**Environment: Mysql**

**Key technique: N/A**

Write a SQL query to find all duplicate emails in a table named `Person`

.

`+----+---------+`

| Id | Email |

+----+---------+

| 1 | a@b.com |

| 2 | c@d.com |

| 3 | a@b.com |

+----+---------+

For example, your query should return the following for the above table:

`+---------+`

| Email |

+---------+

| a@b.com |

+---------+

**Note**: All emails are in lowercase.

**Solution:**

`# Write your MySQL query statement below`

select Email

from Person

group by Email

having count(Email) > 1;

**Submissions:**

**Environment: Python 3.8**

**Key technique: &, set , list**

Given two integer arrays `nums1`

and `nums2`

, return *an array of their intersection*. Each element in the result must be **unique** and you may return the result in **any order**.

**Example 1:**

**Input:** nums1 = [1,2,2,1], nums2 = [2,2]

**Output:** [2]

**Analysis:**

- Use & and set to get Intersection.
- Convert it to a list.

**Solution:**

`class Solution:`

def intersection(self, nums1, nums2):

return list(set(nums1) & set(nums2))

**Submissions:**

**Environment: Python 3.8**

**Key technique: binarytree, Node, collections, deque**

Given the `root`

of a binary tree, return *the average value of the nodes on each level in the form of an array*. Answers within `10-5`

of the actual answer will be accepted.

**Example 1:**

**Input:** root = [3,9,20,null,15,7]

**Output:** [3.00000,14.50000,11.00000]

Explanation: The…

**Environment: Python 3.8**

**Key technique: set, Counters**

Given a string `s`

, return `true`

* if *`s`

* is a **good** string, or *`false`

* otherwise*.

A string `s`

is **good** if **all** the characters that appear in `s`

have the **same** number of occurrences (i.e., the same frequency).

**Example 1:**

**Input:** s = "abacbc"

**Output:** true

**Explanation:** The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s.

**Analysis:**

- Counter it as {‘a’:2, ‘b’:2, ‘c’:2}
- Check value is equal or not.

**Solution:**

`from collections import Counter`

class Solution:

def areOccurrencesEqual(self, s: str) -> bool:

ans=set(Counter(s).values())

return len(ans)==1

**Submissions:**

**Environment: Python 3.8**

**Key technique: List node**

Given the `head`

of a singly linked list, reverse the list, and return *the reversed list*.

**Example 1:**

**Input:** head = [1,2,3,4,5]

**Output:** [5,4,3,2,1]

**Analysis:**

- Use below four steps to revise link.
- temp=head
- head=head.next
- temp.next=ans
- ans=temp

**Environment: Python 3.8**

**Key technique: binarytree, Node**

Given the `root`

of a binary tree, return *the inorder traversal of its nodes' values*.

**Example 1:**

**Input:** root = [1,null,2,3]

**Output:** [1,3,2]

**Analysis:**

- Please see leetcode solution, it is very clear.

**Solution:**

from binarytree import Node

root = Node(1)

root.right = Node(2)

root.right.left=Node(3)

# Getting binary tree

print('Binary tree :', root)#root = [1,null,2,3]

class Solution:

def inorderTraversal(self, root):

if root:

return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)

else:

return []

**Submissions:**

**Environment: Python 3.8**

**Key technique: for, if, !=**

A string is **good** if there are no repeated characters.

Given a string `s`

, return *the number of **good substrings** of length **three **in *`s`

.

Note that if there are multiple occurrences of the same substring, every occurrence should be counted.

A **substring** is a contiguous sequence of characters in a string.

**Example 1:**

**Input:** s = "xyzzaz"

**Output:** 1

**Explanation:** There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz".

The only good substring of length 3 is "xyz".

**Analysis:**

- if s[i] != s[i+1] and s[i] != s[i+2] and s[i+1] !=s[i+2], ans+=1.
- Return ans

**Solution:**

`class Solution:`

def countGoodSubstrings(self, s):

ans=0

for i in range(len(s)-2):

if (s[i] != s[i+1] and s[i] != s[i+2] and s[i+1] !=s[i+2]):

ans+=1

return ans

**Submissions:**

**Environment: Python 3.8**

**Key technique: bin, append, replace**

Given an integer `n`

, return *an array *`ans`

* of length *`n + 1`

* such that for each *`i`

* *(`0 <= i <= n`

)*, *`ans[i]`

* is the **number of *`1`

*'s** in the binary representation of *`i`

.

**Example 1:**

**Input:** n = 2

**Output:** [0,1,1]

**Explanation:**

0 --> 0

1 --> 1

2 --> 10

**Analysis:**

- Use loop and n is loop end. Convert input number to binary such as 1 to ‘0b1’.
- Replace ‘ob’ with ‘’.
- Count all ‘1’ numbers.
- Add them to the list.

**Solution:**

`class Solution:`

def countBits(self, n):

ans=[]

for i in range(n+1):

temp=bin(i).replace('ob', '')

ans.append(temp.count('1'))

return ans

**Submissions:**

**Reference:**

https://leetcode.com/problems/counting-bits/discuss/1348058/python

**Environment: Python 3.8**

**Key technique: ASCII**

The **letter value** of a letter is its position in the alphabet **starting from 0** (i.e. `'a' -> 0`

, `'b' -> 1`

, `'c' -> 2`

, etc.).

The **numerical value** of some string of lowercase English letters `s`

is the **concatenation** of the **letter values**…

**Environment: Python 3.8**

**Key technique: //, %**

Given an integer `n`

(in base `10`

) and a base `k`

, return *the **sum** of the digits of *`n`

* **after** converting *`n`

* from base *`10`

* to base *`k`

.

After converting, each digit should be interpreted as a base `10`

number, and the sum should be returned in base `10`

.

**Example 1:**

**Input:** n = 34, k = 6

**Output:** 9

**Explanation: **34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.

**Analysis:**

- Calculate quotient and remainder between n and k.
- If update n > k, continue step1.
- If update n≤k, stop calculate.
- Summarize the current quotient and remainder.

**Solution:**

`class Solution:`

def sumBase(self, n: int, k: int) -> int:

ans=0

while (n>=k):

ans+=n%k

n//=k

return ans +n

**Submissions:**