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Commit e26a69a

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feat: add solutions to lc problems: No.2956,2957 (#2077)
* No.2956.Find Common Elements Between Two Arrays * No.2957.Remove Adjacent Almost-Equal Characters
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14 files changed

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‎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 -->
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**方法一:哈希表或数组**
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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。
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接下来,我们创建一个长度为 2ドル$ 的数组 $ans,ドル其中 $ans[0]$ 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。
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然后,我们遍历数组 $nums1$ 中的每个元素 $x,ドル如果 $x$ 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,ドル如果 $x$ 在 $s1$ 中出现过,则将 $ans[1]$ 加一。
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最后,我们返回数组 $ans$ 即可。
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时间复杂度 $O(n + m),ドル空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是数组 $nums1$ 和 $nums2$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
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s1, s2 = set(nums1), set(nums2)
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return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] findIntersectionValues(int[] nums1, int[] nums2) {
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int[] s1 = new int[101];
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int[] s2 = new int[101];
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for (int x : nums1) {
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s1[x] = 1;
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}
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for (int x : nums2) {
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s2[x] = 1;
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}
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int[] ans = new int[2];
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for (int x : nums1) {
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ans[0] += s2[x];
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}
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for (int x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
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int s1[101]{};
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int s2[101]{};
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for (int& x : nums1) {
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s1[x] = 1;
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}
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for (int& x : nums2) {
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s2[x] = 1;
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}
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vector<int> ans(2);
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for (int& x : nums1) {
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ans[0] += s2[x];
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}
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for (int& x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findIntersectionValues(nums1 []int, nums2 []int) []int {
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s1 := [101]int{}
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s2 := [101]int{}
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for _, x := range nums1 {
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s1[x] = 1
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}
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for _, x := range nums2 {
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s2[x] = 1
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}
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ans := make([]int, 2)
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for _, x := range nums1 {
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ans[0] += s2[x]
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}
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for _, x := range nums2 {
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ans[1] += s1[x]
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
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const s1: number[] = Array(101).fill(0);
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const s2: number[] = Array(101).fill(0);
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for (const x of nums1) {
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s1[x] = 1;
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}
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for (const x of nums2) {
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s2[x] = 1;
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}
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const ans: number[] = Array(2).fill(0);
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for (const x of nums1) {
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ans[0] += s2[x];
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}
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for (const x of nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2956.Find Common Elements Between Two Arrays/README_EN.md‎

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## Solutions
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**Solution 1: Hash Table or Array**
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We can use two hash tables or arrays $s1$ and $s2$ to record the elements that appear in the two arrays respectively.
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Next, we create an array $ans$ of length 2ドル,ドル where $ans[0]$ represents the number of elements in $nums1$ that appear in $s2,ドル and $ans[1]$ represents the number of elements in $nums2$ that appear in $s1$.
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Then, we traverse each element $x$ in the array $nums1$. If $x$ has appeared in $s2,ドル we increment $ans[0]$. After that, we traverse each element $x$ in the array $nums2$. If $x$ has appeared in $s1,ドル we increment $ans[1]$.
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Finally, we return the array $ans$.
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The time complexity is $O(n + m),ドル and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$ respectively.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
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s1, s2 = set(nums1), set(nums2)
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return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
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```
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### **Java**
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```java
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class Solution {
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public int[] findIntersectionValues(int[] nums1, int[] nums2) {
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int[] s1 = new int[101];
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int[] s2 = new int[101];
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for (int x : nums1) {
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s1[x] = 1;
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}
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for (int x : nums2) {
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s2[x] = 1;
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}
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int[] ans = new int[2];
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for (int x : nums1) {
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ans[0] += s2[x];
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}
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for (int x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
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int s1[101]{};
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int s2[101]{};
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for (int& x : nums1) {
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s1[x] = 1;
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}
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for (int& x : nums2) {
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s2[x] = 1;
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}
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vector<int> ans(2);
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for (int& x : nums1) {
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ans[0] += s2[x];
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}
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for (int& x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findIntersectionValues(nums1 []int, nums2 []int) []int {
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s1 := [101]int{}
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s2 := [101]int{}
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for _, x := range nums1 {
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s1[x] = 1
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}
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for _, x := range nums2 {
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s2[x] = 1
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}
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ans := make([]int, 2)
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for _, x := range nums1 {
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ans[0] += s2[x]
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}
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for _, x := range nums2 {
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ans[1] += s1[x]
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
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const s1: number[] = Array(101).fill(0);
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const s2: number[] = Array(101).fill(0);
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for (const x of nums1) {
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s1[x] = 1;
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}
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for (const x of nums2) {
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s2[x] = 1;
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}
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const ans: number[] = Array(2).fill(0);
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for (const x of nums1) {
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ans[0] += s2[x];
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}
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for (const x of nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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```
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### **...**
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class Solution {
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public:
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vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
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int s1[101]{};
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int s2[101]{};
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for (int& x : nums1) {
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s1[x] = 1;
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}
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for (int& x : nums2) {
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s2[x] = 1;
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}
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vector<int> ans(2);
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for (int& x : nums1) {
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ans[0] += s2[x];
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}
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for (int& x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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};
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func findIntersectionValues(nums1 []int, nums2 []int) []int {
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s1 := [101]int{}
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s2 := [101]int{}
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for _, x := range nums1 {
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s1[x] = 1
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}
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for _, x := range nums2 {
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s2[x] = 1
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}
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ans := make([]int, 2)
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for _, x := range nums1 {
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ans[0] += s2[x]
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}
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for _, x := range nums2 {
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ans[1] += s1[x]
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}
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return ans
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}
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class Solution {
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public int[] findIntersectionValues(int[] nums1, int[] nums2) {
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int[] s1 = new int[101];
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int[] s2 = new int[101];
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for (int x : nums1) {
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s1[x] = 1;
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}
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for (int x : nums2) {
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s2[x] = 1;
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}
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int[] ans = new int[2];
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for (int x : nums1) {
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ans[0] += s2[x];
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}
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for (int x : nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}
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}
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class Solution:
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def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
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s1, s2 = set(nums1), set(nums2)
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return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]
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function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
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const s1: number[] = Array(101).fill(0);
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const s2: number[] = Array(101).fill(0);
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for (const x of nums1) {
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s1[x] = 1;
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}
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for (const x of nums2) {
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s2[x] = 1;
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}
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const ans: number[] = Array(2).fill(0);
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for (const x of nums1) {
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ans[0] += s2[x];
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}
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for (const x of nums2) {
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ans[1] += s1[x];
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}
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return ans;
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}

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