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This approach involves a single pass through the flowerbed array, considering each plot one by one. For each zero found, we check whether it's possible to plant a flower by ensuring that both neighboring plots (if they exist) are not planted. If it's possible, we plant the flower (i.e., change the zero to one) and reduce the count of n. The process stops early if n becomes zero. This greedy method ensures efficiency by trying to plant a flower at each opportunity.
Time Complexity: O(m), where m is the length of the flowerbed array.
Space Complexity: O(1), as no extra space is used except for variables.
1class Solution:
2 def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
3 for i in range(len(flowerbed)):
4 if flowerbed[i] == 0:
5 isPrevEmpty = (i == 0) or (flowerbed[i - 1] == 0)
6 isNextEmpty = (i == len(flowerbed) - 1) or (flowerbed[i + 1] == 0)
7 if isPrevEmpty and isNextEmpty:
8 flowerbed[i] = 1
9 n -= 1
10 if n == 0:
11 return True
12 return n <= 0The Python solution leverages list indexing to easily navigate and modify the flowerbed array. It succinctly checks and modifies the array to plant flowers, aiming to plant 'n' flowers while ensuring no adjacent flowers appear.
This approach calculates the maximum number of flowers that can be planted by counting the stretches of zeros between planted flowers or array boundaries. For each encounter of a continuous stretch of zeros, determine how many flowers can be planted and accumulate this count until it reaches or exceeds n, or if it proves impossible.
Time Complexity: O(m), where m is the length of the flowerbed.
Space Complexity: O(1), only a constant amount of extra space is required.
1
This C solution leverages a counting mechanism, iterating through the flowerbed and adding cushions (zeros) at the boundaries to simplify calculations. The number of available spots for planting is incrementally calculated and checked against n.