To solve this problem, we need to find the maximum number of consecutive 1's in a binary array if we can flip at most k 0's. This problem can be efficiently solved using the sliding window technique which allows us to maintain a window of elements and adjust it dynamically to keep track of the valid subarray (with at most k 0's).

Approach

1. Sliding Window: Use two pointers (left and right) to represent the current window of elements being considered.
2. Zero Count: Track the number of 0's in the current window.
3. Adjust Window: When the number of 0's exceeds k, move the left pointer to the right until the number of 0's in the window is within the allowed limit (k).
4. Max Length: Keep updating the maximum length of the valid window (consecutive 1's with at most k flipped 0's) as we expand the window with the right pointer.

Solution Code

def maxConsecutiveOnes(nums, k):
    left = 0
    max_len = 0
    zero_count = 0

    for right in range(len(nums)):
        # Increment zero count if current element is 0
        if nums[right] == 0:
            zero_count += 1

        # Shrink the window from left if zero count exceeds k
        while zero_count > k:
            if nums[left] == 0:
                zero_count -= 1
            left += 1

        # Update the maximum length of the valid window
        current_len = right - left + 1
        if current_len > max_len:
            max_len = current_len

    return max_len

Explanation

• Initialization: left starts at 0, max_len (to store the result) starts at 0, and zero_count (to track 0's in the window) starts at 0.
• Expand Window: The right pointer moves from left to right. For each element, if it's a 0, increment zero_count.
• Shrink Window: If zero_count exceeds k, move the left pointer to the right. If the element at left is a 0, decrement zero_count (since it's no longer in the window).
• Update Max Length: For each valid window (with ≤ k 0's), calculate its length and update max_len if it's larger than the current value.

This approach runs in O(n) time complexity (since each element is processed at most twice: once by the right pointer and once by the left pointer) and O(1) space complexity (no additional data structures used). It efficiently handles all edge cases, including when all elements are 0's or 1's, and when k is 0.

Example: For nums = [1,1,1,0,0,0,1,1,1,1,0] and k=2, the maximum length is 6 (the subarray from index 5 to 10: 0,1,1,1,1,0 with 2 flipped 0's).

This solution is optimal and widely used for similar subarray problems involving constraints on element counts.

作者声明:本文包含人工智能生成内容。