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feat: add solutions to lc problem: No.2113
No.2113.Elements in Array After Removing and Replacing Elements
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‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：直接计算**
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我们先初始化一个数组 $ans$，长度为 $m$，用于存储答案，初始化所有元素为 $-1$。
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接下来遍历数组 $queries$，对于每个查询，我们先获取当前查询的时间 $t$ 和索引 $i$，先将 $t$ 对 $2n$ 取模，然后判断 $t$ 和 $n$ 的关系：
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- 如果 $t \lt n$，那么 $t$ 时刻的数组元素个数为 $n - t$，并且数组元素是原数组元素整体向左移动 $t$ 个位置后的结果，因此如果 $i \lt n - t$，答案为 $nums[i + t]$；
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- 如果 $t \gt n$，那么 $t$ 时刻的数组元素个数为 $t - n$，并且数组元素是原数组元素的前 $t - n$ 个元素，因此如果 $i \lt t - n$，答案为 $nums[i]$。
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最后返回数组 $ans$ 即可。
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时间复杂度 $O(m)$，其中 $m$ 为数组 $queries$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
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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 elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]:
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n, m = len(nums), len(queries)
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ans = [-1] * m
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for j, (t, i) in enumerate(queries):
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t %= (2 * n)
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if t < n and i < n - t:
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ans[j] = nums[i + t]
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elif t > n and i < t - n:
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ans[j] = nums[i]
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return ans
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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[] elementInNums(int[] nums, int[][] queries) {
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int n = nums.length, m = queries.length;
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int[] ans = new int[m];
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for (int j = 0; j < m; ++j) {
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ans[j] = -1;
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int t = queries[j][0], i = queries[j][1];
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t %= (2 * n);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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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> elementInNums(vector<int>& nums, vector<vector<int>>& queries) {
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int n = nums.size(), m = queries.size();
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vector<int> ans(m, -1);
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for (int j = 0; j < m; ++j) {
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int t = queries[j][0], i = queries[j][1];
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t %= (n * 2);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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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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### **TypeScript**
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### **Go**
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```go
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func elementInNums(nums []int, queries [][]int) []int {
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n, m := len(nums), len(queries)
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ans := make([]int, m)
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for j, q := range queries {
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t, i := q[0], q[1]
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t %= (n * 2)
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ans[j] = -1
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if t < n && i < n-t {
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ans[j] = nums[i+t]
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} else if t > n && i < t-n {
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ans[j] = nums[i]
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}
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}
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return ans
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/README_EN.md

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### **Python3**
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```python
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class Solution:
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def elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]:
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n, m = len(nums), len(queries)
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ans = [-1] * m
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for j, (t, i) in enumerate(queries):
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t %= (2 * n)
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if t < n and i < n - t:
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ans[j] = nums[i + t]
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elif t > n and i < t - n:
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ans[j] = nums[i]
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return ans
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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[] elementInNums(int[] nums, int[][] queries) {
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int n = nums.length, m = queries.length;
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int[] ans = new int[m];
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for (int j = 0; j < m; ++j) {
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ans[j] = -1;
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int t = queries[j][0], i = queries[j][1];
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t %= (2 * n);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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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> elementInNums(vector<int>& nums, vector<vector<int>>& queries) {
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int n = nums.size(), m = queries.size();
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vector<int> ans(m, -1);
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for (int j = 0; j < m; ++j) {
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int t = queries[j][0], i = queries[j][1];
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t %= (n * 2);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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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 elementInNums(nums []int, queries [][]int) []int {
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n, m := len(nums), len(queries)
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ans := make([]int, m)
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for j, q := range queries {
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t, i := q[0], q[1]
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t %= (n * 2)
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ans[j] = -1
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if t < n && i < n-t {
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ans[j] = nums[i+t]
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} else if t > n && i < t-n {
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ans[j] = nums[i]
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}
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}
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return ans
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}
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```
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### **TypeScript**

‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/Solution.cpp

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class Solution {
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public:
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vector<int> elementInNums(vector<int>& nums, vector<vector<int>>& queries) {
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int n = nums.size(), m = queries.size();
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vector<int> ans(m, -1);
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for (int j = 0; j < m; ++j) {
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int t = queries[j][0], i = queries[j][1];
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t %= (n * 2);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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}
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return ans;
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}
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};

‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/Solution.go

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func elementInNums(nums []int, queries [][]int) []int {
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n, m := len(nums), len(queries)
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ans := make([]int, m)
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for j, q := range queries {
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t, i := q[0], q[1]
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t %= (n * 2)
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ans[j] = -1
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if t < n && i < n-t {
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ans[j] = nums[i+t]
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} else if t > n && i < t-n {
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ans[j] = nums[i]
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}
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}
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return ans
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}

‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/Solution.java

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class Solution {
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public int[] elementInNums(int[] nums, int[][] queries) {
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int n = nums.length, m = queries.length;
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int[] ans = new int[m];
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for (int j = 0; j < m; ++j) {
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ans[j] = -1;
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int t = queries[j][0], i = queries[j][1];
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t %= (2 * n);
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if (t < n && i < n - t) {
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ans[j] = nums[i + t];
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} else if (t > n && i < t - n) {
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ans[j] = nums[i];
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}
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}
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return ans;
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}
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}

‎solution/2100-2199/2113.Elements in Array After Removing and Replacing Elements/Solution.py

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class Solution:
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def elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]:
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n, m = len(nums), len(queries)
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ans = [-1] * m
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for j, (t, i) in enumerate(queries):
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t %= (2 * n)
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if t < n and i < n - t:
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ans[j] = nums[i + t]
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elif t > n and i < t - n:
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ans[j] = nums[i]
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return ans

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