feat: add solutions to lc problem: No.1452

No.1452.People Whose List of Favorite Companies Is Not a Subset of
Another List

Author: yanglbme

solution/1400-1499/1452.People Whose List of Favorite Companies Is Not a Subset of Another List/README.md

@@ -52,22 +52,184 @@ favoriteCompanies[3]=["google"] 是 favoriteCompanies[0]=["leetco
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**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。
+
+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。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```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
+                    idx += 1
+            t.append({d[c] for c in v})
+        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:
+                ans.append(i)
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
+        Map<String, Integer> d = new HashMap<>();
+        int idx = 0;
+        int n = favoriteCompanies.size();
+        Set<Integer>[] t = new Set[n];
+        for (int i = 0; i < n; ++i) {
+            var v = favoriteCompanies.get(i);
+            for (var c : v) {
+                if (!d.containsKey(c)) {
+                    d.put(c, idx++);
+                }
+            }
+            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;
+                    }
+                }
+            }
+            if (ok) {
+                ans.add(i);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
+        unordered_map<string, int> d;
+        int idx = 0, n = favoriteCompanies.size();
+        vector<unordered_set<int>> t(n);
+        for (int i = 0; i < n; ++i) {
+            auto v = favoriteCompanies[i];
+            for (auto& c : v) {
+                if (!d.count(c)) {
+                    d[c] = idx++;
+                }
+            }
+            unordered_set<int> s;
+            for (auto& c : v) {
+                s.insert(d[c]);
+            }
+            t[i] = s;
+        }
+        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])) {
+                    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;
+    }
+};
+```
 
+### **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
+				idx++
+			}
+		}
+		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 {
+				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]) {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			ans = append(ans, i)
+		}
+	}
+	return ans
+}
 ```
 
 ### **...**

solution/1400-1499/1452.People Whose List of Favorite Companies Is Not a Subset of Another List/README_EN.md

@@ -54,13 +54,169 @@ Other lists of favorite companies are not a subset of another list, therefore, t
 ### **Python3**
 
 ```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
+                    idx += 1
+            t.append({d[c] for c in v})
+        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:
+                ans.append(i)
+        return ans
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
+        Map<String, Integer> d = new HashMap<>();
+        int idx = 0;
+        int n = favoriteCompanies.size();
+        Set<Integer>[] t = new Set[n];
+        for (int i = 0; i < n; ++i) {
+            var v = favoriteCompanies.get(i);
+            for (var c : v) {
+                if (!d.containsKey(c)) {
+                    d.put(c, idx++);
+                }
+            }
+            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;
+                    }
+                }
+            }
+            if (ok) {
+                ans.add(i);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
+        unordered_map<string, int> d;
+        int idx = 0, n = favoriteCompanies.size();
+        vector<unordered_set<int>> t(n);
+        for (int i = 0; i < n; ++i) {
+            auto v = favoriteCompanies[i];
+            for (auto& c : v) {
+                if (!d.count(c)) {
+                    d[c] = idx++;
+                }
+            }
+            unordered_set<int> s;
+            for (auto& c : v) {
+                s.insert(d[c]);
+            }
+            t[i] = s;
+        }
+        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])) {
+                    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;
+    }
+};
+```
 
+### **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
+				idx++
+			}
+		}
+		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 {
+				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]) {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			ans = append(ans, i)
+		}
+	}
+	return ans
+}
 ```
 
 ### **...**