2463. Minimum Total Distance Traveled
Read the full problem statement on LeetCode.
Difficulty: hard Acceptance: 59% Topics: Array, Dynamic Programming, Sorting
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(mlogm + nlogn + m * n)
# Space: O(n)
import collections
# sort, dp, prefix sum, mono deque
class Solution(object):
def minimumTotalDistance(self, robot, factory):
"""
:type robot: List[int]
:type factory: List[List[int]]
:rtype: int
"""
robot.sort(), factory.sort()
dp = [float("inf")]*(len(robot)+1) # dp[j] at i: min of factory[:i+1] and robot[:j]
dp[0] = 0
for i in xrange(len(factory)):
prefix = 0
dq = collections.deque([(dp[0]-prefix, 0)]) # pattern of min in the sliding window with size (limit+1)
for j in xrange(1, len(robot)+1):
prefix += abs(robot[j-1]-factory[i][0])
if j-dq[0][1] == factory[i][1]+1:
dq.popleft()
while dq and dq[-1][0] >= dp[j]-prefix:
dq.pop()
dq.append((dp[j]-prefix, j))
dp[j] = dq[0][0]+prefix
return dp[-1]
# Time: O(mlogm + nlogn + m * n * l), l is the max limit
# Space: O(n)
import collections
# sort, dp
class Solution2(object):
def minimumTotalDistance(self, robot, factory):
"""
:type robot: List[int]
:type factory: List[List[int]]
:rtype: int
"""
robot.sort(), factory.sort()
dp = [float("inf")]*(len(robot)+1) # dp[j] at i: min of factory[:i+1] and robot[:j]
dp[0] = 0
for i in xrange(len(factory)):
for j in reversed(xrange(1, len(robot)+1)):
curr = 0
for k in xrange(min(factory[i][1], j)+1):
dp[j] = min(dp[j], dp[j-k]+curr)
if (j-1)-k >= 0:
curr += abs(robot[(j-1)-k]-factory[i][0])
return dp[-1]
Solution from kamyu104/LeetCode-Solutions · MIT