Max Consecutive Ones III
Given a binary array and an integer k, return the longest run of 1s you can make by flipping at most k zeros . Treat zeros as a budget inside the window: grow right, and shrink left whenever the zero count exceeds k.
Constraints
- 1 ≤ nums.length ≤ 105
- numsi is either 0 or 1
- 0 ≤ k ≤ nums.length
Example
Input
nums = [1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0], k = 2Output
6Explanation Flip the two 0s at indices 5 and 4 (or nearby) to get a run of six 1s.
Window may hold at most k zeros
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Statusvalid
Two zeros so far; zeros = 2 <= k; best = 4.
What happens in this step
window = [0, 3] [1, 1, 0, 0] before: zeros = 0 (right = 1) right = 2: nums[2] = 0 → zeros = 1 right = 3: nums[3] = 0 → zeros = 2 Both zeros fit inside the k = 2 flip budget — window valid; best becomes 4.
Step 1 of 4
Steps to visualize
- Grow right; if the new value is 0, increment the zero budget used.
- While zeros exceed k, advance left and decrement when leaving a zero.
- The window always represents a stretch fixable with at most k flips.
- Update best with the current window length after each adjustment.
- Continue until right reaches the end of the array.
Walk through the code
Same walkthrough, now with the code. Press Next to move one step and watch which lines run.
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Statusvalid
Two zeros so far; zeros = 2 <= k; best = 4.
What happens in this step
window = [0, 3] [1, 1, 0, 0] before: zeros = 0 (right = 1) right = 2: nums[2] = 0 → zeros = 1 right = 3: nums[3] = 0 → zeros = 2 Both zeros fit inside the k = 2 flip budget — window valid; best becomes 4.
Step 1 of 4
Solution
function longestOnes(nums, k) {
let left = 0;
let zeros = 0;
let best = 0;
for (let right = 0; right < nums.length; right++) {
if (nums[right] === 0) zeros++;
while (zeros > k) {
if (nums[left] === 0) zeros--;
left++;
}
best = Math.max(best, right - left + 1);
}
return best;
}- Time
- O(n)
- Space
- O(1)
Test cases
| Input | Expected | Covers |
|---|---|---|
nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2 | 6 | Docstring example |
nums = [1], k = 0 | 1 | Single one |
nums = [0, 0, 1, 1], k = 2 | 4 | k covers all zeroes |
nums = [1, 1, 1, 1], k = 0 | 4 | All ones |
nums = [0, 0, 0], k = 1 | 1 | All zeroes with k = 1 |
nums = [1, 0, 1, 1, 0, 1], k = 0 | 2 | k = 0 is plain max consecutive ones |
nums = [1,1,0,0,1,1,1,0,1,1], k = 2 | 7 | Walkthrough array best length 7 |