Leetcode with dani

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Repost from A2SV | Africa to Silicon Valley

In-person Conversion Registration is Open! 🚀 We’re excited to welcome students for the A2SV in-person conversion! This is your chance to take your learning to the next level and be part of a vibrant tech community. Open to current students from Addis Ababa University (AAU), Addis Ababa Science and Technology University (AASTU), and Adama Science and Technology University (ASTU). 📅 Registration Dates: March 24 - March 28 ✅ Requirements: - At least 150 solved problems - At least 30 active days 📚 Topics to Cover: Two Pointers, Sorting, Sliding Window, Stack, Queue, Monotonicity, Linked List, Recursion. Don’t miss this opportunity—apply now! 🔗 Apply here: link #A2SV #TechEducation #CodingJourney #LevelUp

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Repost from Codeforces Official

Codeforces Round 1013 (Div. 3) will take place on the 25th of March at 14:35 UTC. Please, join by the link https://codeforces.com/contests/2091?locale=en

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check this it is the same as leetcode

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https://leetcode.com/problems/find-the-winner-of-the-circular-game/description/

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🔵 1823. Find the Winner of the Circular Game 📌 Problem: There are n friends sitting in a circle, numbered 1 to n. Starting from friend 1, count k friends clockwise (including the current one). The k-th friend leaves the circle. Repeat the process, starting from the next friend. The last remaining friend is the winner. 📌 Input: n → Number of friends. k → Step count for elimination. 📌 Output: The winner's number. 📌 Example 1: 🔹 Input: n = 5, k = 2 🔹 Output: 3 🔹 Explanation: Friends leave in this order: 2 → 4 → 1 → 5 → (Winner: 3) 📌 Example 2: 🔹 Input: n = 6, k = 5 🔹 Output: 1 🔹 Explanation: Friends leave in this order: 5 → 4 → 6 → 2 → 3 → (Winner: 1) 🔎 Can you find the last friend standing? 😃

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can solve this question using recursion ?

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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

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▎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

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A2SV__Trees I_2025(Without Code).pdf1.17 MB

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Repost 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

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Repost from Codeforces Official

Codeforces Round 1011 (Div. 2) will take place on the 22nd of March at 14:35 UTC. Please, join by the link https://codeforces.com/contests/2085?locale=en

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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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#Tree second question Answer will be posted at 4 PM

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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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When u create a CV take this as model

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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

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Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM

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Here is our very first leetcode Tree problem (BFS). Answer will be posted at 6:00 PM

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https://10academy.org/kifiya/learn-more