3149. Find the Minimum Cost Array Permutation
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Difficulty: hard Acceptance: 24% Topics: Array, Dynamic Programming, Bit Manipulation, Bitmask
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Reference solution (spoiler · python)
# Time: O((n-1)^2 * 2^(n-1))
# Space: O((n-1) * 2^(n-1))
# dp, backtracing
class Solution(object):
def findPermutation(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
INF = float("inf")
n = len(nums)
dp = [[(INF, -1) for _ in xrange(n-1)] for _ in xrange(1<<(n-1))]
for i in xrange(n-1):
dp[1<<i][i] = (abs((i+1)-nums[0]), -1)
for mask in xrange(1<<(n-1)):
for i in xrange(n-1):
if mask&(1<<i) == 0:
continue
for j in xrange(n-1):
if j == i or mask&(1<<j) == 0:
continue
dp[mask][i] = min(dp[mask][i], (dp[mask^(1<<i)][j][0]+abs((i+1)-nums[j+1]), j))
_, i = min((dp[-1][i][0]+abs(0-nums[i+1]), i) for i in xrange(n-1))
result = [0]
mask = (1<<(n-1))-1
while i != -1:
result.append(i+1)
mask, i = mask^(1<<i), dp[mask][i][1]
return result
Solution from kamyu104/LeetCode-Solutions · MIT
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