605. Can Place Flowers

Updated: 2024-03-12
2 min read
[Algorithms Easy]

LeetCode problem 605

In this problem, we are given a flowerbed represented by an integer array flowerbed, where 0 represents an empty plot and 1 represents a plot with a flower.

We need to determine if we can plant n new flowers in the flowerbed without violating the rule that no two adjacent plots can have flowers.

Naive Solution

To solve this problem, we can use a greedy approach. We iterate through the flowerbed and check each plot. If a plot is empty and its adjacent plots are also empty, we can plant a flower in that plot.

We repeat this process until we have planted all n flowers or we have checked all plots in the flowerbed.


To solve this problem, we will define a helper function check_neighbors that checks if a plot can be planted with a flower and updates the flowerbed and n accordingly.


  1. Initialize a variable i to iterate through the flowerbed.
  2. For each plot in the flowerbed: 2.1. If the current plot is empty and its previous plot is also empty (i.e., flowerbed[i-1] == 0), call the check_neighbors function. 2.2. Otherwise, if the current plot is empty and its next plot is also empty (i.e., flowerbed[i+1] == 0), call the check_neighbors function.
  3. Return True if n is less than or equal to 0 (i.e., all flowers have been planted), otherwise return False.


def canPlaceFlowers(flowerbed, n):
    def check_neighbors(n):
        if i < len(flowerbed) - 1:
            if flowerbed[i+1] == 0:
                flowerbed[i] = 1
                n -= 1
            flowerbed[i] = 1
            n -= 1
        return n

    for i in range(len(flowerbed)):
        if flowerbed[i] == 0:
            if i > 0:
                if flowerbed[i-1] == 0:
                    n = check_neighbors(n)
                n = check_neighbors(n)
    return n <= 0