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Leetcode with dani
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Forwarded from Codeforces Official
Unfortunately, due to technical reasons, the round is postponed by 8.5 hours (to the usual scheduled time). We apologize for any inconvenience caused.
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A big thank you to everyone at this channel and  Alpha ,for a great session! I really enjoyed it.
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For Ethiopians only
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Leetcode with dani
Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM
102. Binary Tree Level Order Traversal

Difficulty: Medium

Problem Statement

Given the root of a binary tree, return the level order traversal of its nodes' values (i.e., from left to right, level by level).

Examples

Example 1:

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

Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]

Output: [[1]]

Example 3:

Input: root = []

Output: []

Constraints

• The number of nodes in the tree is in the range [0, 2000].

-1000 <= Node.val <= 1000
jakes-resume.pdf
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When u create a CV take this as model
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Leetcode with dani
Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:


def helper(qu):
if not qu:
return

i = len(qu)
temp = []
nextsize = 0


while i > 0 :
poped = qu.popleft()
temp.append(poped.val)

if poped.left:
qu.append(poped.left)
nextsize += 1

if poped.right:
qu.append(poped.right)
nextsize += 1


i -= 1

ans.append(temp)
helper(qu)

if not root:
return []

ans =[]
q = deque()
q.append(root)
helper(q)

return ans
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class Solution:
def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:

def helper(qu):
if not qu:
return

i = len(qu)
leng = len(qu)
total = 0

while i > 0:
poped = qu.popleft()
total += poped.val

if poped.left:
qu.append(poped.left)
if poped.right:
qu.append(poped.right)

i -= 1
ans.append(total/leng)
helper(qu)
qu = deque()
qu.append(root)
ans = []

helper(qu)

return ans
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Forwarded from Codeforces Official
Codeforces Round #1012 (Div. 1, Div. 2) will take place on the 23rd of March at 05:35 UTC.
Please, join by the link https://codeforces.com/contests/2089,2090?locale=en
2169. Count Operations to Obtain Zero

Problem

You are given two non-negative integers, num1 and num2. In one operation, do the following:

• If num1 >= num2, subtract num2 from num1.

• If num1 < num2, subtract num1 from num2.

Repeat this until either num1 or num2 becomes zero. Return the total number of operations performed.

Examples

Example 1

Input:
num1 = 2, num2 = 3
Output:
3

Example 2

Input:
num1 = 10, num2 = 10
Output:
1

Constraints

• 0 ≤ num1, num2 ≤ 10⁵
Leetcode with dani
▎2169. Count Operations to Obtain Zero ▎Problem You are given two non-negative integers, num1 and num2. In one operation, do the following: • If num1 >= num2, subtract num2 from num1. • If num1 < num2, subtract num1 from num2. Repeat this until either…
class Solution:
def countOperations(self, num1: int, num2: int) -> int:
count = 0
while num1 and num2:
if num1>=num2:
count += num1//num2
num1 = num1%num2
else:
count += num2//num1
num2 = num2%num1
return count
can solve this question using recursion ?