Problem Description

The problem is about finding the longest sequence of consecutive 1s in a binary array, under the condition that we are allowed to flip at most one 0 to 1. This task tests our ability to manipulate subarrays in a binary context and optimize our approach to account for small alterations in the array to achieve the desired outcome.

Intuition

To find the solution to this problem, we look to a two-pointer approach. Essentially, we're trying to maintain a window that can include up to one 0 to maximize the length of consecutive ones.

Let's think of this as a sliding window that starts from the left end of the array and expands to the right. As we move the right pointer (r) through the array, we keep track of the number of 0s we've included in the window. We initialize k to 1, because we're allowed to flip at most one 0.

When we encounter a 0, we decrement k. If k becomes negative, it means we've encountered more than one zero, and thus, we need to shrink our window from the left by moving the left pointer (l) to the right.

We continue expanding and shrinking the window as needed to always maintain at most one 0 within it. Throughout the process, we keep track of the maximum window size we've seen, which gives us the longest sequence of consecutive 1s with at most one 0 flipped.

When we have finished iterating through the array with our right pointer, the length of the window (r - l) is our maximum number of consecutive 1s, accounting for flipping at most one 0. The code does not explicitly keep a maximum length variable; instead, it relies on the fact that the window size can only increase or stay the same throughout the iteration, because once the window shrinks (when k becomes negative), it will then begin to expand again from a further position in the array.

Learn more about Dynamic Programming and Sliding Window patterns.

Solution Approach

The solution uses a two-pointer technique to efficiently find the maximum length of a subarray with consecutive 1s after at most one 0 has been flipped.

Here is a step-by-step breakdown of the implementation:

1. Initialize two pointers, l and r, to point at the start of the array. These pointers represent the left and right bounds of our sliding window, respectively.
2. Define a variable k with an initial value of 1, which represents how many 0s we can flip. Since we are allowed to flip at most one 0, we start with k = 1.
3. Iterate through the array by moving the right pointer r towards the end. This loop continues until r reaches the end of the array.
4. Within the loop, check if the current number pointed to by r is a 0. If it is, decrement k.
5. If k becomes less than 0, it signifies that our window contains more than one 0, which is not allowed. Thus, we need to increment l, our left pointer, to narrow the window and possibly discard a 0 from the window:
   • Inside another loop, we check if the number at the current l position is a 0. If it is, we increment k because we are "flipping" the 0 back and excluding it from the current window.
   • Then, we move l one step to the right by increasing its value by 1.
6. Increment r each time to examine the next element in the array.
7. After the while loop exits, calculate the length of the current window (r - l). Since the right pointer has reached the end of the array, this value equals the maximum size of the window we found throughout our traversal.

Important notes regarding the code and algorithm:

• There is no need to explicitly keep track of the maximum window size during the loop because the window can only grow or maintain its size. It shrinks only when necessary to exclude an excess 0.

• The solution leverages the problem constraint. Knowing that only one 0 can be flipped, it eliminates the need for complex data structures or algorithms. A straightforward integer (k) is enough to keep track of the condition being met.

• This solution has a time complexity of O(n) where n is the length of the array. This is because each element is checked once by the right pointer r.

• The space complexity of this algorithm is O(1) as it operates with a constant number of extra variables regardless of the input size.

By keeping the algorithm simple and avoiding unnecessary data structures, the solution remains elegant and efficient.

Example Walkthrough

Let’s use a small example to illustrate the solution approach. Consider the binary array [1, 0, 1, 1, 0, 1, 1, 1, 0, 1], where we are allowed to flip at most one 0 to achieve the longest sequence of 1s.

Following the solution steps:

1. Initialize two pointers l and r to 0 (the start of the array).

2. Set k to 1 since we can flip at most one 0.

3. As we start iterating with r, the subarray from l to r initially grows without any issue since the first element is a 1.

4. When r=1, the element is 0. We decrement k to 0 because we used our allowance to flip a 0.

5. We continue to move r. The window now has consecutive 1s and looks like this: [1, 1 (flipped), 1, 1].

6. At r=4, we encounter another 0 and k is already 0. Now we must shrink the window from the left. We move l one step to the right. The subarray does not contain zero at l=1, hence no change in k.

7. Moving l again to 2, we encounter our previously flipped 0. We increment k back to 1.

8. The window is now [1, 1, 0, 1, 1] from l=2 to r=5, and we can flip the 0 at r=4 because k=1.

9. We continue this process, moving r to the end, and each time we hit a 0 with k=0, we shift l to maintain only one flipped 0.

10. At the end, when r is just past the end of the array, the length of the window is calculated by r - l, giving us the longest sequence of consecutive 1s where at most one 0 was flipped.

In this example, the longest subarray we can form after flipping at most one 0 is [1, 1, 0 (flipped), 1, 1, 1] which starts at l=2 and ends just before r=8, resulting in a length of 6.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the input list nums. This efficiency is achieved because the code uses a two-pointer technique that iterates through the list only once. The pointers l (left) and r (right) move through the array without making any nested loops, thus each element is considered only once during the iteration.

The space complexity of the code is O(1), which means it uses constant additional space regardless of input size. No extra data structures that grow with the input size are used; the variables l, r, and k take up a constant amount of space.

Learn more about how to find time and space complexity quickly using problem constraints.