Introduction

In this blog post, we will explore how to solve the LeetCode problem "1004" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown:

• Sliding Window: Use a sliding window to maintain a subarray where the number of 0s is less than or equal to k.
• Expand and Contract: Expand the window to the right. If the number of 0s exceeds k, contract the window from the left until the number of 0s is within the limit again.

• Track Maximum Length: Keep track of the maximum window length encountered during the process.

• Runtime Complexity: O(n), where n is the length of the input array nums. Storage Complexity: O(1).

Code

    def longestOnes(nums, k):
    """
    Finds the maximum number of consecutive 1's in the array if you can flip at most k 0's.

    Args:
        nums: A binary array (list of 0s and 1s).
        k: The maximum number of 0s that can be flipped.

    Returns:
        The maximum number of consecutive 1's after flipping at most k 0's.
    """
    left = 0
    zeros_count = 0
    max_length = 0

    for right in range(len(nums)):
        if nums[right] == 0:
            zeros_count += 1

        while zeros_count > k:
            if nums[left] == 0:
                zeros_count -= 1
            left += 1

        max_length = max(max_length, right - left + 1)

    return max_length