feat(1019-squares-of-a-sorted-array): add py3 solution

.scripts/addnew

@@ -1,6 +1,7 @@
 #!/usr/bin/env zx
 
 import Turndown from "turndown";
+import * as cheerio from "cheerio";
 import { client, gql } from "./graphql.mjs";
 
 const langExt = { python3: "py", nodejs: "js", cpp: "cpp" };
@@ -35,14 +36,12 @@ const questionDir = `${questionIdPadded}_${q.titleSlug}`;
 const solutionDir = `${questionDir}/${language}`;
 const solutionFilePath = `${solutionDir}/solution.${langExt[language]}`;
 
+const $ch = cheerio.load(q.content);
+
+$ch("pre").wrapInner("<code></code>");
+
 const td = new Turndown({});
-td.addRule("pre-code", {
-  filter: "pre",
-  replacement(content) {
-    return "```\n" + content + "```";
-  },
-});
-const mdBody = td.turndown(q.content);
+const mdBody = td.turndown($ch("body").html());
 
 await $`mkdir -p ${solutionDir}`;
 await $`touch ${solutionFilePath}`;

1019_squares-of-a-sorted-array/README.md

@@ -0,0 +1,23 @@
+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].
+    
+
+**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.
+
+**Follow up:** Squaring each element and sorting the new array is very trivial, could you find an `O(n)` solution using a different approach?

1019_squares-of-a-sorted-array/python3/solution.py

@@ -0,0 +1,17 @@
+class Solution:
+    def sortedSquares(self, nums: List[int]) -> List[int]:
+        left = 0
+        right = i = len(nums) - 1
+        result = [None] * len(nums)
+        
+        while i >= 0:
+            if abs(nums[left]) > abs(nums[right]):
+                result[i] = nums[left] ** 2
+                left += 1
+            else:
+                result[i] = nums[right] ** 2
+                right -= 1
+            
+            i -= 1
+        
+        return result