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+# Leetcode Rotate-Array
+
+#### 2022-07-10 08:54
+
+> ##### Algorithms:
+> #algorithm #array_in_place_operation
+> ##### Data structures:
+> #DS #array 
+> ##### Difficulty:
+> #coding_problem #difficulty-medium 
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#array_in_place_operation
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/rotate-array/)
+___
+### Problem
+
+Given an array, rotate the array to the right by `k` steps, where `k` is non-negative.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,5,6,7], k = 3
+**Output:** [5,6,7,1,2,3,4]
+**Explanation:**
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+
+**Example 2:**
+
+**Input:** nums = [-1,-100,3,99], k = 2
+**Output:** [3,99,-1,-100]
+**Explanation:** 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+
+#### Constraints
+
+-   `1 <= nums.length <= 105`
+-   `-231 <= nums[i] <= 231 - 1`
+-   `0 <= k <= 105`
+
+### Thoughts
+
+#### Method 1: Array In-place operation
+
+#### Method 2: Copy
+
+#### Method 3: CPP's STL
+
+### Solution
diff --git a/OJ notes/pages/Leetcode Squares-of-a-Sorted-Array.md b/OJ notes/pages/Leetcode Squares-of-a-Sorted-Array.md
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+# Leetcode Squares-of-a-Sorted-Array
+
+#### 2022-07-10 08:50
+
+> ##### Algorithms:
+> #algorithm #two_pointers 
+> ##### Data structures:
+> #DS #array 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy 
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#two_pointers
+```
+ 
+- [[Leetcode Intersection-of-Two-Arrays-II]]
+- [[Leetcode Merge-Sorted-Array]]
+- [[Leetcode Merge-Two-Sorted-Lists]]
+- [[Two pointers approach]]
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/squares-of-a-sorted-array/)
+___
+### Problem
+
+Given an integer array `nums` sorted in **non-decreasing** order, return _an array of **the squares of each number** sorted in non-decreasing order_.
+
+Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?
+
+#### Examples
+
+**Example 1:**
+
+**Input:** nums = [-4,-1,0,3,10]
+**Output:** [0,1,9,16,100]
+**Explanation:** After squaring, the array becomes [16,1,0,9,100].
+After sorting, it becomes [0,1,9,16,100].
+
+**Example 2:**
+
+**Input:** nums = [-7,-3,2,3,11]
+**Output:** [4,9,9,49,121]
+
+#### Constraints
+
+-   `1 <= nums.length <= 104`
+-   `-104 <= nums[i] <= 104`
+-   `nums` is sorted in **non-decreasing** order.
+
+### Thoughts
+
+> [!summary]
+> This is a #two_pointers 
+
+Using two pointers.
+One from left, one from right end.
+**fill the array in backwards order.**
+
+### Solution
+
+cpp
+```cpp
+class Solution {
+public:
+  vector<int> sortedSquares(vector<int> &nums) {
+    // Double pointers, since array is sorted and access in order.
+    // O(2n)
+
+    int mid;
+    int size = nums.size();
+    int left = 0, right = size - 1;
+    int count = size - 1;
+    vector<int> answer(size);
+
+    for (int i = 0; i < size; i++) {
+      nums[i] = nums[i] * nums[i];
+    }
+
+    while (left <= right) {
+      if (nums[left] >= nums[right]) {
+        answer[count--] = nums[left];
+        left++;
+      } else if (nums[right] > nums[left]) {
+        answer[count--] = nums[right];
+        right--;
+      }
+    }
+
+    return answer;
+  }
+};
+```