LeetCode 605: Can Place Flowers ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2022-08-28

Submissions: Runtime Error ×2 → Wrong Answer → Internal Error → Wrong Answer → Accepted → Wrong Answer ×5 → Accepted

Approach
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Greedy (Greedy): scanning left to right, plant whenever the plot is 0 and neither neighbor is 1 (treating outside the ends as empty), and set it to 1 so the next plot is blocked. Planting as early as possible never hurts, since a flower further left leaves more room on the right. Finally check that the count is >= n.

My Attempts
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I solved it in 2022 and again in 2025, failing 10 times in total:

2022: I first counted “empty plots not next to any flower”, but newly planted flowers block their neighbors too, so adjacent empty plots were both counted: [0, 0], n = 2 returned True instead of False. Padding with zeros and marking each planted plot passed.
2025: tracking prev_plant and next_plant, I first typed pre_plant, so the left neighbor always looked empty; after fixing that, the last line read plots <= n, the comparison reversed.

Solution
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class Solution:
    def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
        bed = [0] + flowerbed + [0]
        count = 0
        for i in range(1, len(bed) - 1):
            if bed[i - 1] == bed[i] == bed[i + 1] == 0:
                bed[i] = 1
                count += 1
        return count >= n

Complexity
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Time O(n), space O(n) for the sentinels; checking the bounds in place gives O(1).

Mistakes to Avoid
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Update the array after planting, or two adjacent empty plots both count.
The comparison is count >= n.

Interview Explanation
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With an empty sentinel at each end I scan left to right, planting and marking any plot with empty neighbors, since planting early never hurts, then check the count is at least n. Time O(n).