1853. Convert Date Format
Problem
Given a table Days with a day column of date type, the goal is to write an SQL query that will convert each date in the table into a string formatted as "day_name, month_name day, year". The result table should be returned in any order, and the output must be case-sensitive.
Example
Consider the following Days table:
1+------------+
2| day |
3+------------+
4| 2022-04-12 |
5| 2021-08-09 |
6| 2020-06-26 |
7+------------+
The expected query result would be:
1+-------------------------+
2| day |
3+-------------------------+
4| Tuesday, April 12, 2022 |
5| Monday, August 9, 2021 |
6| Friday, June 26, 2020 |
7+-------------------------+
Approach
To solve this problem, MySQL's DATE_FORMAT function can be used. This function allows you to format a date value based on a given format pattern. The following format pattern can be used to achieve the required output:
1%W, %M %e, %Y
This pattern will replace %W with the day name, %M with the month name, %e for the day number without leading zeros, and %Y for the year.
Example
Let's walk through the problem example using this approach:
12022-04-12 -> DATE_FORMAT('2022-04-12', '%W, %M %e, %Y') -> Tuesday, April 12, 2022 22021-08-09 -> DATE_FORMAT('2021-08-09', '%W, %M %e, %Y') -> Monday, August 9, 2021 32020-06-26 -> DATE_FORMAT('2020-06-26', '%W, %M %e, %Y') -> Friday, June 26, 2020
Solution
SQL
To achieve the required result, we can write an SQL query as follows:
1SELECT DATE_FORMAT(day, '%W, %M %e, %Y') AS day 2FROM Days;
What's the output of running the following function using input 56?
1KEYBOARD = {
2 '2': 'abc',
3 '3': 'def',
4 '4': 'ghi',
5 '5': 'jkl',
6 '6': 'mno',
7 '7': 'pqrs',
8 '8': 'tuv',
9 '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13 def dfs(path, res):
14 if len(path) == len(digits):
15 res.append(''.join(path))
16 return
17
18 next_number = digits[len(path)]
19 for letter in KEYBOARD[next_number]:
20 path.append(letter)
21 dfs(path, res)
22 path.pop()
23
24 res = []
25 dfs([], res)
26 return res
271private static final Map<Character, char[]> KEYBOARD = Map.of(
2 '2', "abc".toCharArray(),
3 '3', "def".toCharArray(),
4 '4', "ghi".toCharArray(),
5 '5', "jkl".toCharArray(),
6 '6', "mno".toCharArray(),
7 '7', "pqrs".toCharArray(),
8 '8', "tuv".toCharArray(),
9 '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13 List<String> res = new ArrayList<>();
14 dfs(new StringBuilder(), res, digits.toCharArray());
15 return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19 if (path.length() == digits.length) {
20 res.add(path.toString());
21 return;
22 }
23 char next_digit = digits[path.length()];
24 for (char letter : KEYBOARD.get(next_digit)) {
25 path.append(letter);
26 dfs(path, res, digits);
27 path.deleteCharAt(path.length() - 1);
28 }
29}
301const KEYBOARD = {
2 '2': 'abc',
3 '3': 'def',
4 '4': 'ghi',
5 '5': 'jkl',
6 '6': 'mno',
7 '7': 'pqrs',
8 '8': 'tuv',
9 '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13 let res = [];
14 dfs(digits, [], res);
15 return res;
16}
17
18function dfs(digits, path, res) {
19 if (path.length === digits.length) {
20 res.push(path.join(''));
21 return;
22 }
23 let next_number = digits.charAt(path.length);
24 for (let letter of KEYBOARD[next_number]) {
25 path.push(letter);
26 dfs(digits, path, res);
27 path.pop();
28 }
29}
30Solution Implementation
What is the worst case running time for finding an element in a binary tree (not necessarily binary search tree) of size n?
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