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maximum-compatibility-score-sum.py
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maximum-compatibility-score-sum.py
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# Time: O(m^2 * (n + m))
# Space: O(m^2)
import itertools
# weighted bipartite matching solution
class Solution(object):
def maxCompatibilitySum(self, students, mentors):
"""
:type students: List[List[int]]
:type mentors: List[List[int]]
:rtype: int
"""
# Template translated from:
# https://github.com/kth-competitive-programming/kactl/blob/main/content/graph/WeightedMatching.h
def hungarian(a): # Time: O(n^2 * m), Space: O(n + m)
if not a:
return 0, []
n, m = len(a)+1, len(a[0])+1
u, v, p, ans = [0]*n, [0]*m, [0]*m, [0]*(n-1)
for i in xrange(1, n):
p[0] = i
j0 = 0 # add "dummy" worker 0
dist, pre = [float("inf")]*m, [-1]*m
done = [False]*(m+1)
while True: # dijkstra
done[j0] = True
i0, j1, delta = p[j0], None, float("inf")
for j in xrange(1, m):
if done[j]:
continue
cur = a[i0-1][j-1]-u[i0]-v[j]
if cur < dist[j]:
dist[j], pre[j] = cur, j0
if dist[j] < delta:
delta, j1 = dist[j], j
for j in xrange(m):
if done[j]:
u[p[j]] += delta
v[j] -= delta
else:
dist[j] -= delta
j0 = j1
if not p[j0]:
break
while j0: # update alternating path
j1 = pre[j0]
p[j0], j0 = p[j1], j1
for j in xrange(1, m):
if p[j]:
ans[p[j]-1] = j-1
return -v[0], ans # min cost
def score(s, m):
return sum(int(a == b) for a, b in itertools.izip(s, m))
return -hungarian([[-score(s, m) for m in mentors] for s in students])[0]
# Time: O(m * (n + 2^m))
# Space: O(2^m)
# dp solution
class Solution2(object):
def maxCompatibilitySum(self, students, mentors):
"""
:type students: List[List[int]]
:type mentors: List[List[int]]
:rtype: int
"""
def popcount(n): # Time: O(logn) ~= O(1) if n is a 32-bit number
result = 0
while n:
n &= n-1
result += 1
return result
def masks(vvi):
result = []
for vi in vvi:
mask, bit = 0, 1
for i in xrange(len(vi)):
if vi[i]:
mask |= bit
bit <<= 1
result.append(mask)
return result
nums1, nums2 = masks(students), masks(mentors)
dp = [(0, 0)]*(2**len(nums2))
for mask in xrange(len(dp)):
bit = 1
for i in xrange(len(nums2)):
if (mask&bit) == 0:
dp[mask|bit] = max(dp[mask|bit], (dp[mask][0]+(len(students[0])-popcount(nums1[dp[mask][1]]^nums2[i])), dp[mask][1]+1))
bit <<= 1
return dp[-1][0]