feat: add solutions to lc problems: No.2784,2785 (doocs#2442)

* No.2784.Check if Array is Good
* No.2785.Sort Vowels in a String

Author: yanglbme

solution/2700-2799/2784.Check if Array is Good/README.md

@@ -55,19 +55,25 @@
 
 ## 解法
 
-### 方法一
+### 方法一：计数
+
+我们可以用一个哈希表或数组 $cnt$ 记录数组 $nums$ 中每个元素出现的次数。然后我们判断是否满足以下条件：
+
+1. $cnt[n] = 2$，即数组中最大的元素出现了两次；
+2. 对于 $1 \leq i \leq n-1$，均满足 $cnt[i] = 1$，即数组中除了最大的元素外，其他元素都只出现了一次。
+
+如果满足以上两个条件，那么数组 $nums$ 是一个好数组，否则不是。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def isGood(self, nums: List[int]) -> bool:
-        n = len(nums) - 1
         cnt = Counter(nums)
-        cnt[n] -= 2
-        for i in range(1, n):
-            cnt[i] -= 1
-        return all(v == 0 for v in cnt.values())
+        n = len(nums) - 1
+        return cnt[n] == 2 and all(cnt[i] for i in range(1, n))
 ```
 
 ```java
@@ -78,12 +84,11 @@ class Solution {
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            cnt[i] -= 1;
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x != 0) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }
@@ -97,16 +102,15 @@ class Solution {
 public:
     bool isGood(vector<int>& nums) {
         int n = nums.size() - 1;
-        vector<int> cnt(201);
+        int cnt[201]{};
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            --cnt[i];
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }
@@ -122,12 +126,11 @@ func isGood(nums []int) bool {
 	for _, x := range nums {
 		cnt[x]++
 	}
-	cnt[n] -= 2
-	for i := 1; i < n; i++ {
-		cnt[i]--
+	if cnt[n] != 2 {
+		return false
 	}
-	for _, x := range cnt {
-		if x != 0 {
+	for i := 1; i < n; i++ {
+		if cnt[i] != 1 {
 			return false
 		}
 	}
@@ -138,15 +141,40 @@ func isGood(nums []int) bool {
 ```ts
 function isGood(nums: number[]): boolean {
     const n = nums.length - 1;
-    const cnt: number[] = new Array(201).fill(0);
+    const cnt: number[] = Array(201).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }
-    cnt[n] -= 2;
+    if (cnt[n] !== 2) {
+        return false;
+    }
     for (let i = 1; i < n; ++i) {
-        cnt[i]--;
+        if (cnt[i] !== 1) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+```cs
+public class Solution {
+    public bool IsGood(int[] nums) {
+        int n = nums.Length - 1;
+        int[] cnt = new int[201];
+        foreach (int x in nums) {
+            ++cnt[x];
+        }
+        if (cnt[n] != 2) {
+            return false;
+        }
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
+                return false;
+            }
+        }
+        return true;
     }
-    return cnt.every(x => x >= 0);
 }
 ```
 

solution/2700-2799/2784.Check if Array is Good/README_EN.md

@@ -55,19 +55,25 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Counting
+
+We can use a hash table or array $cnt$ to record the number of occurrences of each element in the array $nums$. Then we determine whether the following conditions are met:
+
+1. $cnt[n] = 2$, i.e., the largest element in the array appears twice;
+2. For $1 \leq i \leq n-1$, it holds that $cnt[i] = 1$, i.e., except for the largest element, all other elements appear only once.
+
+If the above two conditions are met, then the array $nums$ is a good array, otherwise it is not.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def isGood(self, nums: List[int]) -> bool:
-        n = len(nums) - 1
         cnt = Counter(nums)
-        cnt[n] -= 2
-        for i in range(1, n):
-            cnt[i] -= 1
-        return all(v == 0 for v in cnt.values())
+        n = len(nums) - 1
+        return cnt[n] == 2 and all(cnt[i] for i in range(1, n))
 ```
 
 ```java
@@ -78,12 +84,11 @@ class Solution {
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            cnt[i] -= 1;
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x != 0) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }
@@ -97,16 +102,15 @@ class Solution {
 public:
     bool isGood(vector<int>& nums) {
         int n = nums.size() - 1;
-        vector<int> cnt(201);
+        int cnt[201]{};
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            --cnt[i];
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }
@@ -122,12 +126,11 @@ func isGood(nums []int) bool {
 	for _, x := range nums {
 		cnt[x]++
 	}
-	cnt[n] -= 2
-	for i := 1; i < n; i++ {
-		cnt[i]--
+	if cnt[n] != 2 {
+		return false
 	}
-	for _, x := range cnt {
-		if x != 0 {
+	for i := 1; i < n; i++ {
+		if cnt[i] != 1 {
 			return false
 		}
 	}
@@ -138,15 +141,40 @@ func isGood(nums []int) bool {
 ```ts
 function isGood(nums: number[]): boolean {
     const n = nums.length - 1;
-    const cnt: number[] = new Array(201).fill(0);
+    const cnt: number[] = Array(201).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }
-    cnt[n] -= 2;
+    if (cnt[n] !== 2) {
+        return false;
+    }
     for (let i = 1; i < n; ++i) {
-        cnt[i]--;
+        if (cnt[i] !== 1) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+```cs
+public class Solution {
+    public bool IsGood(int[] nums) {
+        int n = nums.Length - 1;
+        int[] cnt = new int[201];
+        foreach (int x in nums) {
+            ++cnt[x];
+        }
+        if (cnt[n] != 2) {
+            return false;
+        }
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
+                return false;
+            }
+        }
+        return true;
     }
-    return cnt.every(x => x >= 0);
 }
 ```
 

solution/2700-2799/2784.Check if Array is Good/Solution.cpp

@@ -2,16 +2,15 @@ class Solution {
 public:
     bool isGood(vector<int>& nums) {
         int n = nums.size() - 1;
-        vector<int> cnt(201);
+        int cnt[201]{};
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            --cnt[i];
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }

solution/2700-2799/2784.Check if Array is Good/Solution.cs

@@ -0,0 +1,18 @@
+public class Solution {
+    public bool IsGood(int[] nums) {
+        int n = nums.Length - 1;
+        int[] cnt = new int[201];
+        foreach (int x in nums) {
+            ++cnt[x];
+        }
+        if (cnt[n] != 2) {
+            return false;
+        }
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
+                return false;
+            }
+        }
+        return true;
+    }
+}

solution/2700-2799/2784.Check if Array is Good/Solution.go

@@ -4,12 +4,11 @@ func isGood(nums []int) bool {
 	for _, x := range nums {
 		cnt[x]++
 	}
-	cnt[n] -= 2
-	for i := 1; i < n; i++ {
-		cnt[i]--
+	if cnt[n] != 2 {
+		return false
 	}
-	for _, x := range cnt {
-		if x != 0 {
+	for i := 1; i < n; i++ {
+		if cnt[i] != 1 {
 			return false
 		}
 	}

solution/2700-2799/2784.Check if Array is Good/Solution.java

@@ -5,12 +5,11 @@ public boolean isGood(int[] nums) {
         for (int x : nums) {
             ++cnt[x];
         }
-        cnt[n] -= 2;
-        for (int i = 1; i < n; ++i) {
-            cnt[i] -= 1;
+        if (cnt[n] != 2) {
+            return false;
         }
-        for (int x : cnt) {
-            if (x != 0) {
+        for (int i = 1; i < n; ++i) {
+            if (cnt[i] != 1) {
                 return false;
             }
         }

solution/2700-2799/2784.Check if Array is Good/Solution.py

@@ -1,8 +1,5 @@
 class Solution:
     def isGood(self, nums: List[int]) -> bool:
-        n = len(nums) - 1
         cnt = Counter(nums)
-        cnt[n] -= 2
-        for i in range(1, n):
-            cnt[i] -= 1
-        return all(v == 0 for v in cnt.values())
+        n = len(nums) - 1
+        return cnt[n] == 2 and all(cnt[i] for i in range(1, n))

solution/2700-2799/2784.Check if Array is Good/Solution.ts

@@ -1,12 +1,16 @@
 function isGood(nums: number[]): boolean {
     const n = nums.length - 1;
-    const cnt: number[] = new Array(201).fill(0);
+    const cnt: number[] = Array(201).fill(0);
     for (const x of nums) {
         ++cnt[x];
     }
-    cnt[n] -= 2;
+    if (cnt[n] !== 2) {
+        return false;
+    }
     for (let i = 1; i < n; ++i) {
-        cnt[i]--;
+        if (cnt[i] !== 1) {
+            return false;
+        }
     }
-    return cnt.every(x => x >= 0);
+    return true;
 }

solution/2700-2799/2785.Sort Vowels in a String/README.md

@@ -48,7 +48,13 @@
 
 ## 解法
 
-### 方法一
+### 方法一：排序
+
+我们先将字符串中的所有元音字母存到数组或列表 $vs$ 中，然后我们对 $vs$ 进行排序。
+
+接下来，我们遍历字符串 $s$，保持辅音字母不变，如果是元音字母，则依次按顺序替换为 $vs$ 数组中的字母。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。
 
 <!-- tabs:start -->