6/12 practice

Author: NeoYY

171. Excel Sheet Column Number.cpp

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+/*
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28
+    ...
+Example 1:
+
+Input: "A"
+Output: 1
+Example 2:
+
+Input: "AB"
+Output: 28
+Example 3:
+
+Input: "ZY"
+Output: 701
+
+
+First solution is recursive, second is loop
+*/
+class Solution {
+public:
+    int titleToNumber(string s) {
+        return s == "" ? 0 : ( pow( 26, s.size() - 1 ) * ( s[0] - 64 ) +
+                             titleToNumber( s.substr( 1 ) ) );
+
+    }
+};
+
+
+class Solution {
+public:
+    int titleToNumber(string s) {
+        int res = 0;
+        for ( int i = 0; i < s.size(); ++i )
+        {
+            res *= 26;
+            res += s[i] - 'A' + 1;
+        }
+        return res;
+    }
+};

172. Factorial Trailing Zeroes.cpp

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+/*
+
+Given an integer n, return the number of trailing zeroes in n!.
+
+Example 1:
+
+Input: 3
+Output: 0
+Explanation: 3! = 6, no trailing zero.
+Example 2:
+
+Input: 5
+Output: 1
+Explanation: 5! = 120, one trailing zero.
+Note: Your solution should be in logarithmic time complexity.
+
+*/
+
+class Solution {
+public:
+    int trailingZeroes(int n) {
+        int count = 0;
+        while ( n )
+        {
+            count += n / 5;
+            n = n / 5;
+        }
+        return count;
+    }
+};

175. Combine Two Tables.sql

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+/*
+SQL Schema
+Table: Person
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| PersonId    | int     |
+| FirstName   | varchar |
+| LastName    | varchar |
++-------------+---------+
+PersonId is the primary key column for this table.
+Table: Address
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| AddressId   | int     |
+| PersonId    | int     |
+| City        | varchar |
+| State       | varchar |
++-------------+---------+
+AddressId is the primary key column for this table.
+
+
+Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:
+
+FirstName, LastName, City, State
+
+*/
+
+# Write your MySQL query statement below
+SELECT FirstName, LastName, City, State FROM Person
+LEFT JOIN Address
+ON Person.PersonId = Address.PersonId;

176. Second Highest Salary.sql

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+/*
+Write a SQL query to get the second highest salary from the Employee table.
+
++----+--------+
+| Id | Salary |
++----+--------+
+| 1  | 100    |
+| 2  | 200    |
+| 3  | 300    |
++----+--------+
+For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.
+
++---------------------+
+| SecondHighestSalary |
++---------------------+
+| 200                 |
++---------------------+
+First one create temporary table, second one using IFNULL
+*/
+# Write your MySQL query statement below
+SELECT
+    (SELECT DISTINCT Salary FROM Employee
+        ORDER BY Salary DESC LIMIT 1 OFFSET 1)
+AS SecondHighestSalary
+
+
+# Write your MySQL query statement below
+SELECT
+IFNULL( (SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1 OFFSET 1), NULL) AS SecondHighestSalary

181. Employees Earning More Than Their Managers.sql

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+/*
+The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+
++----+-------+--------+-----------+
+| Id | Name  | Salary | ManagerId |
++----+-------+--------+-----------+
+| 1  | Joe   | 70000  | 3         |
+| 2  | Henry | 80000  | 4         |
+| 3  | Sam   | 60000  | NULL      |
+| 4  | Max   | 90000  | NULL      |
++----+-------+--------+-----------+
+Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+
++----------+
+| Employee |
++----------+
+| Joe      |
++----------+
+
+Solution one is not using join, two is using join
+*/# Write your MySQL query statement below
+SELECT
+    a.Name As Employee
+FROM
+    Employee AS a,
+    Employee AS b
+WHERE
+    a.ManagerId = b.ID
+AND
+    a.Salary > b.Salary
+;
+
+
+
+SELECT
+    a.Name As Employee
+FROM
+    Employee AS a
+JOIN
+    Employee AS b
+ON
+    a.ManagerId = b.ID
+AND
+    a.Salary > b.Salary

182. Duplicate Emails.sql

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+/*
+Write a SQL query to find all duplicate emails in a table named Person.
+
++----+---------+
+| Id | Email   |
++----+---------+
+| 1  | [email protected] |
+| 2  | [email protected] |
+| 3  | [email protected] |
++----+---------+
+For example, your query should return the following for the above table:
+
++---------+
+| Email   |
++---------+
+| [email protected] |
++---------+
+Note: All emails are in lowercase.
+
+First Solution using having, second one use WHERE
+*/
+
+# Write your MySQL query statement below
+    SELECT Email FROM Person GROUP BY Email
+    HAVING COUNT( Email ) > 1;
+
+# Write your MySQL query statement below
+    SELECT Email FROM(
+    SELECT Email, COUNT(Email) AS num FROM Person GROUP BY Email
+        ) AS temp
+    WHERE temp.num > 1;

183. Customers Who Never Order.sql

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+/*
+SQL Schema
+Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
+
+Table: Customers.
+
++----+-------+
+| Id | Name  |
++----+-------+
+| 1  | Joe   |
+| 2  | Henry |
+| 3  | Sam   |
+| 4  | Max   |
++----+-------+
+Table: Orders.
+
++----+------------+
+| Id | CustomerId |
++----+------------+
+| 1  | 3          |
+| 2  | 1          |
++----+------------+
+Using the above tables as example, return the following:
+
++-----------+
+| Customers |
++-----------+
+| Henry     |
+| Max       |
++-----------+
+
+Solution one us NOT IN, two use LEFT JOIN
+*/
+# Write your MySQL query statement below
+SELECT Name AS Customers FROM Customers WHERE Id
+NOT IN ( SELECT CustomerID FROM Orders);
+
+SELECT Name AS Customers 
+FROM Customers
+LEFT JOIN Orders
+ON Customers.Id = Orders.CustomerId
+WHERE Orders.CustomerId IS NULL;

189. Rotate Array.cpp

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+/*
+Given an array, rotate the array to the right by k steps, where k is non-negative.
+
+Follow up:
+
+Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+Could you do it in-place with O(1) extra space?
+
+
+Example 1:
+
+Input: nums = [1,2,3,4,5,6,7], k = 3
+Output: [5,6,7,1,2,3,4]
+Explanation:
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+Example 2:
+
+Input: nums = [-1,-100,3,99], k = 2
+Output: [3,99,-1,-100]
+Explanation:
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+
+
+Constraints:
+
+1 <= nums.length <= 2 * 10^4
+It's guaranteed that nums[i] fits in a 32 bit-signed integer.
+k >= 0
+
+
+
+Solution one using brute force, it does not work in C++
+Two using extra space
+Three using rotation.
+*/
+class Solution {
+public:
+    void rotate(vector<int>& nums, int k) {
+        int size = nums.size();
+        k = k % size;
+        if( k == 0 ) return;
+        int temp, previous;
+        for ( int i = 0; i < k; i++ )
+        {
+            previous = nums[size - 1];
+            for ( int j = 0; j < size; j++ )
+            {
+                temp = nums[j];
+                nums[j] = previous;
+                previous = temp;
+            }
+        }
+    }
+};
+
+class Solution {
+public:
+    void rotate(vector<int>& nums, int k) {
+        int size = nums.size();
+        k = k % size;
+        if ( k == 0 ) return;
+        vector<int> temp( size );
+        for ( int i = 0; i < size; i++ )
+        {
+            temp[ ( i + k ) % size ] = nums[i];
+        }
+        nums = temp;
+    }
+};
+
+class Solution {
+public:
+    void rotate(vector<int>& nums, int k) {
+        int size = nums.size();
+        k = k % size;
+        if( k == 0 ) return;
+        int count = 0;
+        //We need this loop as when ( curr + k ) % size = 0, the array may not be fully rotated.
+        for ( int i = 0; count < size; i++ )
+        {
+            int curr = i;
+            int prev = nums[i];
+            do
+            {
+                int next = ( curr + k ) % size;
+                int temp = nums[next];
+                nums[next] = prev;
+                curr = next;
+                prev = temp;
+                count++;
+            } while( i != curr );
+        }
+    }
+};