Hard
Smallest Sufficient Team — Python
Full explanation · Time O(m * 2^n) · Space O(2^n)
# Time: O(m * 2^n), n is the number of skills
# m is the number of people
# Space: O(2^n)
class Solution(object):
def smallestSufficientTeam(self, req_skills, people):
"""
:type req_skills: List[str]
:type people: List[List[str]]
:rtype: List[int]
"""
lookup = {v: i for i, v in enumerate(req_skills)}
dp = {0: []}
for i, p in enumerate(people):
his_skill_set = 0
for skill in p:
if skill in lookup:
his_skill_set |= 1 << lookup[skill]
for skill_set, people in dp.items():
with_him = skill_set | his_skill_set
if with_him == skill_set: continue
if with_him not in dp or \
len(dp[with_him]) > len(people)+1:
dp[with_him] = people + [i]
return dp[(1<<len(req_skills))-1]