Leetcode with dani

3. Fruit Into Baskets (Easy) Description: You are visiting a farm and have a basket that can hold at most two types of fruits. You want to maximize the number of fruits you collect. Given an array of integers representing the types of fruits in a row, find the maximum number of fruits you can collect in one basket. - Example: For input [1,2,1], the maximum number of fruits collected is 3.

2. Minimum Window Substring (Hard) Description: Given two strings s and t, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return an empty string. - Example: For s = "ADOBECODEBANC" and t = "ABC", the minimum window is "BANC".

1. 3Sum (Hard) Description: Given an integer array nums, return all unique triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. - Example: For input nums = [-1,0,1,2,-1,-4], the output should be [[-1,-1,2],[-1,0,1]].

Python Tips: Two Confusing methods. strip() and split() Methods strip(): This method removes any leading or trailing spaces or specific characters from a string.

  text = "  Hello, World!  "
  print(text.strip())  # Output: "Hello, World!"
  

split(): This method splits a string into a list of substrings based on a specified separator.

  data = "apple,banana,cherry"
  print(data.split(','))  # Output: ['apple', 'banana', 'cherry']
  

To see split() in action, check out the [Compare Version Numbers](https://leetcode.com/problems/compare-version-numbers/) problem on LeetCode, where you'll split version strings using a dot (.) as the separator.

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🌟 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.

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

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).

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

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! 😊

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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

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Maximise the number of toys that can be purchased with amount K Last Updated : 27 Mar, 2024 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.