# Top 10 Algorithms for Coding Interview

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 efforts, this list only serves as an introduction. They are viewed from a Java perspective. The following concepts will be covered:
1. String
3. Tree
4. Graph
5. Sorting
6. Recursion vs. Iteration
7. Dynamic Programming
8. Bit Manipulation
9. Probability
10. Combinations and Permutations
1. String
Without code auto-completion of any IDE, the following methods should be remembered.
 `toCharyArray() //get char array of a StringArrays.sort() //sort an arrayArrays.toString(char[] a) //convert to stringcharAt(int x) //get a char at the specific indexlength() //string lengthlength //array size`
Also in Java a String is not a char array. A String contains a char array and other fields and methods.
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; } }}`
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:
1. Binary Search Tree: for all nodes, left children <= current node <= right children
2. Balanced vs. Unbalanced: In a balanced tree, the depth of the left and right subtrees of every node differ by 1 or less.
3. Full Binary Tree: every node other than the leaves has two children.
4. 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.
5. 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
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
5. Sorting
Time complexity of different sorting algorithms. You can go to wiki to see basic idea of them.
 Algorithm Average Time Worst Time Space Bubble sort n^2 n^2 1 Selection sort n^2 n^2 1 Counting Sort n+k n+k n+k Insertion sort n^2 n^2 Quick sort n log(n) n^2 Merge sort n log(n) n log(n) depends
In addition, here are some implementations/demos: Counting sort, 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:
1. An instance is solved using the solutions for smaller instances.
2. The solution for a smaller instance might be needed multiple times.
3. The solutions to smaller instances are stored in a table, so that each smaller instance is solved only once.
4. 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; 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];}`
8. Bit Manipulation
Bit operators:
 OR (|) AND (&) XOR (^) Left Shift (<<) Right Shift (>>) Not (~) 1|0=1 1&0=0 1^0=1 0010<<2=1000 1100>>2=0011 ~1=0
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<
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;
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
calculateProbability(50) = 0.97
10. Combinations and Permutations
The difference between combination and permutation is whether order matters.
Please leave your comment if you think any other problem should be here.
References/Recommmended Materials:
1. Binary tree
2. Introduction to Dynamic Programming
3. UTSA Dynamic Programming slides