there are too many zeros

Record the length

Frequently asked questions

Why does flipping k zeros become a window problem?

Why shrink only until one zero leaves?

What does k = 0 do?

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