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Mar 14, 2024
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feat: add solutions to lc problems: No.2784,2785 #2442

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80 changes: 54 additions & 26 deletions solution/2700-2799/2784.Check if Array is Good/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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
Expand All @@ -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;
}
}
Expand All @@ -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;
}
}
Expand All @@ -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
}
}
Expand All @@ -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);
}
```

Expand Down
80 changes: 54 additions & 26 deletions solution/2700-2799/2784.Check if Array is Good/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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
Expand All @@ -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;
}
}
Expand All @@ -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;
}
}
Expand All @@ -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
}
}
Expand All @@ -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);
}
```

Expand Down
11 changes: 5 additions & 6 deletions solution/2700-2799/2784.Check if Array is Good/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -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;
}
}
Expand Down
18 changes: 18 additions & 0 deletions solution/2700-2799/2784.Check if Array is Good/Solution.cs
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Original file line number Diff line number Diff line change
@@ -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;
}
}
9 changes: 4 additions & 5 deletions solution/2700-2799/2784.Check if Array is Good/Solution.go
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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
}
}
Expand Down
9 changes: 4 additions & 5 deletions solution/2700-2799/2784.Check if Array is Good/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -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;
}
}
Expand Down
7 changes: 2 additions & 5 deletions solution/2700-2799/2784.Check if Array is Good/Solution.py
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Original file line number Diff line number Diff line change
@@ -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))
12 changes: 8 additions & 4 deletions solution/2700-2799/2784.Check if Array is Good/Solution.ts
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Original file line number Diff line number Diff line change
@@ -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;
}
8 changes: 7 additions & 1 deletion solution/2700-2799/2785.Sort Vowels in a String/README.md
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Expand Up @@ -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 -->

Expand Down
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