.scripts/addnew

@@ -4,7 +4,7 @@ import Turndown from "turndown";
 import * as cheerio from "cheerio";
 import { client, gql } from "./graphql.mjs";
 
-const langExt = { python3: "py", nodejs: "js", cpp: "cpp", sql: "sql" };
+const langExt = { python3: "py", nodejs: "js", cpp: "cpp" };
 
 const link = await question("Enter leetcode problem link: ");
 const titleSlug = path.basename(new URL(link).pathname);

1308_smallest-string-with-swaps/README.md

@@ -1,42 +0,0 @@
-You are given a string `s`, and an array of pairs of indices in the string `pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the string.
-
-You can swap the characters at any pair of indices in the given `pairs` **any number of times**.
-
-Return the lexicographically smallest string that `s` can be changed to after using the swaps.
-
-**Example 1:**
-
-    Input: s = "dcab", pairs = [[0,3],[1,2]]
-    Output: "bacd"
-    Explaination: 
-    Swap s[0] and s[3], s = "bcad"
-    Swap s[1] and s[2], s = "bacd"
-    
-
-**Example 2:**
-
-    Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
-    Output: "abcd"
-    Explaination: 
-    Swap s[0] and s[3], s = "bcad"
-    Swap s[0] and s[2], s = "acbd"
-    Swap s[1] and s[2], s = "abcd"
-
-**Example 3:**
-
-    Input: s = "cba", pairs = [[0,1],[1,2]]
-    Output: "abc"
-    Explaination: 
-    Swap s[0] and s[1], s = "bca"
-    Swap s[1] and s[2], s = "bac"
-    Swap s[0] and s[1], s = "abc"
-    
-
-**Constraints:**
-
-*   `1 <= s.length <= 10^5`
-*   `0 <= pairs.length <= 10^5`
-*   `0 <= pairs[i][0], pairs[i][1] < s.length`
-*   `s` only contains lower case English letters.
-
-https://leetcode.com/problems/smallest-string-with-swaps

1308_smallest-string-with-swaps/python3/solution.py

@@ -1,52 +0,0 @@
-from collections import defaultdict
-
-class Solution:
-    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
-        '''
-        Key idea: if (a, b) indexes can be swapped and (b, c) indexes can
-        be swapped, then (a, c) can be swapped so infinite swap across (a, b, c).
-        Can be expanded to cover any other letters that are swappable.
-        '''
-        
-        # Make the adjacency list
-        swapmap = defaultdict(list)
-        
-        for i, j in pairs:
-            swapmap[i].append(j)
-            swapmap[j].append(i)
-            
-        
-        def dfs(s, vertex, chars, vertices):
-            chars.append(s[vertex])
-            vertices.append(vertex)
-            
-            visited.add(vertex)
-            
-            for neighbor in swapmap[vertex]:
-                if neighbor not in visited:
-                    dfs(s, neighbor, chars, vertices)
-        
-        
-        output = [''] * len(s)
-        visited = set()
-        
-        for vertex, _ in enumerate(s):
-            if vertex not in visited:
-                connected_chars = []
-                connected_vertices = []
-                
-                dfs(s, vertex, connected_chars, connected_vertices)
-                
-                # This might seem confusing but remember that if we find one character
-                # with the `vertex` connected (i.e swappable) with other characters, then
-                # they can all be swapped with each other infinite times. Our goal is to
-                # get the lexicographically smallest string (i.e most sorted) from doing
-                # these allowed swaps. Hence, we can go ahead and safely sort the chars
-                # obtained from DFS of vertex.
-                connected_chars.sort()
-                connected_vertices.sort()
-                
-                for i, connected_vertex in enumerate(connected_vertices):
-                    output[connected_vertex] = connected_chars[i]
-        
-        return ''.join(output)

1541_top-travellers/README.md

@@ -1,78 +0,0 @@
-Table: `Users`
-
-    +---------------+---------+
-    | Column Name   | Type    |
-    +---------------+---------+
-    | id            | int     |
-    | name          | varchar |
-    +---------------+---------+
-    id is the primary key for this table.
-    name is the name of the user.
-    
-
-Table: `Rides`
-
-    +---------------+---------+
-    | Column Name   | Type    |
-    +---------------+---------+
-    | id            | int     |
-    | user_id       | int     |
-    | distance      | int     |
-    +---------------+---------+
-    id is the primary key for this table.
-    user_id is the id of the user who traveled the distance "distance".
-    
-
-Write an SQL query to report the distance traveled by each user.
-
-Return the result table ordered by `travelled_distance` in **descending order**, if two or more users traveled the same distance, order them by their `name` in **ascending order**.
-
-The query result format is in the following example.
-
-**Example 1:**
-
-    Input: 
-    Users table:
-    +------+-----------+
-    | id   | name      |
-    +------+-----------+
-    | 1    | Alice     |
-    | 2    | Bob       |
-    | 3    | Alex      |
-    | 4    | Donald    |
-    | 7    | Lee       |
-    | 13   | Jonathan  |
-    | 19   | Elvis     |
-    +------+-----------+
-    Rides table:
-    +------+----------+----------+
-    | id   | user_id  | distance |
-    +------+----------+----------+
-    | 1    | 1        | 120      |
-    | 2    | 2        | 317      |
-    | 3    | 3        | 222      |
-    | 4    | 7        | 100      |
-    | 5    | 13       | 312      |
-    | 6    | 19       | 50       |
-    | 7    | 7        | 120      |
-    | 8    | 19       | 400      |
-    | 9    | 7        | 230      |
-    +------+----------+----------+
-    Output: 
-    +----------+--------------------+
-    | name     | travelled_distance |
-    +----------+--------------------+
-    | Elvis    | 450                |
-    | Lee      | 450                |
-    | Bob      | 317                |
-    | Jonathan | 312                |
-    | Alex     | 222                |
-    | Alice    | 120                |
-    | Donald   | 0                  |
-    +----------+--------------------+
-    Explanation: 
-    Elvis and Lee traveled 450 miles, Elvis is the top traveler as his name is alphabetically smaller than Lee.
-    Bob, Jonathan, Alex, and Alice have only one ride and we just order them by the total distances of the ride.
-    Donald did not have any rides, the distance traveled by him is 0.
-
-https://leetcode.com/problems/top-travellers/

1541_top-travellers/sql/solution.sql

@@ -1,6 +0,0 @@
-# Write your MySQL query statement below
-SELECT Users.name as name, COALESCE(SUM(distance), 0) as travelled_distance
-FROM Users
-LEFT JOIN Rides ON Users.id = Rides.user_id
-GROUP BY Users.id
-ORDER BY travelled_distance DESC, name ASC;