Short Technical Notes

Given the  head  of a linked list, remove the  n th  node from the end of the list and return its head.   Example 1: Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5] Example 2: Input: head = [1], n = 1 Output: [] Example 3: Input: head = [1,2], n = 1 Output: [1]   Constraints: The number of nodes in the list is  sz . 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz   Follow up:  Could you do this in one pass? Solution: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeNthFromEnd ( ListNode head , int n ) { ListNode fast = head; ListNode slow = head; for ( int i = 0 ; i < n; i++){ fast = fast . next ; } i...

You are given an array  prices  where  prices[i]  is the price of a given stock on the  i th  day. You want to maximize your profit by choosing a  single day  to buy one stock and choosing a  different day in the future  to sell that stock. Return  the maximum profit you can achieve from this transaction . If you cannot achieve any profit, return  0 .   Example 1: Input: prices = [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell. Example 2: Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transactions are done and the max profit = 0.   Constraints: 1 <= prices.length <= 10 5 0 <= prices[i] <= 10 4 Solution:  class Solution { public int maxProfit ( int [] prices ) { if ( prices . length < 2){ ...

Given an integer array  nums , return  an array   answer   such that   answer[i]   is equal to the product of all the elements of   nums   except   nums[i] . The product of any prefix or suffix of  nums  is  guaranteed  to fit in a  32-bit  integer. You must write an algorithm that runs in  O(n)  time and without using the division operation.   Example 1: Input: nums = [1,2,3,4] Output: [24,12,8,6] Example 2: Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0]   Constraints: 2 <= nums.length <= 10 5 -30 <= nums[i] <= 30 The product of any prefix or suffix of  nums  is  guaranteed  to fit in a  32-bit  integer. Solution: class Solution { public int [] productExceptSelf ( int [] nums ) { int [] pr = new int [ nums . length ]; int [] sf = new int [ nums . length ]; int [] res = new int [ nums . length ]; int prc = 0 ;...

A virtual environment is the most used tool by the developers to isolate the dependencies for different projects. Suppose you have two projects say porj1 and proj2 . proj1 needs the Django dependency with version 3.2 but your proj2 needs the Django dependency with version 2.2. In this situation you need a virtual environment to keep the both version on your system separately.  How to create virtual environment in python:  Decide a directory where you want to create the virtual environment. You can use your project directory or any other directory as per your wish.  Run the below command. Here` awesome_proj_env ` is the folder where virtual environment will be created. if the folder does not exists then it will be created automatically. python3 -m venv awesome_proj_env    Activate the virtual environment: On Linux/Mac OSX: source awesome_proj_env/bin/activate  On Windows: awesome_proj_env \Scripts\activate.bat Deactivate the virtual environment in Pyth...

public class NumberPattern { public static void main(String a []) { for ( int i = 1; i <=4; i ++) { for ( int j =0; j <4; j ++) { int sum = i + j ; // if sum is greater than 4 then subtract 4 from the sum otherwise print the sum System. out .print( sum > 4 ? sum - 4+ " " : sum + " " ); } System. out .println(); } } }