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feat: add solutions to lc problems: No.2978,2979 (#2162)
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‎solution/2900-2999/2978.Symmetric Coordinates/README.md

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# [2978. Symmetric Coordinates](https://leetcode.cn/problems/symmetric-coordinates)
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[English Version](/solution/2900-2999/2978.Symmetric%20Coordinates/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <font face="monospace"><code>Pairs</code></font></p>
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<pre>
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+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| X | int |
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| Y | int |
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+-------------+------+
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Each row includes X and Y, where both are integers. Table may contain duplicate values.
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</pre>
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<p>Two coordindates <code>(X1, Y1)</code> and <code>(X2, Y2)</code> are said to be <strong>symmetric</strong> coordintes if <code>X1 == Y2</code> and <code>X2 == Y1</code>.</p>
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<p>Write a solution that outputs, among all these <strong>symmetric</strong> <strong>coordintes</strong>, only those <strong>unique</strong> coordinates that satisfy the condition <code>X1 &lt;= Y1</code>.</p>
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<p>Return <em>the result table ordered by </em><code>X</code> <em>and </em> <code>Y</code> <em>(respectively)</em> <em>in <strong>ascending order</strong></em>.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Coordinates table:
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+----+----+
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| X | Y |
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+----+----+
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| 20 | 20 |
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| 20 | 20 |
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| 20 | 21 |
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| 23 | 22 |
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| 22 | 23 |
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| 21 | 20 |
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+----+----+
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<strong>Output:</strong>
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+----+----+
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| x | y |
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+----+----+
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| 20 | 20 |
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| 20 | 21 |
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| 22 | 23 |
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+----+----+
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<strong>Explanation:</strong>
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- (20, 20) and (20, 20) are symmetric coordinates because, X1 == Y2 and X2 == Y1. This results in displaying (20, 20) as a distinctive coordinates.
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- (20, 21) and (21, 20) are symmetric coordinates because, X1 == Y2 and X2 == Y1. However, only (20, 21) will be displayed because X1 &lt;= Y1.
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- (23, 22) and (22, 23) are symmetric coordinates because, X1 == Y2 and X2 == Y1. However, only (22, 23) will be displayed because X1 &lt;= Y1.
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The output table is sorted by X and Y in ascending order.
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：窗口函数 + 自连接**
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我们可以使用窗口函数 `ROW_NUMBER()` 来为每一行添加一个自增的序号，然后再自连接两张表，连接条件为 `p1.x = p2.y AND p1.y = p2.x AND p1.x <= p1.y AND p1.id != p2.id`，最后再排序去重即可。
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT
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ROW_NUMBER() OVER () AS id,
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x,
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y
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FROM Coordinates
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)
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SELECT DISTINCT
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p1.x,
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p1.y
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FROM
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P AS p1
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JOIN P AS p2 ON p1.x = p2.y AND p1.y = p2.x AND p1.x <= p1.y AND p1.id != p2.id
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2978.Symmetric Coordinates/README_EN.md

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# [2978. Symmetric Coordinates](https://leetcode.com/problems/symmetric-coordinates)
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[中文文档](/solution/2900-2999/2978.Symmetric%20Coordinates/README.md)
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## Description
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<p>Table: <font face="monospace"><code>Pairs</code></font></p>
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<pre>
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+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| X | int |
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| Y | int |
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+-------------+------+
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Each row includes X and Y, where both are integers. Table may contain duplicate values.
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</pre>
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<p>Two coordindates <code>(X1, Y1)</code> and <code>(X2, Y2)</code> are said to be <strong>symmetric</strong> coordintes if <code>X1 == Y2</code> and <code>X2 == Y1</code>.</p>
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<p>Write a solution that outputs, among all these <strong>symmetric</strong> <strong>coordintes</strong>, only those <strong>unique</strong> coordinates that satisfy the condition <code>X1 &lt;= Y1</code>.</p>
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<p>Return <em>the result table ordered by </em><code>X</code> <em>and </em> <code>Y</code> <em>(respectively)</em> <em>in <strong>ascending order</strong></em>.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Coordinates table:
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+----+----+
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| X | Y |
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+----+----+
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| 20 | 20 |
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| 20 | 20 |
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| 20 | 21 |
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| 23 | 22 |
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| 22 | 23 |
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| 21 | 20 |
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+----+----+
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<strong>Output:</strong>
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+----+----+
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| x | y |
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+----+----+
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| 20 | 20 |
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| 20 | 21 |
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| 22 | 23 |
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+----+----+
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<strong>Explanation:</strong>
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- (20, 20) and (20, 20) are symmetric coordinates because, X1 == Y2 and X2 == Y1. This results in displaying (20, 20) as a distinctive coordinates.
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- (20, 21) and (21, 20) are symmetric coordinates because, X1 == Y2 and X2 == Y1. However, only (20, 21) will be displayed because X1 &lt;= Y1.
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- (23, 22) and (22, 23) are symmetric coordinates because, X1 == Y2 and X2 == Y1. However, only (22, 23) will be displayed because X1 &lt;= Y1.
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The output table is sorted by X and Y in ascending order.
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</pre>
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## Solutions
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**Solution 1: Window Function + Self Join**
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We can use the window function `ROW_NUMBER()` to add an auto-incrementing sequence number to each row. Then, we perform a self join on the two tables, with the join conditions being `p1.x = p2.y AND p1.y = p2.x AND p1.x <= p1.y AND p1.id != p2.id`. Finally, we sort and remove duplicates.
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT
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ROW_NUMBER() OVER () AS id,
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x,
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y
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FROM Coordinates
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)
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SELECT DISTINCT
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p1.x,
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p1.y
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FROM
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P AS p1
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JOIN P AS p2 ON p1.x = p2.y AND p1.y = p2.x AND p1.x <= p1.y AND p1.id != p2.id
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2978.Symmetric Coordinates/Solution.sql

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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT
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ROW_NUMBER() OVER () AS id,
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x,
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y
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FROM Coordinates
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)
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SELECT DISTINCT
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p1.x,
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p1.y
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FROM
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P AS p1
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JOIN P AS p2 ON p1.x = p2.y AND p1.y = p2.x AND p1.x <= p1.y AND p1.id != p2.id
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ORDER BY 1, 2;

‎solution/2900-2999/2979.Most Expensive Item That Can Not Be Bought/README.md

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# [2979. Most Expensive Item That Can Not Be Bought](https://leetcode.cn/problems/most-expensive-item-that-can-not-be-bought)
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[English Version](/solution/2900-2999/2979.Most%20Expensive%20Item%20That%20Can%20Not%20Be%20Bought/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given two <strong>distinct</strong> <strong>prime</strong> numbers <code>primeOne</code> and <code>primeTwo</code>.</p>
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<p>Alice and Bob are visiting a market. The market has an <strong>infinite</strong> number of items, for <strong>any</strong> positive integer <code>x</code> there exists an item whose price is <code>x</code>. Alice wants to buy some items from the market to gift to Bob. She has an <strong>infinite</strong> number of coins in the denomination <code>primeOne</code> and <code>primeTwo</code>. She wants to know the <strong>most expensive</strong> item she can <strong>not</strong> buy to gift to Bob.</p>
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<p>Return <em>the price of the <strong>most expensive</strong> item which Alice can not gift to Bob</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> primeOne = 2, primeTwo = 5
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> The prices of items which cannot be bought are [1,3]. It can be shown that all items with a price greater than 3 can be bought using a combination of coins of denominations 2 and 5.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> primeOne = 5, primeTwo = 7
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<strong>Output:</strong> 23
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<strong>Explanation:</strong> The prices of items which cannot be bought are [1,2,3,4,6,8,9,11,13,16,18,23]. It can be shown that all items with a price greater than 23 can be bought.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt; primeOne, primeTwo &lt; 10<sup>4</sup></code></li>
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<li><code>primeOne</code>, <code>primeTwo</code> are prime numbers.</li>
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<li><code>primeOne * primeTwo &lt; 10<sup>5</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：Chicken McNugget 定理**
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根据 Chicken McNugget 定理，两个互质的正整数 $a$ 和 $b$，最大不能表示的数为 $ab - a - b$。
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时间复杂度 $O(1)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def mostExpensiveItem(self, primeOne: int, primeTwo: int) -> int:
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return primeOne * primeTwo - primeOne - primeTwo
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int mostExpensiveItem(int primeOne, int primeTwo) {
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return primeOne * primeTwo - primeOne - primeTwo;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int mostExpensiveItem(int primeOne, int primeTwo) {
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return primeOne * primeTwo - primeOne - primeTwo;
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}
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};
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```
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### **Go**
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```go
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func mostExpensiveItem(primeOne int, primeTwo int) int {
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return primeOne*primeTwo - primeOne - primeTwo
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}
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```
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### **TypeScript**
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```ts
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function mostExpensiveItem(primeOne: number, primeTwo: number): number {
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return primeOne * primeTwo - primeOne - primeTwo;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn most_expensive_item(prime_one: i32, prime_two: i32) -> i32 {
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prime_one * prime_two - prime_one - prime_two
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}
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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