K Empty Slots

Hard

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ArraysBinary SearchSliding Windows

You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are off. We turn on exactly one bulb everyday until all bulbs are on.

You are given an array bulbs of length n where bulbs[i] = x means that on the i^th day, we will turn on the bulb at position x where i is 0-indexed and x is 1-indexed.

Given an integer k, return the minimum day number such that there exist two turned-on bulbs that have exactly k bulbs between them that are all off.

If there isn't such day, return -1.

Example 1:

Input: bulbs = [1,3,2], k = 1
Output: 2
Explanation:
On the first day: bulbs[0] = 1, first bulb is turned on: [1,0,0]
On the second day: bulbs[1] = 3, third bulb is turned on: [1,0,1]
On the third day: bulbs[2] = 2, second bulb is turned on: [1,1,1]
We return the second day because there were two on bulbs with one off bulb between them.

Example 2:

Input: bulbs = [1,2,3], k = 1
Output: -1

Constraints:

n == bulbs.length
1 <= n <= 2 * 10⁴
1 <= bulbs[i] <= n
bulbs is a permutation of numbers from 1 to n.
0 <= k <= 2 * 10⁴

Try coding it in LeetCode

K Empty Slots

Hard

Asked by:

3 views

Topics:

ArraysBinary SearchSliding Windows

You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are off. We turn on exactly one bulb everyday until all bulbs are on.

You are given an array bulbs of length n where bulbs[i] = x means that on the i^th day, we will turn on the bulb at position x where i is 0-indexed and x is 1-indexed.

Given an integer k, return the minimum day number such that there exist two turned-on bulbs that have exactly k bulbs between them that are all off.

If there isn't such day, return -1.

Example 1:

Input: bulbs = [1,3,2], k = 1
Output: 2
Explanation:
On the first day: bulbs[0] = 1, first bulb is turned on: [1,0,0]
On the second day: bulbs[1] = 3, third bulb is turned on: [1,0,1]
On the third day: bulbs[2] = 2, second bulb is turned on: [1,1,1]
We return the second day because there were two on bulbs with one off bulb between them.

Example 2:

Input: bulbs = [1,2,3], k = 1
Output: -1

Constraints:

n == bulbs.length
1 <= n <= 2 * 10⁴
1 <= bulbs[i] <= n
bulbs is a permutation of numbers from 1 to n.
0 <= k <= 2 * 10⁴

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the integers representing the flower planting days?
Is it possible for the input array to be empty or null?
If no 'k empty slots' exist, what should I return?
Is the input array guaranteed to contain only positive integers representing planting days?
Are the planting days in the input array distinct, or can a flower be planted on the same day as another?

Brute Force Solution

Approach

The goal is to find the earliest day where there are two flowers blooming with exactly 'k' empty spots between them. The brute force approach involves checking every possible consecutive set of days to see if the condition is met.

Here's how the algorithm would work step-by-step:

Start by considering the first two flower bloom days.
Check if the number of empty spots between them is exactly 'k'. If so, we have a potential answer.
Now, consider the first three flower bloom days, then the first four, and so on, checking every possible consecutive subset of days.
For each subset, verify that the flowers blooming on the edges of the subset have exactly 'k' empty spots between them.
Also, for each subset, confirm that no other flowers bloomed earlier than the edge flowers, within the empty slots.
If we find a subset that meets both conditions, record the last bloom day in the subset as a possible answer.
After checking all possible subsets of days, select the earliest bloom day from our recorded possible answers. This is the earliest day that fulfills the requirements.

Code Implementation

def k_empty_slots_brute_force(flowers, k_empty):
    number_of_days = len(flowers)
    earliest_day = float('inf')

    for start_day in range(number_of_days - 1):
        for end_day in range(start_day + 1, number_of_days):
            # Iterate through possible subsets of days
            subset_days = flowers[start_day:end_day + 1]
            first_flower_position = flowers[start_day]
            last_flower_position = flowers[end_day]

            if abs(first_flower_position - last_flower_position) != k_empty + 1:
                continue

            valid_subset = True
            # Check if no flower bloomed earlier in the k_empty slots
            for day_index in range(start_day + 1, end_day):
                flower_position = flowers[day_index]

                if (first_flower_position < flower_position < last_flower_position and
                        (flower_position < min(first_flower_position, last_flower_position) or
                         flower_position > max(first_flower_position, last_flower_position))):
                    valid_subset = False
                    break

                if (first_flower_position < last_flower_position and
                        first_flower_position < flower_position < last_flower_position):
                    valid_subset = False
                    break
                elif (first_flower_position > last_flower_position and
                      first_flower_position > flower_position > last_flower_position):
                    valid_subset = False
                    break

            if valid_subset:
                # Consider the last bloom day as a candidate
                earliest_day = min(earliest_day, end_day + 1)

    if earliest_day == float('inf'):
        return -1
    else:
        return earliest_day

Big(O) Analysis

Time Complexity

O(n^3) – The outer loop iterates through all possible starting bloom days, resulting in approximately n iterations. The inner loop checks every consecutive subset of bloom days from the starting bloom day to the end, taking up to n iterations. Inside the inner loop, verifying that the empty slots between the edge flowers remain empty from other bloom days requires iterating through the 'k' empty slots, which can be as large as n in the worst case. Therefore the total complexity approximates n * n * n, resulting in O(n^3).

Space Complexity

O(1) – The brute force approach, as described, primarily involves iterating and comparing values. While the explanation mentions considering subsets and recording 'possible answers', the key aspect is that we are looking for the earliest day. This implies that we would only need to store a single variable, or a constant number of variables, to keep track of the current earliest day that satisfies the condition and potentially a few indices. The algorithm does not create auxiliary data structures like arrays or hash maps whose sizes depend on the input size N. Therefore, the auxiliary space complexity is constant, O(1).

Optimal Solution

Approach

We want to find the earliest day where there's a group of 'k' empty flower slots between two blooming flowers. Instead of checking every single day, we use a sliding window approach to efficiently track bloom statuses.

Here's how the algorithm would work step-by-step:

Think of the flower slots as points on a line, and the blooming flowers as walls that divide the line into segments.
Imagine a 'window' of size 'k+2' sliding across the line. The window always includes two blooming flowers, one at each end.
As the window slides, we check if all the flower slots inside the window (excluding the ends) are empty (not yet bloomed).
To avoid checking all flowers every time the window moves, we keep track of when each flower blooms.
When the window slides, we only need to consider flowers that bloomed before the flowers at the ends of the window.
If we find a window where all the flowers inside bloomed *after* the flowers at the ends, we've found a segment of 'k' empty slots between two blooming flowers.
We want the *earliest* such day, so we keep track of the minimum possible bloom day when we find the window.

Code Implementation

def k_empty_slots(flowers, number_of_empty_slots):
    number_of_flowers = len(flowers)
    bloom_days = [0] * number_of_flowers

    for day, flower in enumerate(flowers):
        bloom_days[flower - 1] = day + 1

    minimum_day = float('inf')

    left_flower = 0
    right_flower = number_of_empty_slots + 1

    # Slide the window across all possible flower positions
    while right_flower < number_of_flowers:
        
        valid_range = True

        # Check flowers inside window
        for flower_to_check in range(left_flower + 1, right_flower):

            # Validate if bloom day is before window bounds
            if bloom_days[flower_to_check] < bloom_days[left_flower] or \
               bloom_days[flower_to_check] < bloom_days[right_flower]:
                valid_range = False
                break

        # If all empty slots bloomed after the window bounds
        if valid_range:
            minimum_day = min(minimum_day, max(bloom_days[left_flower], bloom_days[right_flower]))

        left_flower += 1
        right_flower += 1

    if minimum_day == float('inf'):
        return -1
    else:
        return minimum_day

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the 'days' array once, effectively simulating a sliding window of size k+2. For each window position, it checks if all flowers within that window bloomed after the boundary flowers of that window. The check for the flowers in the window is bounded by the size of the input array. Because each flower is visited only a constant number of times and the sliding window itself takes O(n) time. The total number of operations is thus proportional to the size of the 'days' array, n. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm keeps track of when each flower blooms using an auxiliary array of size N (where N is the number of flower slots). This array stores the bloom day for each flower, allowing the algorithm to quickly check the bloom status of flowers within the sliding window. The space used by this bloom day array is directly proportional to the number of flower slots. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty flowers array	Return -1 immediately since no flowers can bloom to satisfy the condition.
k is zero	The problem is defined for k>0, treat k=0 the same as the base case where a single valid slot always exists, returning 1 if n >=2, or -1 if n<2
k is larger than array length - 2	Return -1 since it's impossible to find k empty slots between two blooming flowers.
flowers array contains duplicate days	Only the earliest bloom day should be considered, as the flower cannot bloom again, so replace the old bloom day with the new one if it's earlier.
Integer overflow when calculating indices with large array size or k	Ensure indices are calculated using appropriate data types (e.g., long) to prevent overflow if the flower array or k is excessively large.
No valid solution exists	If the entire array is iterated and no k empty slots are found, return -1.
Flowers bloom out of order	The solution must correctly determine the first day when k consecutive empty slots exist, regardless of the initial order of bloom days.
Large flowers array exceeding memory constraints	Consider using a more memory-efficient data structure like an interval tree if the size of the flowers array is extremely large and memory becomes a constraint.