You are given a 0-indexed binary string floor
, which represents the colors of tiles on a floor:
-
floor[i] = '0'
denotes that theith
tile of the floor is colored black. - On the other hand,
floor[i] = '1'
denotes that theith
tile of the floor is colored white.
You are also given numCarpets
and carpetLen
. You have numCarpets
black carpets, each of length carpetLen
tiles. Cover the tiles with the given carpets such that the number of white tiles still visible is minimum. Carpets may overlap one another.
Return the minimum number of white tiles still visible.
Example 1:
Input: floor = "10110101", numCarpets = 2, carpetLen = 2
Output: 2
Explanation:
The figure above shows one way of covering the tiles with the carpets such that only 2 white tiles are visible.
No other way of covering the tiles with the carpets can leave less than 2 white tiles visible.
Example 2:
Input: floor = "11111", numCarpets = 2, carpetLen = 3
Output: 0
Explanation:
The figure above shows one way of covering the tiles with the carpets such that no white tiles are visible.
Note that the carpets are able to overlap one another.
Constraints:
-
1 <= carpetLen <= floor.length <= 1000
-
floor[i]
is either'0'
or'1'
. -
1 <= numCarpets <= 1000
SOLUTION:
class Solution:
def minTiles(self, floor: str, numCarpets: int, carpetLen: int, i, n) -> int:
key = (i, numCarpets)
if i >= n:
return 0
if key in self.cache:
return self.cache[key]
if numCarpets <= 0:
val = floor[i:].count("1")
self.cache[key] = val
return val
a = float('inf')
if floor[i] == "1":
a = self.minTiles(floor, numCarpets - 1, carpetLen, i + carpetLen, n)
b = self.minTiles(floor, numCarpets, carpetLen, i + 1, n)
if floor[i] == "1":
b += 1
val = min(a, b)
self.cache[key] = val
return val
def minimumWhiteTiles(self, floor: str, numCarpets: int, carpetLen: int, i = 0) -> int:
self.cache = {}
return self.minTiles(floor, numCarpets, carpetLen, 0, len(floor))
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