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Aug 3, 2024
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feat: add solutions to lc problem: No.1452 #3357

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195 changes: 110 additions & 85 deletions ...ople Whose List of Favorite Companies Is Not a Subset of Another List/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -29,7 +29,7 @@ tags:
<p><strong>示例 1：</strong></p>

<pre><strong>输入：</strong>favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;,&quot;microsoft&quot;],[&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;],[&quot;amazon&quot;]]
<strong>输出：</strong>[0,1,4]
<strong>输出：</strong>[0,1,4]
<strong>解释：</strong>
favoriteCompanies[2]=[&quot;google&quot;,&quot;facebook&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 的子集。
favoriteCompanies[3]=[&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 和 favoriteCompanies[1]=[&quot;google&quot;,&quot;microsoft&quot;] 的子集。
Expand All @@ -39,7 +39,7 @@ favoriteCompanies[3]=[&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetco
<p><strong>示例 2：</strong></p>

<pre><strong>输入：</strong>favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;leetcode&quot;,&quot;amazon&quot;],[&quot;facebook&quot;,&quot;google&quot;]]
<strong>输出：</strong>[0,1]
<strong>输出：</strong>[0,1]
<strong>解释：</strong>favoriteCompanies[2]=[&quot;facebook&quot;,&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 的子集，因此，答案为 [0,1] 。
</pre>

Expand Down Expand Up @@ -70,9 +70,9 @@ favoriteCompanies[3]=[&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetco

### 方法一：哈希表

将每个 `company` 字符串列表都转换为一个整数类型的集合。然后遍历每个集合，判断其是否是其他集合的子集，如果不是，则将其下标加入结果集。
我们可以将每个公司映射到一个唯一的整数，然后对于每个人，我们将他们收藏的公司转换为整数集合，最后判断是否存在一个人的收藏公司是另一个人的子集。

时间复杂度 $O(n^2 \times m)$，其中 $n$ 为 `favoriteCompanies` 的长度，$m$ 为 `favoriteCompanies[i]` 的最大长度。
时间复杂度 $(n \times m \times k + n^2 \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别是 `favoriteCompanies` 的长度和每个公司清单的平均长度，而 $k$ 是每个公司的平均长度。

<!-- tabs:start -->

Expand All @@ -81,25 +81,19 @@ favoriteCompanies[3]=[&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetco
```python
class Solution:
def peopleIndexes(self, favoriteCompanies: List[List[str]]) -> List[int]:
d = {}
idx = 0
t = []
for v in favoriteCompanies:
for c in v:
if c not in d:
d[c] = idx
d = {}
n = len(favoriteCompanies)
nums = [set() for _ in range(n)]
for i, ss in enumerate(favoriteCompanies):
for s in ss:
if s not in d:
d[s] = idx
idx += 1
t.append({d[c] for c in v})
nums[i].add(d[s])
ans = []
for i, nums1 in enumerate(t):
ok = True
for j, nums2 in enumerate(t):
if i == j:
continue
if not (nums1 - nums2):
ok = False
break
if ok:
for i in range(n):
if not any(i != j and (nums[i] & nums[j]) == nums[i] for j in range(n)):
ans.append(i)
return ans
```
Expand All @@ -109,32 +103,26 @@ class Solution:
```java
class Solution {
public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
int n = favoriteCompanies.size();
Map<String, Integer> d = new HashMap<>();
int idx = 0;
int n = favoriteCompanies.size();
Set<Integer>[] t = new Set[n];
Set<Integer>[] nums = new Set[n];
Arrays.setAll(nums, i -> new HashSet<>());
for (int i = 0; i < n; ++i) {
var v = favoriteCompanies.get(i);
for (var c : v) {
if (!d.containsKey(c)) {
d.put(c, idx++);
var ss = favoriteCompanies.get(i);
for (var s : ss) {
if (!d.containsKey(s)) {
d.put(s, idx++);
}
nums[i].add(d.get(s));
}
Set<Integer> s = new HashSet<>();
for (var c : v) {
s.add(d.get(c));
}
t[i] = s;
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < n; ++i) {
boolean ok = true;
for (int j = 0; j < n; ++j) {
if (i != j) {
if (t[j].containsAll(t[i])) {
ok = false;
break;
}
for (int j = 0; j < n && ok; ++j) {
if (i != j && nums[j].containsAll(nums[i])) {
ok = false;
}
}
if (ok) {
Expand All @@ -152,95 +140,132 @@ class Solution {
class Solution {
public:
vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
int n = favoriteCompanies.size();
unordered_map<string, int> d;
int idx = 0, n = favoriteCompanies.size();
vector<unordered_set<int>> t(n);
int idx = 0;
vector<unordered_set<int>> nums(n);

for (int i = 0; i < n; ++i) {
auto v = favoriteCompanies[i];
for (auto& c : v) {
if (!d.count(c)) {
d[c] = idx++;
for (const auto& s : favoriteCompanies[i]) {
if (!d.contains(s)) {
d[s] = idx++;
}
nums[i].insert(d[s]);
}
unordered_set<int> s;
for (auto& c : v) {
s.insert(d[c]);
}
t[i] = s;
}

auto check = [](const unordered_set<int>& a, const unordered_set<int>& b) {
for (int x : a) {
if (!b.contains(x)) {
return false;
}
}
return true;
};

vector<int> ans;
for (int i = 0; i < n; ++i) {
bool ok = true;
for (int j = 0; j < n; ++j) {
if (i == j) continue;
if (check(t[i], t[j])) {
for (int j = 0; j < n && ok; ++j) {
if (i != j && check(nums[i], nums[j])) {
ok = false;
break;
}
}
if (ok) {
ans.push_back(i);
}
}
return ans;
}

bool check(unordered_set<int>& nums1, unordered_set<int>& nums2) {
for (int v : nums1) {
if (!nums2.count(v)) {
return false;
}
}
return true;
return ans;
}
};
```

#### Go

```go
func peopleIndexes(favoriteCompanies [][]string) []int {
d := map[string]int{}
idx, n := 0, len(favoriteCompanies)
t := make([]map[int]bool, n)
for i, v := range favoriteCompanies {
for _, c := range v {
if _, ok := d[c]; !ok {
d[c] = idx
func peopleIndexes(favoriteCompanies [][]string) (ans []int) {
n := len(favoriteCompanies)
d := make(map[string]int)
idx := 0
nums := make([]map[int]struct{}, n)

for i := 0; i < n; i++ {
nums[i] = make(map[int]struct{})
for _, s := range favoriteCompanies[i] {
if _, ok := d[s]; !ok {
d[s] = idx
idx++
}
nums[i][d[s]] = struct{}{}
}
s := map[int]bool{}
for _, c := range v {
s[d[c]] = true
}
t[i] = s
}
ans := []int{}
check := func(nums1, nums2 map[int]bool) bool {
for v, _ := range nums1 {
if _, ok := nums2[v]; !ok {

check := func(a, b map[int]struct{}) bool {
for x := range a {
if _, ok := b[x]; !ok {
return false
}
}
return true
}
for i := 0; i < n; i++ {
ok := true
for j := 0; j < n; j++ {
if i == j {
continue
}
if check(t[i], t[j]) {
for j := 0; j < n && ok; j++ {
if i != j && check(nums[i], nums[j]) {
ok = false
break
}
}
if ok {
ans = append(ans, i)
}
}
return ans

return
}
```

#### TypeScript

```ts
function peopleIndexes(favoriteCompanies: string[][]): number[] {
const n = favoriteCompanies.length;
const d: Map<string, number> = new Map();
let idx = 0;
const nums: Set<number>[] = Array.from({ length: n }, () => new Set<number>());

for (let i = 0; i < n; i++) {
for (const s of favoriteCompanies[i]) {
if (!d.has(s)) {
d.set(s, idx++);
}
nums[i].add(d.get(s)!);
}
}

const check = (a: Set<number>, b: Set<number>): boolean => {
for (const x of a) {
if (!b.has(x)) {
return false;
}
}
return true;
};

const ans: number[] = [];
for (let i = 0; i < n; i++) {
let ok = true;
for (let j = 0; j < n && ok; j++) {
if (i !== j && check(nums[i], nums[j])) {
ok = false;
}
}
if (ok) {
ans.push(i);
}
}

return ans;
}
```

Expand Down
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