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feat: add solutions to lc problems: No.2956,2957 (#2077)
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* No.2956.Find Common Elements Between Two Arrays
* No.2957.Remove Adjacent Almost-Equal Characters
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@yanglbme
yanglbme authored Dec 10, 2023
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100 changes: 97 additions & 3 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/README.md
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<!-- 这里可写通用的实现逻辑 -->

**方法一：哈希表或数组**

我们可以用两个哈希表或数组 $s1$ 和 $s2$ 分别记录两个数组中出现的元素。

接下来，我们创建一个长度为 $2$ 的数组 $ans$，其中 $ans[0]$ 表示 $nums1$ 中出现在 $s2$ 中的元素个数，$ans[1]$ 表示 $nums2$ 中出现在 $s1$ 中的元素个数。

然后，我们遍历数组 $nums1$ 中的每个元素 $x$，如果 $x$ 在 $s2$ 中出现过，则将 $ans[0]$ 加一。接着，我们遍历数组 $nums2$ 中的每个元素 $x$，如果 $x$ 在 $s1$ 中出现过，则将 $ans[1]$ 加一。

最后，我们返回数组 $ans$ 即可。

时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是数组 $nums1$ 和 $nums2$ 的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
s1, s2 = set(nums1), set(nums2)
return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public int[] findIntersectionValues(int[] nums1, int[] nums2) {
int[] s1 = new int[101];
int[] s2 = new int[101];
for (int x : nums1) {
s1[x] = 1;
}
for (int x : nums2) {
s2[x] = 1;
}
int[] ans = new int[2];
for (int x : nums1) {
ans[0] += s2[x];
}
for (int x : nums2) {
ans[1] += s1[x];
}
return ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
int s1[101]{};
int s2[101]{};
for (int& x : nums1) {
s1[x] = 1;
}
for (int& x : nums2) {
s2[x] = 1;
}
vector<int> ans(2);
for (int& x : nums1) {
ans[0] += s2[x];
}
for (int& x : nums2) {
ans[1] += s1[x];
}
return ans;
}
};
```
### **Go**
```go
func findIntersectionValues(nums1 []int, nums2 []int) []int {
s1 := [101]int{}
s2 := [101]int{}
for _, x := range nums1 {
s1[x] = 1
}
for _, x := range nums2 {
s2[x] = 1
}
ans := make([]int, 2)
for _, x := range nums1 {
ans[0] += s2[x]
}
for _, x := range nums2 {
ans[1] += s1[x]
}
return ans
}
```

### **TypeScript**

```ts
function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
const s1: number[] = Array(101).fill(0);
const s2: number[] = Array(101).fill(0);
for (const x of nums1) {
s1[x] = 1;
}
for (const x of nums2) {
s2[x] = 1;
}
const ans: number[] = Array(2).fill(0);
for (const x of nums1) {
ans[0] += s2[x];
}
for (const x of nums2) {
ans[1] += s1[x];
}
return ans;
}
```

### **...**
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100 changes: 97 additions & 3 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/README_EN.md
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## Solutions

**Solution 1: Hash Table or Array**

We can use two hash tables or arrays $s1$ and $s2$ to record the elements that appear in the two arrays respectively.

Next, we create an array $ans$ of length $2$, where $ans[0]$ represents the number of elements in $nums1$ that appear in $s2$, and $ans[1]$ represents the number of elements in $nums2$ that appear in $s1$.

Then, we traverse each element $x$ in the array $nums1$. If $x$ has appeared in $s2$, we increment $ans[0]$. After that, we traverse each element $x$ in the array $nums2$. If $x$ has appeared in $s1$, we increment $ans[1]$.

Finally, we return the array $ans$.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$ respectively.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
s1, s2 = set(nums1), set(nums2)
return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
```

### **Java**

```java

class Solution {
public int[] findIntersectionValues(int[] nums1, int[] nums2) {
int[] s1 = new int[101];
int[] s2 = new int[101];
for (int x : nums1) {
s1[x] = 1;
}
for (int x : nums2) {
s2[x] = 1;
}
int[] ans = new int[2];
for (int x : nums1) {
ans[0] += s2[x];
}
for (int x : nums2) {
ans[1] += s1[x];
}
return ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
int s1[101]{};
int s2[101]{};
for (int& x : nums1) {
s1[x] = 1;
}
for (int& x : nums2) {
s2[x] = 1;
}
vector<int> ans(2);
for (int& x : nums1) {
ans[0] += s2[x];
}
for (int& x : nums2) {
ans[1] += s1[x];
}
return ans;
}
};
```
### **Go**
```go
func findIntersectionValues(nums1 []int, nums2 []int) []int {
s1 := [101]int{}
s2 := [101]int{}
for _, x := range nums1 {
s1[x] = 1
}
for _, x := range nums2 {
s2[x] = 1
}
ans := make([]int, 2)
for _, x := range nums1 {
ans[0] += s2[x]
}
for _, x := range nums2 {
ans[1] += s1[x]
}
return ans
}
```

### **TypeScript**

```ts
function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
const s1: number[] = Array(101).fill(0);
const s2: number[] = Array(101).fill(0);
for (const x of nums1) {
s1[x] = 1;
}
for (const x of nums2) {
s2[x] = 1;
}
const ans: number[] = Array(2).fill(0);
for (const x of nums1) {
ans[0] += s2[x];
}
for (const x of nums2) {
ans[1] += s1[x];
}
return ans;
}
```

### **...**
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21 changes: 21 additions & 0 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.cpp
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class Solution {
public:
vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
int s1[101]{};
int s2[101]{};
for (int& x : nums1) {
s1[x] = 1;
}
for (int& x : nums2) {
s2[x] = 1;
}
vector<int> ans(2);
for (int& x : nums1) {
ans[0] += s2[x];
}
for (int& x : nums2) {
ans[1] += s1[x];
}
return ans;
}
};
18 changes: 18 additions & 0 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
func findIntersectionValues(nums1 []int, nums2 []int) []int {
s1 := [101]int{}
s2 := [101]int{}
for _, x := range nums1 {
s1[x] = 1
}
for _, x := range nums2 {
s2[x] = 1
}
ans := make([]int, 2)
for _, x := range nums1 {
ans[0] += s2[x]
}
for _, x := range nums2 {
ans[1] += s1[x]
}
return ans
}
20 changes: 20 additions & 0 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
class Solution {
public int[] findIntersectionValues(int[] nums1, int[] nums2) {
int[] s1 = new int[101];
int[] s2 = new int[101];
for (int x : nums1) {
s1[x] = 1;
}
for (int x : nums2) {
s2[x] = 1;
}
int[] ans = new int[2];
for (int x : nums1) {
ans[0] += s2[x];
}
for (int x : nums2) {
ans[1] += s1[x];
}
return ans;
}
}
4 changes: 4 additions & 0 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.py
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@@ -0,0 +1,4 @@
class Solution:
def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
s1, s2 = set(nums1), set(nums2)
return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
18 changes: 18 additions & 0 deletions solution/2900-2999/2956.Find Common Elements Between Two Arrays/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
const s1: number[] = Array(101).fill(0);
const s2: number[] = Array(101).fill(0);
for (const x of nums1) {
s1[x] = 1;
}
for (const x of nums2) {
s2[x] = 1;
}
const ans: number[] = Array(2).fill(0);
for (const x of nums1) {
ans[0] += s2[x];
}
for (const x of nums2) {
ans[1] += s1[x];
}
return ans;
}
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