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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

Hey everyone! 🌟 We're on the lookout for a admin to help out with our programming channel! If you know at least one programming language and love sharing knowledge, we’d love to have you on board. We’re looking for someone who can post regularly and connect with our awesome community. If you’re interested, drop us a message! 😊

Count distinct elements in every window of size k

Given an array of size N and an integer K, return the count of distinct numbers in all windows of size K.

Examples:

Input: arr[] = {1, 2, 1, 3, 4, 2, 3}, K = 4
Output: 3 4 4 3
Explanation:
First window is {1, 2, 1, 3}, count of distinct numbers is 3
Second window is {2, 1, 3, 4} count of distinct numbers is 4
Third window is {1, 3, 4, 2} count of distinct numbers is 4
Fourth window is {3, 4, 2, 3} count of distinct numbers is 3

Input: arr[] = {1, 2, 4, 4}, K = 2
Output: 2 2 1
Explanation:
First window is {1, 2}, count of distinct numbers is 2
First window is {2, 4}, count of distinct numbers is 2
First window is {4, 4}, count of distinct numbers is 1

Question: Find the majority element in an array, which is defined as the element that appears more than n/2 times. Use an efficient algorithm to solve this in linear time O(n) and constant space O(1).

Can you solve this? With in 20 min. Challenge ur self

🌟 Finding the Majority Element in an Array 🌟

Have you ever wondered how to identify the element that appears more than half the time in an array? 🤔 Let’s dive into a clever algorithm that does just that: The Boyer-Moore Voting Algorithm! 🗳

▎🔍 How It Works:

1. Initialization:
- Start with two variables:
- candidate = None
- count = 0

2. Candidate Selection:
- Loop through each element in the array:
- If count is 0, set the current element as the new candidate.
- If the current element matches the candidate, increase count.
- If it doesn’t match, decrease count.

3. Verification (Optional):
- After the first pass, you can run through the array again to confirm that your candidate is indeed the majority element!

▎📈 Why Use This Algorithm?
- Time Complexity: \\( O(n) \\) – Efficiently processes the array in linear time!
- Space Complexity: \\( O(1) \\) – Uses only a constant amount of extra space!

▎🚀 Example:
Consider the array: [2, 2, 1, 1, 1, 2, 2].
- The algorithm will help you find that 2 is the majority element! 🎉

This method is not only efficient but also elegant.