Squares-of-a-Sorted-Array

Author: Shikha-code36

README.md

@@ -33,3 +33,6 @@ Check the notes for the explaination - [Notes](https://stingy-shallot-4ea.notion
 -[x][Contains Duplicate](Arrays/217-Contains-duplicate.py)
 -[x][Product Of Array Except Self](Arrays/238-product-of-array-except-self.py)
 
+-[x][Two Pointers](Two-Pointers)
+-[x][Squares of a Sorted Array](Two-Pointers/977-Squares-of-a-Sorted-Array.py)
+

Two-Pointers/977-Squares-of-a-Sorted-Array.py

@@ -0,0 +1,42 @@
+"Leetcode- https://leetcode.com/problems/squares-of-a-sorted-array/ "
+'''
+Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
+
+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].
+'''
+
+# Solution 1
+defsortedSquares(self, nums):
+ans= []
+
+foriinrange(len(nums)):
+ans.append(nums[i]*nums[i])
+i+=1
+
+ans.sort()
+returnans
+
+#T: O(NLOGN)
+#S: O(N)
+
+# Solution 2 - Two Pointer
+defsortedSquares(self, nums):
+i=0
+j=len(nums)-1
+temp= []
+while(i<=j):
+ifnums[i]*nums[i] >nums[j]*nums[j]:
+temp.append(nums[i]*nums[i])
+i+=1
+else:
+temp.append(nums[j]*nums[j])
+j-=1
+returntemp.sort()
+
+#T: O(N)
+#S: O(N)