max-consecutive-ones-iii

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1004. Max Consecutive Ones III (Medium)

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

 

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

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
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.
• 0 <= k <= nums.length

Related Topics

[Array] [Binary Search] [Prefix Sum] [Sliding Window]

Similar Questions

1. Longest Substring with At Most K Distinct Characters (Medium)
2. Longest Repeating Character Replacement (Medium)
3. Max Consecutive Ones (Easy)
4. Max Consecutive Ones II (Medium)

Hints

Hint 1

One thing's for sure, we will only flip a zero if it extends an existing window of 1s. Otherwise, there's no point in doing it, right? Think Sliding Window!

Hint 2

Since we know this problem can be solved using the sliding window construct, we might as well focus in that direction for hints. Basically, in a given window, we can never have > K zeros, right?

Hint 3

We don't have a fixed size window in this case. The window size can grow and shrink depending upon the number of zeros we have (we don't actually have to flip the zeros here!).

Hint 4

The way to shrink or expand a window would be based on the number of zeros that can still be flipped and so on.