Merge pull request codeharborhub#3552 from ImmidiSivani/leetcode-1929

added solution for 1929

dsa-solutions/lc-solutions/0400-0499/0493-Reverse-Pairs.md

@@ -318,7 +318,7 @@ public:
 <h2>Authors:</h2>
 
 <div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
-{['Ishitamukherjee2004'].map(username => (
+{['Ishitamukherjee2004', 'ImmidiSivani'].map(username => (
  <Author key={username} username={username} />
 ))}
-</div>
+</div>

dsa-solutions/lc-solutions/1900-1999/1929-concatenation-of-array.md

@@ -0,0 +1,178 @@
+---
+
+id: concatenation-of-array
+title: Concatenation of Array
+sidebar_label: 1929-Concatenation of Array
+tags:
+  - Concatenation
+  - Arrays
+  - Brute Force
+  - Optimized
+  - LeetCode
+  - Python
+  - Java
+  - C++
+description: "This is a solution to the Concatenate Array problem on LeetCode."
+sidebar_position: 2
+
+---
+
+In this tutorial, we will solve the Concatenate Array problem using a simple approach and an optimized approach. We will provide the implementation of the solution in C++, Java, and Python.
+
+## Problem Description
+
+Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (0-indexed).
+
+Specifically, `ans` is the concatenation of two `nums` arrays.
+
+Return the array `ans`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: nums = [1,2,1]
+Output: [1,2,1,1,2,1]
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,3,2,1]
+Output: [1,3,2,1,1,3,2,1]
+```
+
+### Constraints
+
+- `n == nums.length`
+- ` 1 <= n <= 1000`
+
+
+
+---
+
+## Solution for Concatenate Array Problem
+
+### Approach 1: Simple Concatenation
+
+The approach is straightforward: iterate through the array `nums` and copy each element twice, once to the first half and once to the second half of the result array `ans`.
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    vector<int> getConcatenation(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> ans(2 * n);
+        for (int i = 0; i < n; ++i) {
+            ans[i] = nums[i];
+            ans[i + n] = nums[i];
+        }
+        return ans;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public int[] getConcatenation(int[] nums) {
+        int n = nums.length;
+        int[] ans = new int[2 * n];
+        for (int i = 0; i < n; i++) {
+            ans[i] = nums[i];
+            ans[i + n] = nums[i];
+        }
+        return ans;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def getConcatenation(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * (2 * n)
+        for i in range(n):
+            ans[i] = nums[i]
+            ans[i + n] = nums[i]
+        return ans
+```
+
+</TabItem>
+</Tabs>
+
+### Approach 2: Optimized Concatenation
+
+An optimized approach for concatenation in some languages can take advantage of built-in functions to concatenate the array directly.
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    vector<int> getConcatenation(vector<int>& nums) {
+        vector<int> ans = nums;
+        ans.insert(ans.end(), nums.begin(), nums.end());
+        return ans;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+import java.util.*;
+
+class Solution {
+    public int[] getConcatenation(int[] nums) {
+        int n = nums.length;
+        int[] ans = Arrays.copyOf(nums, 2 * n);
+        System.arraycopy(nums, 0, ans, n, n);
+        return ans;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def getConcatenation(self, nums: List[int]) -> List[int]:
+        return nums + nums
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity:** $O(n)$, where `n` is the length of the input array `nums`. We iterate through the array once.
+- **Space Complexity:** $O(n)$, since we are using a new array `ans` of size `2n`.
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>