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+# 0977. Squares of a Sorted Array
+
+* Difficulty: easy
+* Link: https://leetcode.com/problems/squares-of-a-sorted-array/
+* Topics: Array-String
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT: List[int]
+    - OUTPUT: List[int]
+
+# Naive Solution
+
+### Thought Process
+
+1. squares of each number
+2. sort it
+- Implement
+    
+    ```python
+    class Solution:
+        def sortedSquares(self, nums: List[int]) -> List[int]:
+            """
+            1. squares of each number
+            2. sort it
+            """
+            for idx, num in enumerate(nums):
+                nums[idx] = num ** 2
+            nums.sort()
+            return nums
+    ```
+    
+
+### Complexity
+
+- Time complexity:$O(nlogn)$
+    
+    ![Untitled](./Untitled.png)
+    
+- Space complexity:$O(1)$
+
+### Problems & Improvement
+
+- **Follow up:** Squaring each element and sorting the new array is very trivial, could you find an `O(n)` solution using a different approach?
+
+# Improvement
+
+### Thought Process
+
+- `nums` is sorted in **non-decreasing** order.
+1. split the array into two array
+    1. negative array
+    2. positive array
+    
+    ```python
+    [-4,-1,0,3,10]
+     ^^^^^ ^^^^^^
+    ```
+    
+2. Reverse the negative array
+3. begin to traverse two array
+    1. square the num and check which element is larger, insert the larger one
+- Implement
+    
+    ```python
+    class Solution:
+        def sortedSquares(self, nums: List[int]) -> List[int]:
+            # split the array into two array (by linked list)
+            positive = []
+            negative = nums
+            for idx, num in enumerate(nums):
+                if num >= 0:
+                    positive = nums[idx:]
+                    negative = nums[:idx]
+                    break
+    
+            # reverse the negative array
+            negative = negative[::-1]
+    
+            # traverse the array
+            result = []
+            p, n = 0, 0
+            while p < len(positive) and n < len(negative):
+                if abs(negative[n]) < abs(positive[p]):
+                    result.append(negative[n] ** 2)
+                    n += 1
+                else:
+                    result.append(positive[p] ** 2)
+                    p += 1
+            for i in range(p, len(positive)):
+                result.append(positive[i] ** 2)
+            for i in range(n, len(negative)):
+                result.append(negative[i] ** 2)
+    
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity: $O(n)$
+    
+    ![Untitled](./Untitled%201.png)
+    
+- Space complexity:$O(n)$
+
+# Improvement
+
+### Thought Process
+
+- 可以不用 Reverse Array，因為 **non-decreasing** order，兩側最大，從兩側往內夾擊
+- implement
+    
+    ```python
+    class Solution:
+        def sortedSquares(self, nums: List[int]) -> List[int]:
+            # split the array into two array (by linked list)
+            right = len(nums) - 1
+            left = 0
+            # traverse the array
+            result = [0] * len(nums)
+            resultIdx = len(nums) - 1
+            while resultIdx >=0 :
+                if abs(nums[left]) > abs(nums[right]):
+                    result[resultIdx] = nums[left] ** 2
+                    left += 1
+                else:
+                    result[resultIdx] = nums[right] ** 2
+                    right -= 1
+                resultIdx -=1
+    
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity: $O(n)$
+    
+    ![Untitled](./Untitled%202.png)
+    
+- Space complexity:$O(1)$
+
+# Note
+
+- **[Python: A comparison of lots of approaches! [Sorting, two pointers, deque, iterator, generator]](https://leetcode.com/problems/squares-of-a-sorted-array/discuss/310865/Python%3A-A-comparison-of-lots-of-approaches!-Sorting-two-pointers-deque-iterator-generator)**
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