🌟 Daily Algorithm Challenge! 🌟

Starting today, I’ll be posting a new question every day to help us prepare for A2SV in November! 💻✨

🗓 Challenge yourself and share your solutions!

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the answer is 042

Topic of This Week Linked List

we will work on this algorithm for the coming 7 days.

What is linked list https://youtu.be/XGSoQ0J5An4?si=-FmTaU2tI0LNP2tg

it is really a good podcast with a graduate from AAU in Information system degree

21. Merge Two Sorted Lists (Easy)

Problem:

You are given the heads of two sorted linked lists, list1 and list2.

Your task is to merge these two lists into one sorted linked list. The merged list should be formed by splicing together the nodes of list1 and list2.

Return: The head of the newly merged sorted linked list.

Examples:

Example 1:

Input: list1 = [1, 2, 4], list2 = [1, 3, 4]
Output: [1, 1, 2, 3, 4, 4]
Example 2:

Input: list1 = [], list2 = []
Output: []
Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

try this question

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Find Middle of the Linked List
#Q19
Given a Singly Linked List, the task is to find the middle of the linked list. If the number of nodes are even, then there would be two middle nodes, so return the second middle node.

Example:

Input: linked list = 1 -> 2 -> 3 -> 4 -> 5
Output: 3
Explanation: There are 5 nodes in the linked list and there is one middle node whose value is 3.

Maximise the number of toys that can be purchased with amount K

Given an array consisting of the cost of toys. Given an integer K depicting the amount of money available to purchase toys. Write a program to find the maximum number of toys one can buy with the amount K.

Note: One can buy only 1 quantity of a particular toy.

Examples:

Input: N = 10, K = 50, cost = { 1, 12, 5, 111, 200, 1000, 10, 9, 12, 15 }
Output: 6
Explanation: Toys with amount 1, 5, 9, 10, 12, and 12 can be purchased resulting in a total amount of 49. Hence, maximum number of toys is 6.

how many of u can solve this problem with in 45min. share ur answer

answer:

def max_toys(cost, K):
    cost.sort()  
    count = 0
    total_cost = 0
    
    for price in cost:
        if total_cost + price <= K:
            total_cost += price
            count += 1
        else:
            break
            
    return count

N = 10
K = 50
cost = [1, 12, 5, 111, 200, 1000, 10, 9, 12, 15]
print(max_toys(cost, K))  # Output: 6

i could not post in the previous days for some reason, and I apologize for that. I will start posting from now on