feat(1019-squares-of-a-sorted-array): add py3 solution
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#!/usr/bin/env zx
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import Turndown from "turndown";
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import * as cheerio from "cheerio";
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import { client, gql } from "./graphql.mjs";
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const langExt = { python3: "py", nodejs: "js", cpp: "cpp" };
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@ -35,14 +36,12 @@ const questionDir = `${questionIdPadded}_${q.titleSlug}`;
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const solutionDir = `${questionDir}/${language}`;
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const solutionFilePath = `${solutionDir}/solution.${langExt[language]}`;
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const $ch = cheerio.load(q.content);
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$ch("pre").wrapInner("<code></code>");
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const td = new Turndown({});
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td.addRule("pre-code", {
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filter: "pre",
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replacement(content) {
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return "```\n" + content + "```";
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},
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});
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const mdBody = td.turndown(q.content);
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const mdBody = td.turndown($ch("body").html());
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await $`mkdir -p ${solutionDir}`;
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await $`touch ${solutionFilePath}`;
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1019_squares-of-a-sorted-array/README.md
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1019_squares-of-a-sorted-array/README.md
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Given an integer array `nums` sorted in **non-decreasing** order, return _an array of **the squares of each number** sorted in non-decreasing order_.
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**Example 1:**
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Input: nums = [-4,-1,0,3,10]
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Output: [0,1,9,16,100]
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Explanation: After squaring, the array becomes [16,1,0,9,100].
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After sorting, it becomes [0,1,9,16,100].
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**Example 2:**
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Input: nums = [-7,-3,2,3,11]
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Output: [4,9,9,49,121]
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**Constraints:**
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* `1 <= nums.length <= 104`
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* `-104 <= nums[i] <= 104`
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* `nums` is sorted in **non-decreasing** order.
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**Follow up:** Squaring each element and sorting the new array is very trivial, could you find an `O(n)` solution using a different approach?
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1019_squares-of-a-sorted-array/python3/solution.py
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1019_squares-of-a-sorted-array/python3/solution.py
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class Solution:
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def sortedSquares(self, nums: List[int]) -> List[int]:
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left = 0
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right = i = len(nums) - 1
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result = [None] * len(nums)
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while i >= 0:
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if abs(nums[left]) > abs(nums[right]):
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result[i] = nums[left] ** 2
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left += 1
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else:
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result[i] = nums[right] ** 2
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right -= 1
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i -= 1
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return result
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