feat: add solutions to lc problem: No.0781 (doocs#4104)

No.0781.Rabbits in Forest

Author: yanglbme

solution/0700-0799/0781.Rabbits in Forest/README.md

@@ -31,10 +31,10 @@ tags:
 <strong>输入：</strong>answers = [1,1,2]
 <strong>输出：</strong>5
 <strong>解释：</strong>
-两只回答了 "1" 的兔子可能有相同的颜色，设为红色。 
+两只回答了 "1" 的兔子可能有相同的颜色，设为红色。
 之后回答了 "2" 的兔子不会是红色，否则他们的回答会相互矛盾。
-设回答了 "2" 的兔子为蓝色。 
-此外，森林中还应有另外 2 只蓝色兔子的回答没有包含在数组中。 
+设回答了 "2" 的兔子为蓝色。
+此外，森林中还应有另外 2 只蓝色兔子的回答没有包含在数组中。
 因此森林中兔子的最少数量是 5 只：3 只回答的和 2 只没有回答的。
 </pre>
 
@@ -60,7 +60,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：贪心 + 哈希表
+
+根据题目描述，回答相同的兔子，可能属于同一种颜色，而回答不同的兔子，不可能属于同一种颜色。
+
+因此，我们用一个哈希表 $\textit{cnt}$ 记录每种回答出现的次数。对于每种回答 $x$ 及其出现次数 $v$，我们按照每种颜色有 $x + 1$ 只兔子的原则，计算出兔子的最少数量，并将其加入答案。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{answers}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -69,26 +75,83 @@ tags:
 ```python
 class Solution:
     def numRabbits(self, answers: List[int]) -> int:
-        counter = Counter(answers)
-        return sum([math.ceil(v / (k + 1)) * (k + 1) for k, v in counter.items()])
+        cnt = Counter(answers)
+        ans = 0
+        for x, v in cnt.items():
+            group = x + 1
+            ans += (v + group - 1) // group * group
+        return ans
 ```
 
 #### Java
 
 ```java
 class Solution {
     public int numRabbits(int[] answers) {
-        Map<Integer, Integer> counter = new HashMap<>();
-        for (int e : answers) {
-            counter.put(e, counter.getOrDefault(e, 0) + 1);
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int x : answers) {
+            cnt.merge(x, 1, Integer::sum);
         }
-        int res = 0;
-        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
-            int answer = entry.getKey(), count = entry.getValue();
-            res += (int) Math.ceil(count / ((answer + 1) * 1.0)) * (answer + 1);
+        int ans = 0;
+        for (var e : cnt.entrySet()) {
+            int group = e.getKey() + 1;
+            ans += (e.getValue() + group - 1) / group * group;
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int numRabbits(vector<int>& answers) {
+        unordered_map<int, int> cnt;
+        for (int x : answers) {
+            ++cnt[x];
         }
-        return res;
+        int ans = 0;
+        for (auto& [x, v] : cnt) {
+            int group = x + 1;
+            ans += (v + group - 1) / group * group;
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func numRabbits(answers []int) (ans int) {
+	cnt := map[int]int{}
+	for _, x := range answers {
+		cnt[x]++
+	}
+	for x, v := range cnt {
+		group := x + 1
+		ans += (v + group - 1) / group * group
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function numRabbits(answers: number[]): number {
+    const cnt = new Map<number, number>();
+    for (const x of answers) {
+        cnt.set(x, (cnt.get(x) || 0) + 1);
+    }
+    let ans = 0;
+    for (const [x, v] of cnt.entries()) {
+        const group = x + 1;
+        ans += Math.floor((v + group - 1) / group) * group;
     }
+    return ans;
 }
 ```
 

solution/0700-0799/0781.Rabbits in Forest/README_EN.md

@@ -58,7 +58,13 @@ The smallest possible number of rabbits in the forest is therefore 5: 3 that ans
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Hash Map
+
+According to the problem description, rabbits that give the same answer may belong to the same color, while rabbits that give different answers cannot belong to the same color.
+
+Therefore, we use a hash map $\textit{cnt}$ to record the number of occurrences of each answer. For each answer $x$ and its occurrence $v$, we calculate the minimum number of rabbits based on the principle that each color has $x + 1$ rabbits, and add it to the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $\textit{answers}$.
 
 <!-- tabs:start -->
 
@@ -67,26 +73,83 @@ The smallest possible number of rabbits in the forest is therefore 5: 3 that ans
 ```python
 class Solution:
     def numRabbits(self, answers: List[int]) -> int:
-        counter = Counter(answers)
-        return sum([math.ceil(v / (k + 1)) * (k + 1) for k, v in counter.items()])
+        cnt = Counter(answers)
+        ans = 0
+        for x, v in cnt.items():
+            group = x + 1
+            ans += (v + group - 1) // group * group
+        return ans
 ```
 
 #### Java
 
 ```java
 class Solution {
     public int numRabbits(int[] answers) {
-        Map<Integer, Integer> counter = new HashMap<>();
-        for (int e : answers) {
-            counter.put(e, counter.getOrDefault(e, 0) + 1);
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int x : answers) {
+            cnt.merge(x, 1, Integer::sum);
         }
-        int res = 0;
-        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
-            int answer = entry.getKey(), count = entry.getValue();
-            res += (int) Math.ceil(count / ((answer + 1) * 1.0)) * (answer + 1);
+        int ans = 0;
+        for (var e : cnt.entrySet()) {
+            int group = e.getKey() + 1;
+            ans += (e.getValue() + group - 1) / group * group;
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int numRabbits(vector<int>& answers) {
+        unordered_map<int, int> cnt;
+        for (int x : answers) {
+            ++cnt[x];
         }
-        return res;
+        int ans = 0;
+        for (auto& [x, v] : cnt) {
+            int group = x + 1;
+            ans += (v + group - 1) / group * group;
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func numRabbits(answers []int) (ans int) {
+	cnt := map[int]int{}
+	for _, x := range answers {
+		cnt[x]++
+	}
+	for x, v := range cnt {
+		group := x + 1
+		ans += (v + group - 1) / group * group
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function numRabbits(answers: number[]): number {
+    const cnt = new Map<number, number>();
+    for (const x of answers) {
+        cnt.set(x, (cnt.get(x) || 0) + 1);
+    }
+    let ans = 0;
+    for (const [x, v] of cnt.entries()) {
+        const group = x + 1;
+        ans += Math.floor((v + group - 1) / group) * group;
     }
+    return ans;
 }
 ```
 

solution/0700-0799/0781.Rabbits in Forest/Solution.cpp

@@ -0,0 +1,15 @@
+class Solution {
+public:
+    int numRabbits(vector<int>& answers) {
+        unordered_map<int, int> cnt;
+        for (int x : answers) {
+            ++cnt[x];
+        }
+        int ans = 0;
+        for (auto& [x, v] : cnt) {
+            int group = x + 1;
+            ans += (v + group - 1) / group * group;
+        }
+        return ans;
+    }
+};

solution/0700-0799/0781.Rabbits in Forest/Solution.go

@@ -0,0 +1,11 @@
+func numRabbits(answers []int) (ans int) {
+	cnt := map[int]int{}
+	for _, x := range answers {
+		cnt[x]++
+	}
+	for x, v := range cnt {
+		group := x + 1
+		ans += (v + group - 1) / group * group
+	}
+	return
+}

solution/0700-0799/0781.Rabbits in Forest/Solution.java

@@ -1,14 +1,14 @@
 class Solution {
     public int numRabbits(int[] answers) {
-        Map<Integer, Integer> counter = new HashMap<>();
-        for (int e : answers) {
-            counter.put(e, counter.getOrDefault(e, 0) + 1);
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int x : answers) {
+            cnt.merge(x, 1, Integer::sum);
         }
-        int res = 0;
-        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
-            int answer = entry.getKey(), count = entry.getValue();
-            res += (int) Math.ceil(count / ((answer + 1) * 1.0)) * (answer + 1);
+        int ans = 0;
+        for (var e : cnt.entrySet()) {
+            int group = e.getKey() + 1;
+            ans += (e.getValue() + group - 1) / group * group;
         }
-        return res;
+        return ans;
     }
-}
+}

solution/0700-0799/0781.Rabbits in Forest/Solution.py

@@ -1,4 +1,8 @@
 class Solution:
     def numRabbits(self, answers: List[int]) -> int:
-        counter = Counter(answers)
-        return sum([math.ceil(v / (k + 1)) * (k + 1) for k, v in counter.items()])
+        cnt = Counter(answers)
+        ans = 0
+        for x, v in cnt.items():
+            group = x + 1
+            ans += (v + group - 1) // group * group
+        return ans

solution/0700-0799/0781.Rabbits in Forest/Solution.ts

@@ -0,0 +1,12 @@
+function numRabbits(answers: number[]): number {
+    const cnt = new Map<number, number>();
+    for (const x of answers) {
+        cnt.set(x, (cnt.get(x) || 0) + 1);
+    }
+    let ans = 0;
+    for (const [x, v] of cnt.entries()) {
+        const group = x + 1;
+        ans += Math.floor((v + group - 1) / group) * group;
+    }
+    return ans;
+}

solution/0700-0799/0790.Domino and Tromino Tiling/README.md

@@ -95,28 +95,16 @@ tags:
 ```python
 class Solution:
     def numTilings(self, n: int) -> int:
-        @cache
-        def dfs(i, j):
-            if i > n or j > n:
-                return 0
-            if i == n and j == n:
-                return 1
-            ans = 0
-            if i == j:
-                ans = (
-                    dfs(i + 2, j + 2)
-                    + dfs(i + 1, j + 1)
-                    + dfs(i + 2, j + 1)
-                    + dfs(i + 1, j + 2)
-                )
-            elif i > j:
-                ans = dfs(i, j + 2) + dfs(i + 1, j + 2)
-            else:
-                ans = dfs(i + 2, j) + dfs(i + 2, j + 1)
-            return ans % mod
-
+        f = [1, 0, 0, 0]
         mod = 10**9 + 7
-        return dfs(0, 0)
+        for i in range(1, n + 1):
+            g = [0] * 4
+            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod
+            g[1] = (f[2] + f[3]) % mod
+            g[2] = (f[1] + f[3]) % mod
+            g[3] = f[0]
+            f = g
+        return f[0]
 ```
 
 #### Java
@@ -184,31 +172,4 @@ func numTilings(n int) int {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def numTilings(self, n: int) -> int:
-        f = [1, 0, 0, 0]
-        mod = 10**9 + 7
-        for i in range(1, n + 1):
-            g = [0] * 4
-            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod
-            g[1] = (f[2] + f[3]) % mod
-            g[2] = (f[1] + f[3]) % mod
-            g[3] = f[0]
-            f = g
-        return f[0]
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->