Embedded Files
Preface:
Preface:
My currently plan is in the next 2 years join my dream company ---- Google,inc! This blog is used for practicing my coding skill to meet this goal !~ Good Luck !~
About me:
About me:
My name is Chuck Zhang, who is a software developer currently working at eBay.inc. I'm really interested in Algorithm and coding, welcome to discuss with me for any algorithm question.
Below is my LinkedIn:
www.linkedin.com/in/ucichuck
^_^ Dream come true--- Googler!!! ^_^
Top 10 Algorithms for Coding
Top 10 Algorithms for Coding
The following are top 10 algorithms related concepts in coding interview. I will try to illustrate those concepts though some simple examples. As understanding those concepts requires much more effort, this list only serves as an introduction. They are viewed from a Java perspective and the following concepts will be covered:
String
Linked List
Tree
Graph
Sorting
Recursion vs. Iteration
Dynamic Programming
Bit Manipulation
Probability
Combinations and Permutations
I will keep updating this list to add more classic problems and problems from Leetcode.
1. String
Without code auto-completion of any IDE, the following methods should be remembered.
toCharArray() //get char array of a StringArrays.sort() //sort an arrayArrays.toString(char[] a) //convert to string charAt(int x) //get a char at the specific index length() //string length length //array size
Also unlike in C++, a String is not a char array in Java. It is a class that contains a char array and other fields and methods.
Classic problems: Longest Palindromic Substring, Word Break.
Classic problems:
1) Evaluate Reverse Polish Notation
2) Longest Palindromic Substring
5) Median of Two Sorted Arrays
6) Regular Expression Matching
17) Longest Consecutive Sequence
23) Distinct Subsequences Total
25) Remove Duplicates from Sorted Array
26) Remove Duplicates from Sorted Array II
27) Longest Substring Without Repeating Characters
28) Longest Substring that contains 2 unique characters
2. Linked List
The implementation of a linked list is pretty simple in Java. Each node has a value and a link to next node.
class Node { int val; Node next; Node(int x) { val = x; next = null; }}
Two popular applications of linked list are stack and queue.
Stack
class Stack{ Node top; public Node peek(){ if(top != null){ return top; } return null; } public Node pop(){ if(top == null){ return null; }else{ Node temp = new Node(top.val); top = top.next; return temp; } } public void push(Node n){ if(n != null){ n.next = top; top = n; } }}
Queue
class Queue{ Node first, last; public void enqueue(Node n){ if(first == null){ first = n; last = first; }else{ last.next = n; last = n; } } public Node dequeue(){ if(first == null){ return null; }else{ Node temp = new Node(first.val); first = first.next; return temp; } }}
It is worth to mention that Java standard library already contains a class called “Stack“, and LinkedListcan be used as a Queue. (LinkedList implements the Queue interface) If you directly need a stack or queue to solve problems in your interview, you can directly use them.
Classic Problems: Add Two Numbers, Reorder List.
Classic Problems:
4) Copy List with Random Pointer
7) Remove Duplicates from Sorted List
3. Tree
Tree here is normally binary tree. Each node contains a left node and right node like the following:
class TreeNode{ int value; TreeNode left; TreeNode right;}
Here are some concepts related with trees:
Binary Search Tree: for all nodes, left children <= current node <= right children
Balanced vs. Unbalanced: In a balanced tree, the depth of the left and right subtrees of every node differ by 1 or less.
Full Binary Tree: every node other than the leaves has two children.
Perfect Binary Tree: a full binary tree in which all leaves are at the same depth or same level, and in which every parent has two children.
Complete Binary Tree: a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible
Classic problems: Binary Tree Preorder Traversal , Binary Tree Inorder Traversal, Binary Tree Postorder Traversal.
Classic problems:
1) Binary Tree Preorder Traversal
2) Binary Tree Inorder Traversal
3) Binary Tree Postorder Traversal
5) Validate Binary Search Tree
6) Flatten Binary Tree to Linked List
8) Construct Binary Tree from Inorder and Postorder Traversal
9) Convert Sorted Array to Binary Search Tree
10) Convert Sorted List to Binary Search Tree
11) Minimum Depth of Binary Tree
12) Binary Tree Maximum Path Sum *
4. Graph
Graph related questions mainly focus on depth first search and breath first search.
Below is a simple implementation of a graph and breath first search.
1) Define a GraphNode
class GraphNode{ int val; GraphNode next; GraphNode[] neighbors; boolean visited; GraphNode(int x) { val = x; } GraphNode(int x, GraphNode[] n){ val = x; neighbors = n; } public String toString(){ return "value: "+ this.val; }}
2) Define a Queue
class Queue{ GraphNode first, last; public void enqueue(GraphNode n){ if(first == null){ first = n; last = first; }else{ last.next = n; last = n; } } public GraphNode dequeue(){ if(first == null){ return null; }else{ GraphNode temp = new GraphNode(first.val, first.neighbors); first = first.next; return temp; } }}
3) Breath First Search uses a Queue
public class GraphTest { public static void main(String[] args) { GraphNode n1 = new GraphNode(1); GraphNode n2 = new GraphNode(2); GraphNode n3 = new GraphNode(3); GraphNode n4 = new GraphNode(4); GraphNode n5 = new GraphNode(5); n1.neighbors = new GraphNode[]{n2,n3,n5}; n2.neighbors = new GraphNode[]{n1,n4}; n3.neighbors = new GraphNode[]{n1,n4,n5}; n4.neighbors = new GraphNode[]{n2,n3,n5}; n5.neighbors = new GraphNode[]{n1,n3,n4}; breathFirstSearch(n1, 5); } public static void breathFirstSearch(GraphNode root, int x){ if(root.val == x) System.out.println("find in root"); Queue queue = new Queue(); root.visited = true; queue.enqueue(root); while(queue.first != null){ GraphNode c = (GraphNode) queue.dequeue(); for(GraphNode n: c.neighbors){ if(!n.visited){ System.out.print(n + " "); n.visited = true; if(n.val == x) System.out.println("Find "+n); queue.enqueue(n); } } } }}
Output:
value: 2 value: 3 value: 5 Find value: 5
value: 4
Classic Problems:
5. Sorting
Time complexity of different sorting algorithms. You can go to wiki to see basic idea of them.
* BinSort, Radix Sort and CountSort use different set of assumptions than the rest, and so they are not “general” sorting methods. (Thanks to Fidel for pointing this out)
In addition, here are some implementations/demos: Mergesort, Quicksort, InsertionSort.
6. Recursion vs. Iteration
Recursion should be a built-in thought for programmers. It can be demonstrated by a simple example.
Question: there are n stairs, each time one can climb 1 or 2. How many different ways to climb the stairs.
Step 1: Finding the relationship before n and n-1.
To get n, there are only two ways, one 1-stair from n-1 or 2-stairs from n-2. If f(n) is the number of ways to climb to n, then f(n) = f(n-1) + f(n-2)
Step 2: Make sure the start condition is correct.
f(0) = 0;
f(1) = 1;
public static int f(int n){ if(n <= 2) return n; int x = f(n-1) + f(n-2); return x;}
The time complexity of the recursive method is exponential to n. There are a lot of redundant computations.
f(5)
f(4) + f(3)
f(3) + f(2) + f(2) + f(1)
f(2) + f(1) + f(2) + f(2) + f(1)
It should be straightforward to convert the recursion to iteration.
public static int f(int n) { if (n <= 2){ return n; } int first = 1, second = 2; int third = 0; for (int i = 3; i <= n; i++) { third = first + second; first = second; second = third; } return third;}
For this example, iteration takes less time. You may also want to check out Recursion vs Iteration.
7. Dynamic Programming
Dynamic programming is a technique for solving problems with the following properties:
An instance is solved using the solutions for smaller instances.
The solution for a smaller instance might be needed multiple times.
The solutions to smaller instances are stored in a table, so that each smaller instance is solved only once.
Additional space is used to save time.
.
The problem of climbing steps perfectly fit those 4 properties. Therefore, it can be solve by using dynamic programming.
public static int[] A = new int[100]; public static int f3(int n) { if (n <= 2) A[n]= n; if(A[n] > 0) return A[n]; else A[n] = f3(n-1) + f3(n-2);//store results so only calculate once! return A[n];}
Get bit i for a give number n. (i count from 0 and starts from right)
public static boolean getBit(int num, int i){ int result = num & (1<<i); if(result == 0){ return false; }else{ return true; }}
For example, get second bit of number 10.
i=1, n=10
1<<1= 10
1010&10=10
10 is not 0, so return true;
Classic Problems:
9. Probability
Solving probability related questions normally requires formatting the problem well. Here is just a simple example of such kind of problems.
There are 50 people in a room, what’s the probability that two people have the same birthday? (Ignoring the fact of leap year, i.e., 365 day every year)
Very often calculating probability of something can be converted to calculate the opposite. In this example, we can calculate the probability that all people have unique birthdays. That is: 365/365 * 364/365 * 363/365 * … * 365-n/365 * … * 365-49/365. And the probability that at least two people have the same birthday would be 1 – this value.
public static double caculateProbability(int n){ double x = 1; for(int i=0; i<n; i++){ x *= (365.0-i)/365.0; } double pro = Math.round((1-x) * 100); return pro/100;}
calculateProbability(50) = 0.97
10. Combinations and Permutations
The difference between combination and permutation is whether order matters.
Class Problems:
Some other problems need us to use observations to form rules to solve them:
Please leave your comment if you think any other problem should be here.
Page updated
Google Sites
Report abuse