How do you solve Can Place Flowers on LeetCode?

Can Place Flowers (LeetCode #605) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the constraints of adjacent plots is crucial for determining valid planting positions.

What is the time complexity of Can Place Flowers?

The optimal solution for Can Place Flowers runs in O(n) time and uses O(1) space. This approach runs in O(n) time because we only make a single pass through the flowerbed array, checking each plot once.

What is the brute force solution for Can Place Flowers?

The brute force approach for Can Place Flowers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible position in the flowerbed to see if we can plant a flower. It tries to place a flower in each empty spot and checks if the adjacent spots are empty, which can be slow. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Can Place Flowers?

Can Place Flowers uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Can Place Flowers?

Always clarify constraints and edge cases with the interviewer. Explain your thought process as you code, especially when transitioning from brute-force to optimal solutions. Practice dry runs of your solution to ensure you can explain it clearly.

What are common mistakes when solving Can Place Flowers?

Not considering edge cases where the flowerbed starts or ends with empty plots. Failing to update the last planted position correctly can lead to incorrect counts.

#605

Can Place Flowers

Easy

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the flowerbed, counting how many flowers can be planted without checking adjacent plots repeatedly. This is more efficient and straightforward.

⚙️

Algorithm

6 steps

1Step 1: Initialize a count for flowers planted and a variable to track the last position of a planted flower.
2Step 2: Iterate through the flowerbed array.
3Step 3: For each empty plot, check if it can be planted based on the last planted position.
4Step 4: If it can be planted, update the last planted position and increment the count.
5Step 5: If the count of planted flowers reaches n, return true.
6Step 6: After the loop, return false if n is not met.

solution.py11 lines

1def canPlaceFlowers(flowerbed, n):
2    count = 0
3    last_position = -2
4    for i in range(len(flowerbed)):
5        if flowerbed[i] == 0 and i - last_position > 1:
6            flowerbed[i] = 1
7            last_position = i
8            count += 1
9            if count >= n:
10                return True
11    return count >= n

ℹ

Complexity note: This approach runs in O(n) time because we only make a single pass through the flowerbed array, checking each plot once.

1Understanding the constraints of adjacent plots is crucial for determining valid planting positions.
2Efficiently tracking the last planted position can reduce unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.