# 1125. Smallest Sufficient Team

## Description

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 skills people[0], people[1], and people[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 <= 16
• 1 <= req_skills[i].length <= 16
• req_skills[i] consists of lowercase English letters.
• All the strings of req_skills are unique.
• 1 <= people.length <= 60
• 0 <= people[i].length <= 16
• 1 <= people[i][j].length <= 16
• people[i][j] consists of lowercase English letters.
• All the strings of people[i] are unique.
• Every skill in people[i] is a skill in req_skills.
• It is guaranteed a sufficient team exists.

## Solutions

Solution 1: State Compression Dynamic Programming

We notice that the length of req_skills does not exceed $16$, so we can use a binary number of length no more than $16$ to represent whether each skill is mastered. Let’s denote the length of req_skills as $m$ and the length of people as $n$.

First, we map each skill in req_skills to a number, i.e., $d[s]$ represents the number of skill $s$. Then, we traverse each person in people and represent the skills they master with a binary number, i.e., $p[i]$ represents the skills mastered by the person with number $i$.

Next, we define the following three arrays:

• Array $f[i]$ represents the minimum number of people to master the skill set $i$, where each bit of the binary representation of $i$ is $1$, indicating that the corresponding skill is mastered. Initially, $f[0] = 0$, and all other positions are infinity.
• Array $g[i]$ represents the number of the last person when the skill set $i$ is mastered by the minimum number of people.
• Array $h[i]$ represents the previous skill set state when the skill set $i$ is mastered by the minimum number of people.

We enumerate each skill set in the range of $[0,..2^m-1]$, for each skill set $i$:

 We enumerate each person $j$ in people. If $f[i] + 1 \lt f[i p[j]]$, it means that $f[i p[j]]$ can be transferred from $f[i]$. At this time, we update $f[i p[j]]$ to $f[i] + 1$, and update $g[i p[j]]$ to $j$, and update $h[i p[j]]$ to $i$. That is, when the current skill set state is $i p[j]$, the number of the last person is $j$, and the previous skill set state is $i$. Here, the symbol  represents bitwise OR operation.

Finally, we start from the skill set $i=2^m-1$, find the number of the last person at this time $g[i]$, add it to the answer, then update $i$ to $h[i]$, and keep backtracking until $i=0$, to get the personnel numbers in the smallest necessary team.

The time complexity is $O(2^m \times n)$, and the space complexity is $O(2^m)$. Here, $m$ and $n$ are the lengths of req_skills and people, respectively.

• class Solution {
public int[] smallestSufficientTeam(String[] req_skills, List<List<String>> people) {
Map<String, Integer> d = new HashMap<>();
int m = req_skills.length;
int n = people.size();
for (int i = 0; i < m; ++i) {
d.put(req_skills[i], i);
}
int[] p = new int[n];
for (int i = 0; i < n; ++i) {
for (var s : people.get(i)) {
p[i] |= 1 << d.get(s);
}
}
int[] f = new int[1 << m];
int[] g = new int[1 << m];
int[] h = new int[1 << m];
final int inf = 1 << 30;
Arrays.fill(f, inf);
f[0] = 0;
for (int i = 0; i < 1 << m; ++i) {
if (f[i] == inf) {
continue;
}
for (int j = 0; j < n; ++j) {
if (f[i] + 1 < f[i | p[j]]) {
f[i | p[j]] = f[i] + 1;
g[i | p[j]] = j;
h[i | p[j]] = i;
}
}
}
List<Integer> ans = new ArrayList<>();
for (int i = (1 << m) - 1; i != 0; i = h[i]) {
}
return ans.stream().mapToInt(Integer::intValue).toArray();
}
}

• class Solution {
public:
vector<int> smallestSufficientTeam(vector<string>& req_skills, vector<vector<string>>& people) {
unordered_map<string, int> d;
int m = req_skills.size(), n = people.size();
for (int i = 0; i < m; ++i) {
d[req_skills[i]] = i;
}
int p[n];
memset(p, 0, sizeof(p));
for (int i = 0; i < n; ++i) {
for (auto& s : people[i]) {
p[i] |= 1 << d[s];
}
}
int f[1 << m];
int g[1 << m];
int h[1 << m];
memset(f, 63, sizeof(f));
f[0] = 0;
for (int i = 0; i < 1 << m; ++i) {
if (f[i] == 0x3f3f3f3f) {
continue;
}
for (int j = 0; j < n; ++j) {
if (f[i] + 1 < f[i | p[j]]) {
f[i | p[j]] = f[i] + 1;
g[i | p[j]] = j;
h[i | p[j]] = i;
}
}
}
vector<int> ans;
for (int i = (1 << m) - 1; i; i = h[i]) {
ans.push_back(g[i]);
}
return ans;
}
};

• class Solution:
def smallestSufficientTeam(
self, req_skills: List[str], people: List[List[str]]
) -> List[int]:
d = {s: i for i, s in enumerate(req_skills)}
m, n = len(req_skills), len(people)
p = [0] * n
for i, ss in enumerate(people):
for s in ss:
p[i] |= 1 << d[s]
f = [inf] * (1 << m)
g = [0] * (1 << m)
h = [0] * (1 << m)
f[0] = 0
for i in range(1 << m):
if f[i] == inf:
continue
for j in range(n):
if f[i] + 1 < f[i | p[j]]:
f[i | p[j]] = f[i] + 1
g[i | p[j]] = j
h[i | p[j]] = i
i = (1 << m) - 1
ans = []
while i:
ans.append(g[i])
i = h[i]
return ans


• func smallestSufficientTeam(req_skills []string, people [][]string) (ans []int) {
d := map[string]int{}
for i, s := range req_skills {
d[s] = i
}
m, n := len(req_skills), len(people)
p := make([]int, n)
for i, ss := range people {
for _, s := range ss {
p[i] |= 1 << d[s]
}
}
const inf = 1 << 30
f := make([]int, 1<<m)
g := make([]int, 1<<m)
h := make([]int, 1<<m)
for i := range f {
f[i] = inf
}
f[0] = 0
for i := range f {
if f[i] == inf {
continue
}
for j := 0; j < n; j++ {
if f[i]+1 < f[i|p[j]] {
f[i|p[j]] = f[i] + 1
g[i|p[j]] = j
h[i|p[j]] = i
}
}
}
for i := 1<<m - 1; i != 0; i = h[i] {
ans = append(ans, g[i])
}
return
}

• function smallestSufficientTeam(req_skills: string[], people: string[][]): number[] {
const d: Map<string, number> = new Map();
const m = req_skills.length;
const n = people.length;
for (let i = 0; i < m; ++i) {
d.set(req_skills[i], i);
}
const p: number[] = new Array(n).fill(0);
for (let i = 0; i < n; ++i) {
for (const s of people[i]) {
p[i] |= 1 << d.get(s)!;
}
}
const inf = 1 << 30;
const f: number[] = new Array(1 << m).fill(inf);
const g: number[] = new Array(1 << m).fill(0);
const h: number[] = new Array(1 << m).fill(0);
f[0] = 0;
for (let i = 0; i < 1 << m; ++i) {
if (f[i] === inf) {
continue;
}
for (let j = 0; j < n; ++j) {
if (f[i] + 1 < f[i | p[j]]) {
f[i | p[j]] = f[i] + 1;
g[i | p[j]] = j;
h[i | p[j]] = i;
}
}
}
const ans: number[] = [];
for (let i = (1 << m) - 1; i; i = h[i]) {
ans.push(g[i]);
}
return ans;
}