LeetCode 683: K Empty Slots Solution in Python – A Step-by-Step Guide

What Is LeetCode 683: K Empty Slots?

In LeetCode 683: K Empty Slots, you’re given an array flowers of length n, where flowers[i] is the position (1-based) where a flower blooms on day i+1, and an integer k. Your task is to find the earliest day when there are exactly k empty slots between two bloomed flowers. Return this day as an integer, or -1 if no such day exists. For example, with flowers = [1, 3, 2] and k = 1, on day 2 (after blooming at 1 and 3), there’s 1 empty slot (position 2) between 1 and 3, so return 2. The problem tests your ability to analyze positional sequences efficiently.

Problem Statement

• Input:
• flowers: List of integers (blooming positions, 1-based).
• k: Integer (target number of empty slots).
• Output: An integer—earliest day with k empty slots between two flowers, or -1 if impossible.
• Rules:
• flowers[i]: Position where flower blooms on day i+1.
• Positions are a permutation of 1 to n.
• Find day when two bloomed positions have exactly k empty slots between them.
• Return earliest such day or -1.

Constraints

• 1 ≤ flowers.length ≤ 2 × 10⁴.
• flowers[i] is a permutation of [1, 2, ..., n].
• 0 ≤ k ≤ n - 1.

Examples

• Input: flowers = [1, 3, 2], k = 1
• Output: 2 (Day 2: 1 and 3 bloom, 2 empty, 1 slot between)
• Input: flowers = [1, 2, 3], k = 1
• Output: -1 (No gaps of 1 slot)
• Input: flowers = [3, 1, 5, 4, 2], k = 2
• Output: 4 (Day 4: 1, 3, 4, 5 bloom, 2 slots between 1 and 4)

These examples set the garden—let’s solve it!

Understanding the Problem: Finding K Empty Slots

To solve LeetCode 683: K Empty Slots in Python, we need to determine the earliest day when two bloomed positions in the flowers array have exactly k empty slots between them. A brute-force approach—checking all pairs each day—would be O(n² * n) for n days and n² pairs, inefficient for n ≤ 2 × 10⁴. Since blooms occur sequentially and we’re looking for gaps, a sliding window or position tracking can optimize this. We’ll use:

• Best Solution (Sliding Window with Position Array): O(n) time, O(n) space—fast and clever.
• Alternative Solution (Brute-Force with Daily Check): O(n²) time, O(n) space—simple but slow.

Let’s start with the sliding window solution, breaking it down for beginners.

Best Solution: Sliding Window with Position Array

Why This Is the Best Solution

The sliding window with position array approach is the top choice for LeetCode 683 because it’s highly efficient—O(n) time with O(n) space—and uses a brilliant insight: track bloom days in a position array and slide a window of size k+2 to find adjacent blooms with exactly k empty slots, leveraging the fact that only boundary blooms matter. It fits constraints (n ≤ 2 × 10⁴) perfectly and is intuitive once you see the window logic. It’s like sliding a ruler across the garden to measure gaps between flowers!

How It Works

Think of this as a gap slider:

• Step 1: Build Position Array:
• days[pos-1] = day: Day when position pos blooms.
• Step 2: Slide Window:
• Window size = k + 2 (2 blooms + k slots).
• For each window [left, right]:
• Check if left and right bloom before any middle position.
• If so, gap = right - left - 1 = k, record day = max(days[left], days[right]).
• Step 3: Find Earliest Day:
• Return min day found, or -1 if none.
• Step 4: Return Result:
• Return earliest valid day.

It’s like a window measurer—slide and check!

Step-by-Step Example

Example: flowers = [1, 3, 2], k = 1

• Step 1: Days array:
• flowers[0] = 1, days[0] = 1.
• flowers[1] = 3, days[2] = 2.
• flowers[2] = 2, days[1] = 3.
• days = [1, 3, 2].
• Step 2: Window size = k + 2 = 3:
• Window [0, 2]: left=0 (day 1), right=2 (day 2), middle=1 (day 3).
• Days: 1, 3, 2.
• Left (1) and right (2) < middle (3), gap = 2 - 0 - 1 = 1 = k.
• Day = max(1, 2) = 2.
• Step 3: Min day = 2.
• Step 4: Return 2.
• Output: 2.

Example: flowers = [1, 2, 3], k = 1

• Step 1: days = [1, 2, 3].
• Step 2: Window size = 3:
• Window [0, 2]: 1, 2, 3.
• Middle (2) < right (3), no k=1 gap (gap = 2).
• Step 3: No valid day, return -1.
• Output: -1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def kEmptySlots(self, flowers: List[int], k: int) -> int:
        # Step 1: Build position array (days when each position blooms)
        n = len(flowers)
        days = [0] * n
        for day, pos in enumerate(flowers):
            days[pos - 1] = day + 1

        # Step 2: Slide window of size k + 2
        min_day = float('inf')
        left = 0
        right = k + 1

        while right < n:
            # Check if left and right are min/max with k slots between
            valid = True
            for i in range(left + 1, right):
                if days[i] < days[left] or days[i] < days[right]:
                    valid = False
                    break
            if valid:
                min_day = min(min_day, max(days[left], days[right]))

            # Slide window: Move to next earliest bloom
            if right + 1 < n and days[left] < days[right]:
                left += 1
            else:
                left = right
                right = left + k + 1

        # Step 3: Return earliest day or -1
        return min_day if min_day != float('inf') else -1

• Lines 4-8: Build: days[pos-1] = day + 1 for 1-based positions.
• Lines 11-14: Init: Window size k+2, track min day.
• Lines 16-29: Slide:
• Check window validity, update min_day, slide based on earliest bloom.
• Line 32: Return: Min day or -1 if none found.
• Time Complexity: O(n)—single pass with window sliding.
• Space Complexity: O(n)—days array.

This is like a garden gapper—slide and spot!

Alternative Solution: Brute-Force with Daily Check

Why an Alternative Approach?

The brute-force with daily check approach simulates each day—O(n²) time, O(n) space. It’s simpler, building the garden state daily and checking all pairs, but inefficient for large n. It’s like watching the garden grow day-by-day and measuring gaps!

How It Works

Picture this as a daily gardener:

• Step 1: Build bloom state array.
• Step 2: Process each day:
• Mark bloom position, check all pairs for k slots.
• Step 3: Return earliest day or -1.

It’s like a bloom watcher—grow and measure!

Step-by-Step Example

Example: Same as above

• Step 1: Blooms = [0, 0, 0].
• Step 2: Process:
• Day 1: [1, 0, 0], no pairs.
• Day 2: [1, 0, 1], gap = 1, return 2.
• Output: 2.

Code for Brute-Force Approach

Copy codeclass Solution:
    def kEmptySlots(self, flowers: List[int], k: int) -> int:
        # Step 1: Initialize bloom state
        n = len(flowers)
        blooms = [0] * n

        # Step 2: Process each day
        for day, pos in enumerate(flowers, 1):
            blooms[pos - 1] = 1
            # Check all pairs
            for i in range(n):
                if blooms[i]:
                    for j in range(i + 1, n):
                        if blooms[j]:
                            gap = j - i - 1
                            if gap == k:
                                return day
        # Step 3: No valid day found
        return -1

• Lines 4-6: Init: Bloom array.
• Lines 9-20: Process:
• Mark bloom, check pairs for k slots, return day if found.
• Line 23: Return -1 if no match.
• Time Complexity: O(n²)—n days, O(n) pair checks.
• Space Complexity: O(n)—bloom array.

It’s a daily checker—bloom and scan!

Comparing the Two Solutions

• Sliding Window (Best):
• Pros: O(n) time, O(n) space, fast and elegant.
• Cons: Window logic less obvious.
• Brute-Force (Alternative):
• Pros: O(n²) time, O(n) space, straightforward.
• Cons: Slower, less efficient.

Sliding window wins for speed.

Additional Examples and Edge Cases

• Input: flowers = [1], k = 0
• Output: -1.
• Input: flowers = [1, 2], k = 0
• Output: 1.

Sliding window handles these well.

Complexity Breakdown

• Sliding Window: Time O(n), Space O(n).
• Brute-Force: Time O(n²), Space O(n).

Sliding window is optimal.

Key Takeaways

• Sliding Window: Gap sliding—smart!
• Brute-Force: Daily checking—clear!
• Gardens: Blooming is fun.
• Python Tip: Windows rock—see Python Basics.

Final Thoughts: Tend That Garden

LeetCode 683: K Empty Slots in Python is a fun gardening challenge. Sliding window with position array offers speed and elegance, while brute-force provides a clear baseline. Want more? Try LeetCode 239: Sliding Window Maximum or LeetCode 76: Minimum Window Substring. Ready to plant? Head to Solve LeetCode 683 on LeetCode and find those slots today!