Algorithm
Problem Name: 1125. Smallest Sufficient Team
In a project, you have a list of required skills req_skills, and a list of people. The ith person people[i] contains a list of skills that the person has.
Consider a sufficient team: a set of people such that for every required skill in req_skills, there is at least one person in the team who has that skill. We can represent these teams by the index of each person.
- For example,
team = [0, 1, 3]represents the people with skillspeople[0],people[1], andpeople[3].
Return any sufficient team of the smallest possible size, represented by the index of each person. You may return the answer in any order.
It is guaranteed an answer exists.
Example 1:
Input: req_skills = ["java","nodejs","reactjs"], people = [["java"],["nodejs"],["nodejs","reactjs"]] Output: [0,2]
Example 2:
Input: req_skills = ["algorithms","math","java","reactjs","csharp","aws"], people = [["algorithms","math","java"],["algorithms","math","reactjs"],["java","csharp","aws"],["reactjs","csharp"],["csharp","math"],["aws","java"]] Output: [1,2]
Constraints:
1 <= req_skills.length <= 161 <= req_skills[i].length <= 16req_skills[i]consists of lowercase English letters.- All the strings of
req_skillsare unique. 1 <= people.length <= 600 <= people[i].length <= 161 <= people[i][j].length <= 16people[i][j]consists of lowercase English letters.- All the strings of
people[i]are unique. - Every skill in
people[i]is a skill inreq_skills. - It is guaranteed a sufficient team exists.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const smallestSufficientTeam = function(req_skills, people) {
const m = req_skills.length
const n = people.length
const skill2Bitmap = req_skills
.map((x, i) => [x, i])
.reduce((dict, cur) => {
dict[cur[0]] = 1 << cur[1]
return dict
}, {})
const newPeople = people.map(x => {
return x.reduce((acc, cur) => {
const y = skill2Bitmap[cur]
if (y !== undefined) {
acc |= y
}
return acc
}, 0)
})
const all = (1 << m) - 1
const dp = {}
for (let j = 0; j < n; j++) {
if (newPeople[j] > 0) {
dp[newPeople[j]] = new Set([j])
}
}
if (dp[all]) {
return Array.from(dp[all]).sort()
}
for (let k = 0; k < n; k++) {
for (let s in dp) {
for (let j = 0; j < n; j++) {
if (newPeople[j] === 0 || dp[s].has(j)) continue
const newIdx = s | newPeople[j]
if (dp[newIdx] === undefined) {
dp[newIdx] = new Set([...dp[s], j])
if (newIdx === all) {
return Array.from(dp[all]).sort()
}
}
}
}
}
return []
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def smallestSufficientTeam(self, req_skills: List[str], people: List[List[str]]) -> List[int]:
n, m = len(req_skills), len(people)
key = {v: i for i, v in enumerate(req_skills)}
dp = {0: []}
for i, p in enumerate(people):
his_skill = 0
for skill in p:
if skill in key:
his_skill |= 1 << key[skill]
for skill_set, need in list(dp.items()):
with_him = skill_set | his_skill
if with_him == skill_set: continue
if with_him not in dp or len(dp[with_him]) > len(need) + 1:
dp[with_him] = need + [i]
return dp[(1 << n) - 1]
Input
Output