Santiago Valdarrama’s Post

Day 444 of Daily LeetCode | 2485. Find the Pivot Integer (Easy) https://lnkd.in/e6P9YNq5 The code defines a function pivotInteger that takes an integer n and calculates the sum of the first n natural numbers. It then checks if the sum is a perfect square by comparing the square of the floor value of its square root to the sum itself. If it is a perfect square, the function returns the square root; otherwise, it returns -1.

Day 578 of Daily LeetCode | 912. Sort an Array (Medium) https://lnkd.in/gaudCM9X The code defines a sortArray method that sorts a list of integers using the merge sort algorithm. It contains two internal sorting functions: heapSort, which sorts the list using a heap, and mergeSort, which uses the merge sort algorithm. The mergeSort function is used to return the sorted array.

#javascripty State pattern The state pattern allows an object to alter its behavior when its internal state changes. In this example, we create a simple state pattern with an Order class that will update the status with the next() method. const ORDER_STATUS = { waitingForPayment: 'Waiting for payment', shipping: 'Shipping', delivered: 'Delivered', }; class OrderStatus { constructor(name, nextStatus) { this.name = name; this.nextStatus = nextStatus; } next() { return new this.nextStatus(); } } class WaitingForPayment extends OrderStatus { constructor() { super(ORDER_STATUS.waitingForPayment, Shipping); } } class Shipping extends OrderStatus { constructor() { super(ORDER_STATUS.shipping, Delivered); } } class Delivered extends OrderStatus { constructor() { super(ORDER_STATUS.delivered, Delivered); } } class Order { constructor() { this.state = new WaitingForPayment(); } next() { this.state = this.state.next(); } } export { Order }; A complete example is here 👉 https://lnkd.in/gXFneU5k Conclusion Use this pattern when the object’s behavior depends on its state, and its behavior changes in runtime depending on that state.

🚀 Day 20 of Daily LeetCode: Problem Type: Array,BitManipulation Level: Easy Today,I solved the "SINGLE NUMBER" problem .In this I was given a non-empty array of integers nums, every element appears twice except for one and i need to find that single one. I was supposed to implement a solution with a linear runtime complexity and use only constant extra space. BREAKDOWN OF CODE: 1->Firstly, I initialized 'result' to 0 2->Now I XOR each number with result. Since XORing a number with itself results in 0, duplicate numbers will cancel each other out. 3->Then at the end of the loop, result will be the single number that appears only once in 'nums' 4->In the end 'Result' is returned. #dailyleetcode #learninginpublic

🎯 Leetcode Daily My Initial Approach 1. Using Complements to Reach :  Initial approach was a 2sum like approach. For each element x, you compute the complement k-x and aim to pair it with another number from the array.   2. Building a Sums Array: You proposed creating a sums list to store adjusted values of array elements based on k. Then, you would iterate through the list to check if all elements became zero or divisible by k. Final approach(Correct) - I referred hints for this 1. Using Frequency Array for Remainders: 2. Validating Pairing Conditions: The algorithm checks two conditions--   ->Elements with remainder `0` must appear in pairs, i.e., `freq[0]` must be even.     ->For every remainder ( i) , its complement ( k - i ) must have the same frequency (`freq[i] == freq[k - i]`), ensuring valid pairings. 3. Handling Edge Cases: Special handling is required for remainders where ( i == k - i ) (such as ( k = 6 ), remainder 3 pairing with itself) and for remainders of 0.   4. Complexity: The solution runs in (O(n) ) time, making it highly efficient, especially for large arrays, with space complexity (O(k) ) due to the freq array.

useActionState The useActionState() Hook accepts three parameters: An "action" function, which is executed when the form action is triggered. An initial state object, that sets the starting state of the form before any user interaction. [Optional] A permalink that refers to the unique page URL that this form modifies. And returns three values in a tuple: The current state of the form. A function to trigger the form action. A boolean indicating whether the action is pending.

I just solved a code problems: Its says; write a function findNeedle() that takes an array full of junk but containing one "needle" After your function finds the needle it should return a message (as a string) that says: "found the needle at position " plus the index it found the needle, so: This was how I handle it: 1: I looped over the array using for...let looping method. 2: I then gave an if condition to the loop to return the array that has only the word "needle" and its position. Then result was return correctly: Codewars

Day 440 of Daily LeetCode | 2540. Minimum Common Value (Easy) https://lnkd.in/ehNSVPq6 The code defines a function getCommon that takes two sorted arrays nums1 and nums2 as input. It iterates through nums1 and uses two pointers, i and j, to compare elements in both arrays. If a common element is found, it is returned; otherwise, the function returns -1.

🏆 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝐖𝐞𝐞𝐤𝐥𝐲 𝐂𝐨𝐧𝐭𝐞𝐬𝐭 𝟒𝟎𝟕 💡 Here's a quick dive into the approach I used: 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝗔 - 𝐀𝐜𝐜𝐞𝐩𝐭𝐞𝐝   By comparing each corresponding bit of the integers n and k, I checked if a change was necessary. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐁- 𝐀𝐜𝐜𝐞𝐩𝐭𝐞𝐝  Alice will always win if given string has at least 1 vowel. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐂- 𝐀𝐜𝐜𝐞𝐩𝐭𝐞𝐝  • 𝑃𝑟𝑒𝑓𝑖𝑥 𝐶𝑜𝑢𝑛𝑡 𝑜𝑓 '1'𝑠: Compute the number of '1's up to each position in the string using a prefix count array. • 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑠: Iterate through the string, counting the number of '1's and handling transitions between '0's and '1's to calculate the maximum operations. • 𝐻𝑎𝑛𝑑𝑙𝑒 𝑇𝑟𝑎𝑖𝑙𝑖𝑛𝑔 𝑍𝑒𝑟𝑜𝑠: Adjust the result for any trailing '0's to ensure all possible operations are counted. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐃- 𝐑𝐞𝐣𝐞𝐜𝐭𝐞𝐝 (470/550 test passed) • 𝐶𝑜𝑚𝑝𝑢𝑡𝑒 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠: Calculate the differences between corresponding elements in two arrays. • 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 𝑆𝑒𝑔𝑚𝑒𝑛𝑡𝑠: Handle positive and negative segments separately to determine the maximum required operations for each. • 𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒 𝑅𝑒𝑠𝑢𝑙𝑡𝑠: Sum up the required operations based on the maximum values in positive and negative segments. #leetcode #contest

Moore's Algorithm for Majority Element 📌 I have solved the Leetcode question 169 for the Majority element by the Moore's algorithm which has the O(n) time complexity. Here is an explanation for you 👇 This algorithm is known as the Boyer-Moore Voting Algorithm. It works by maintaining a candidate for the majority element (ans) and a count (freq). When freq reaches zero, it picks a new candidate. If the current number is the candidate, it increases the count; otherwise, it decreases it. By the end of the loop, the candidate that remains is the majority element, provided that such an element exists. Dry run for this code 📃 👇 Initialization: n = nums.size() = 3 freq = 0 ans = 0 Iteration through the array: Iteration 1 (i = 0): nums[i] = 3 Since freq == 0, set ans = nums[i] which means ans = 3. Now check if ans == nums[i] (i.e., 3 == 3), which is true. So, increment freq: freq = 1. Iteration 2 (i = 1): nums[i] = 2 ans is currently 3. Check if ans == nums[i] (i.e., 3 == 2), which is false. So, decrement freq: freq = 0. Iteration 3 (i = 2): nums[i] = 3 Since freq == 0, set ans = nums[i] which means ans = 3. Check if ans == nums[i] (i.e., 3 == 3), which is true. So, increment freq: freq = 1. End of Loop: The loop has finished, and the final value of ans is 3. Return Value: The function returns ans, which is 3. I hope this may help you 😊