Merge pull request #1183 from parikhitritgithub/dev2

Adding Solution of a problem 442

Author: ajay-dhangar

dsa-solutions/lc-solutions/0400-0499/0442-Find-All-Duplicates-in-an-Array.md

@@ -0,0 +1,159 @@
+---
+id: find-all-duplicates-in-an-array
+title: find-all-duplicates-in-an-array
+sidebar_label: 0442 find-all-duplicates-in-an-array
+tags:
+  - hashmap
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: This is a solution to the. Find All Duplicates in an Array problem on LeetCode
+---
+
+## Problem Description
+
+Given an integer array nums of length n where all the integers of nums are in the range `[1, n]` and each integer appears once or twice, return an array of all the integers that appears twice.
+
+You must write an algorithm that runs in $O(n)$ time and uses only constant extra space.
+
+### Examples
+
+**Example 1:**
+
+```
+
+Input: nums = [4,3,2,7,8,2,3,1]
+Output: [2,3]
+
+```
+
+**Example 2:**
+
+
+```
+Input: nums = [1,1,2]
+Output: [1]
+```
+
+**Example 3:**
+
+
+```
+Input: nums = [1]
+Output: []
+```
+
+
+### Constraints
+
+-  $1 \leq \text{nums.length} \leq  105$
+- Each element in nums appears once or twice.
+
+
+---
+
+## Solution for Find All Duplicates in an Array
+
+### Intuition
+
+   - Since each element in the array represents a valid index `(1 to n)`, we can leverage this information.
+   - We might be able to sort the array and then compare adjacent elements to find duplicates efficiently.
+
+
+
+### Approach
+ 
+   - Utilize existing order: We know that each element should appear at most twice and all values are in the range `[1, n]`. This suggests some inherent order in the array.
+   - Leverage sorting: By sorting the array, we can group duplicate elements together, making them easier to identify.
+   - Iterate and compare: After sorting, iterating through the array, we can compare adjacent elements. If two consecutive elements are identical, that element is a duplicate.
+
+
+
+
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@parikhitkurmi"/>
+    
+   ```python
+//python
+
+class Solution:
+    def findDuplicates(self, nums: List[int]) -> List[int]:
+        ans =[]
+        n=len(nums)
+        for x in nums:
+            x = abs(x)
+            if nums[x-1]<0:
+                ans.append(x)
+            nums[x-1] *= -1
+        return ans
+
+```
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@parikhitkurmi"/>
+
+   ```java
+//java
+
+
+class Solution {
+    public List<Integer> findDuplicates(int[] nums) {
+        List<Integer> ans = new ArrayList<>();
+        int n = nums.length;
+        for (int i = 0; i < n; i++) {
+            int x = Math.abs(nums[i]);
+            if (nums[x - 1] < 0) {
+                ans.add(x);
+            }
+            nums[x - 1] *= -1;
+        }
+        return ans;
+    }
+}
+
+
+```
+</TabItem>
+<TabItem value="C++" label="C++">
+<SolutionAuthor name="@parikhitkurmi"/>
+
+   ```cpp
+//cpp
+
+class Solution {
+public:
+    vector<int> findDuplicates(vector<int>& nums) {
+        vector<int>ans;
+        int n=size(nums);
+        for(int i=0;i<n;i++){
+            int x=abs(nums[i]);
+            if(nums[x-1]<0){
+                
+                ans.push_back(x);
+            }
+            nums[x-1]*=-1;
+        }
+        return ans;
+    }
+};
+
+```
+
+  </TabItem>
+</Tabs>
+
+
+
+
+
+## References
+
+- **LeetCode Problem:** [Continuous Subarray Sum](https://leetcode.com/problems/find-all-duplicates-in-an-array/)
+- **Solution Link:** [Continuous Subarray Sum](https://leetcode.com/problems/find-all-duplicates-in-an-array/submissions/)
+- **Authors GeeksforGeeks Profile:** [parikhit kurmi](https://www.geeksforgeeks.org/user/sololeveler673/)
+- **Authors Leetcode:** [parikhit kurmi](https://leetcode.com/u/parikhitkurmi14/)