Can Place Flowers Leetcode Problem 605 [Python Solution]

Welcome back!

Today, we're going to tackle the Can Place Flowers problem from LeetCode.

This problem is categorized as "Easy," but it's a bit more challenging than it appears at first glance.

We'll break it down step by step and provide you with an efficient Python solution.

By the end of this guide, you'll have a clear understanding of how to solve this problem.

Problem Overview

Question: You have a long flowerbed in which some of the plots are planted with flowers, while others are empty.

However, you can't plant flowers in adjacent plots.

Given an integer array flowerbed containing 0s (empty) and 1s (not empty), and an integer n, your task is to determine if you can plant n new flowers in the flowerbed without violating the rule of not planting flowers adjacent to each other.

Return True if it's possible, and False otherwise.

Example 1:

Input: flowerbed = [1, 0, 0, 0, 1], n = 1
Output: True

Example 2:

Input: flowerbed = [1, 0, 0, 0, 1], n = 2
Output: False

Constraints:

• 1 <= flowerbed.length <= 2 * 10^4
• flowerbed[i] is either 0 or 1.
• There are no two adjacent flowers in the flowerbed.
• 0 <= n <= flowerbed.length

Now, let's delve into understanding the problem and its constraints.

Understanding Constraints

The constraints play a crucial role in guiding our approach to solving the problem.

Let's break down the key constraints:

1. flowerbed.length can be as large as 20,000, which means our solution should be efficient and avoid unnecessary iterations.
2. The flowerbed array contains only 0s and 1s.

0 represents an empty plot, and 1 represents a plot with a flower.
3. There are no adjacent flowers in the flowerbed, which simplifies our task as we won't have to consider cases where we can't plant any flowers due to adjacent flowers.

Now that we've covered the problem's basics, it's time to explore different approaches to solving it.

Can Place Flowers LeetCode Problem Solution

We'll present multiple approaches to solve this problem, each with a different level of space complexity.

Let's dive in!

1. Brute Force Approach

In this approach, we'll iterate through the flowerbed array, check each position, and try to plant flowers following the given rules.

We'll use a Python code snippet to illustrate this approach:

def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
    empty = 0 if flowerbed[0] == 0 else 1
    
    for f in flowerbed:
        if f == 1:
            n -= (empty - 1) // 2  # Integer division, rounding toward zero
            empty = 0
        else:
            empty += 1
    
    n -= (empty) // 2
    return n <= 0

In this code, we initialize a variable empty to keep track of contiguous empty spots.

We start by checking the first position in the flowerbed.

If it's empty (0), we set empty to 1. Then, we iterate through the flowerbed.

If we encounter a flower (1), we calculate how many flowers we can plant in the contiguous empty spots (using integer division) and update the empty count to 0. If it's an empty spot, we increment the empty count.

Finally, we check if we were able to plant all required flowers (n <= 0) and return True or False accordingly.

This approach has a space complexity of O(1) since it doesn't require any additional data structures.

2. Efficient Approach with Iteration

Here's another efficient approach with O(1) space complexity, which directly iterates through the flowerbed:

def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
    for i in range(len(flowerbed)):
        if n == 0:
            return True
        if (
            (i == 0 or flowerbed[i - 1] == 0)
            and flowerbed[i] == 0
            and (i == len(flowerbed) - 1 or flowerbed[i + 1] == 0)
        ):
            flowerbed[i] = 1
            n -= 1

    return n == 0

In this code, we directly iterate through the flowerbed array and check each position.

We maintain a count n to track how many flowers we need to plant.

If n reaches 0, we return True because we've successfully planted all the required flowers.

The core logic checks if the current position and its adjacent positions are suitable for planting a flower.

If they are, we plant a flower and decrement n.

This approach also has O(1) space complexity because we don't use any additional data structures.

3. Solution with O(n) Space Complexity

In this approach, we'll use an additional list f to simplify the problem.

It has O(n) space complexity because we create a new list based on the original flowerbed.

Here's the code:

def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
    f = [0] + flowerbed + [0]

    for i in range(1, len(f) - 1):  # Skipping the first and last positions
        if f[i - 1] == 0 and f[i] == 0 and f[i + 1] == 0:
            f[i] = 1
            n -= 1
    return n <= 0

In this approach, we create a new list f by adding 0 at both the beginning and end of the original flowerbed.

This simplifies the problem because it ensures that we can plant flowers at the edges without violating the adjacent flowers rule.

We then iterate through f, checking each position to see if it's suitable for planting a flower.

If it is, we plant a flower and decrement n.

Finally, we check if we were able to plant all required flowers (n <= 0) and return True or False accordingly.

While this approach uses more space, it provides a straightforward solution to the problem.

Time and Space Complexity

Now, let's analyze the time and space complexity of our solutions:

Brute Force Approach:

• Time Complexity: O(n) – We iterate through the flowerbed array once, where n is the length of the array.
• Space Complexity: O(1) – We use a constant amount of space to store variables.

Efficient Approach with Iteration:

• Time Complexity: O(n) – We iterate through the flowerbed array once, where n is the length

  of the array.

• Space Complexity: O(1) – We use a constant amount of space to store variables.

Solution with O(n) Space Complexity:

• Time Complexity: O(n) – We iterate through the modified f array once, where n is the length of the array.
• Space Complexity: O(n) – We create an additional list f based on the original flowerbed.

Reasoning Behind Our Approach

Our solutions are based on the observation that we can simplify the problem by adding artificial empty spots at the beginning and end of the flowerbed.

This allows us to treat all positions as if they have adjacent empty spots, making it easier to determine where we can plant flowers.

The brute force approach and the efficient approach with iteration demonstrate different ways to track the contiguous empty spots and calculate the number of flowers that can be planted in each position.

These solutions ensure that we adhere to the no-adjacent-flowers rule while efficiently finding the answer.

The solution with O(n) space complexity provides a clear and concise way to approach the problem by creating a modified list f where the boundaries are treated as empty spots.

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Conclusion

In this guide, we've explored the Can Place Flowers problem on LeetCode, categorized as an "Easy" problem.

We've provided multiple solutions, each with its own space complexity:

• A brute force approach with O(1) space complexity that efficiently tracks contiguous empty spots.
• An efficient approach with O(1) space complexity that iterates through the flowerbed.
• A solution with O(n) space complexity that uses a modified list to simplify the problem.

We hope this guide has helped you understand the problem and its constraints and provided you with clear, efficient solutions.

If you have any questions, suggestions, or comments, please feel free to share them below.

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For the original problem statement and to practice solving it on LeetCode, follow this link: Can Place Flowers LeetCode Problem.

Happy coding!