mding5692

Easy parallel python with concurrent.futures

As of version 3.3, python includes the very promising concurrent.futures module, with elegant context managers for running tasks concurrently. Thanks to the simple and consistent interface you can use both threads and processes with minimal effort.

For most CPU bound tasks - anything that is heavy number crunching - you want your program to use all the CPUs in your PC. The simplest way to get a CPU bound task to run in parallel is to use the ProcessPoolExecutor, which will create enough sub-processes to keep all your CPUs busy.

We use the context manager thusly:

with concurrent.futures.ProcessPoolExecutor() as executor:

mding5692

Studying for a Tech Interview Sucks, so Here's a Cheat Sheet to Help

This list is meant to be a both a quick guide and reference for further research into these topics. It's basically a summary of that comp sci course you never took or forgot about, so there's no way it can cover everything in depth. It also will be available as a gist on Github for everyone to edit and add to.

Data Structure Basics

###Array ####Definition:

• Stores data elements based on an sequential, most commonly 0 based, index.
• Based on tuples from set theory.

mding5692

Node Reverse(Node head) {
   if (head == null || head.next == null) {
       return head;
   }
    Node curr = null;
    Node nextNode = null;
    while (true) {
       nextNode = head.next;
       head.next = curr;
       curr = head;

mding5692

void Inorder(Node root) {
    if (root == null) { return;}
    Inorder(root.left);
    System.out.print(root.data + " ");
    Inorder(root.right);
}

mding5692

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {

mding5692

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        
        if (s.length() == 0 ) {
            return 0;
        }
        
        if (s.length() == 1) {
            return 1;
        }

mding5692

    public ListNode reverseBetween(ListNode head, int m, int n) {
    
    if(m==n) {
        return head;
    }
    
    ListNode curr = head;
    ListNode tempM = null; 
    ListNode tempAfterN = null; 
    ListNode tempBeforeM = null; 

mding5692

public class Solution {
    public int sumNumbers(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int sum = 0;
        return dfs(root,sum);
    }
    
    public int dfs(TreeNode root, int sum) {

mding5692

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {

mding5692

Node newNode = new Node(yourString);
if (table[hashLocation] == null ) {
  table[hashLocation] = newNode;
} else {
  // iterate through nodes
  currNode = table[hashLocation];
  while (currNode.next != null ) {
    currNode = currNode.next;
  }
  currNode.next = newNode;