Max Consecutive Ones III Interview Question for Amazon

def longestOnes(nums, k):
    left = 0
    zeros = 0
    max_length = 0

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

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

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

    return max_length

Explanation:

The function longestOnes takes a binary array nums and an integer k as input. It returns the maximum number of consecutive 1s in the array if you can flip at most k 0s.

The algorithm uses a sliding window approach. The left pointer represents the left boundary of the window, and the right pointer represents the right boundary of the window. The zeros variable keeps track of the number of 0s in the current window.

For each element in the array, we move the right pointer to the right. If the current element is 0, we increment the zeros counter. If the number of 0s in the window exceeds k, we move the left pointer to the right until the number of 0s in the window is less than or equal to k. During this process, if the element at the left pointer is 0, we decrement the zeros counter.

At each step, we update the max_length variable to be the maximum of the current max_length and the length of the current window (right - left + 1).

Finally, we return the max_length.

Example:

nums = [1,1,1,0,0,0,1,1,1,1,0]
k = 2
print(longestOnes(nums, k)) # Output: 6

nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1]
k = 3
print(longestOnes(nums, k)) # Output: 10

Brute Force Solution:

A brute force solution would be to iterate through all possible subarrays and, for each subarray, count the number of 0s. If the number of 0s is less than or equal to k, we update the max_length variable.

def longestOnes_brute_force(nums, k):
    max_length = 0
    for i in range(len(nums)):
        for j in range(i, len(nums)):
            subarray = nums[i:j+1]
            zeros = subarray.count(0)
            if zeros <= k:
                max_length = max(max_length, len(subarray))
    return max_length

Big(O) Time Complexity Analysis:

The time complexity of the sliding window solution is O(n), where n is the length of the input array. This is because we iterate through the array once with the right pointer, and the left pointer can only move from left to right once in the array. Thus, both left and right pointers each visit each element a single time.

The time complexity of the brute force solution is O(n^3), where n is the length of the input array. This is because we iterate through all possible subarrays (O(n^2)), and for each subarray, we count the number of 0s (O(n)).

Big(O) Space Complexity Analysis:

The space complexity of the sliding window solution is O(1), because we only use a constant amount of extra space for the left, right, zeros, and max_length variables.

The space complexity of the brute force solution is also O(1), because we only use a constant amount of extra space for the max_length variable. The subarray does create a slice of nums, but in terms of asymptotic complexity, it is bounded by O(n), but is dominated by the O(n^3) time complexity.

Edge Cases:

Empty array: If the input array is empty, the function should return 0.
k is 0: If k is 0, the function should return the length of the longest subarray of 1s.
k is greater than the number of 0s in the array: If k is greater than the number of 0s in the array, the function should return the length of the array.
Array contains only 1s: If the array contains only 1s, the function should return the length of the array.
Array contains only 0s: If the array contains only 0s, the function should return k if k is less than the length of the array, otherwise, it returns the length of the array.