Top 10 Algorithms for Coding Interview
Lastest Upate: 3/20/2015 (PDF Version)
The following are the common topics for coding interviews. As understanding those concepts requires much more effort, this tutorial only serves as an introduction. The topics that are covered include: 1) String/Array/Matrix, 2) Linked List, 3) Tree, 4) Heap, 5) Graph, 6) Sorting, 7) Recursion vs. Iteration, 8) Dynamic Programming, 9) Bit Manipulation, 10) Probability, 11) Combinations and Permutations, and other interesting problems. I highly recommend you to read "Simple Java" first, if you need a brief review of Java basics. If you want to see examples/projects using a library, you can use JavaSED.com.
Common Approaches to solve problems: sorting, binary search, hash table, heap, tree, depthfirst search, dynamic programming.
1. String/Array/Matrix
String in Java is a class that contains a char array and other fields and methods. Without autocompletion from any IDE, the following methods should be remembered.
toCharArray() //get char array of a String charAt(int x) //get a char at the specific index length() //string length length //array size substring(int beginIndex) substring(int beginIndex, int endIndex) Integer.valueOf()//string to integer String.valueOf()/integer to string Arrays.sort() //sort an array Arrays.toString(char[] a) //convert to string Arrays.copyOf(T[] original, int newLength) System.arraycopy(Object src, int srcPos, Object dest, int destPos, int length) 
Strings/arrays are easy to understand, but the interview problems often require advanced algorithm to solve, such as dynamic programming, recursion, etc.
Classic problems:
0) Rotate Array
1) Evaluate Reverse Polish Notation (Stack)
2) Longest Palindromic Substring (DP)
3) Word Break (DP)
3) Word Break II (DP, DFS)
4) Word Ladder (Queue, BFS)
5) Median of Two Sorted Arrays
6) Regular Expression Matching
7) Merge Intervals
8) Insert Interval
9) Two Sum
9) Two Sum II – Input array is sorted
9) Two Sum III  Data structure design
9) 3Sum
9) 4Sum
10) 3Sum Closest
11) String to Integer
12) Merge Sorted Array
13) Valid Parentheses
14) Implement strStr()
15) Set Matrix Zeroes
16) Search Insert Position
17) Longest Consecutive Sequence
18) Valid Palindrome
19) Spiral Matrix
20) Search a 2D Matrix
21) Rotate Image [Palantir]
22) Triangle
23) Distinct Subsequences Total
24) Maximum Subarray [Palantir, LinkedIn]
24) Maximum Product Subarray [LinkedIn]
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 [Google]
29) Palindrome Partitioning
29) Palindrome Partitioning II
30) Reverse Words in a String
31) Find Minimum in Rotated Sorted Array
31) Find Minimum in Rotated Sorted Array II
32) Find Peak Element
33) Min Stack
34) Majority Element
35) Combination Sum (DFS)
35) Combination Sum II (DFS)
36) Best Time to Buy and Sell Stock
36) Best Time to Buy and Sell Stock II
36) Best Time to Buy and Sell Stock III (DP)
36) Best Time to Buy and Sell Stock IV (DP)
37) Longest Common Prefix [Google]
38) Largest Number
39) Combinations (DFS)
40) Compare Version Numbers
41) Gas Station
42) Candy [Google]
43) Jump Game
44) Pascal's Triangle
44) Pascal’s Triangle II
45) Container With Most Water
46) Count and Say
47) Repeated DNA Sequences
48) House Robber
49) Dungeon Game (DP)
50) Number of Islands (DFS/BFS)
51) Surrounded Regions (BFS)
52) Max Points on a Line
53) Letter Combinations of a Phone Number (DFS)
54) Remove Element
55) Anagrams
56) Search for a Range
57) Simplify Path
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 LinkedList can be used as a Queue (add() and remove()). (LinkedList implements the Queue interface.) If you need a stack or queue to solve problems during your interview, you can directly use them.
Classic Problems:
1) Add Two Numbers
2) Reorder List
3) Linked List Cycle
4) Copy List with Random Pointer
5) Merge Two Sorted Lists
6) Merge k Sorted Lists *
7) Remove Duplicates from Sorted List
8) Partition List
9) LRU Cache
10) Intersection of Two Linked Lists
11) Remove Linked List Elements
12) Swap Nodes in Pairs
3. Tree & Heap
A tree normally refers to a 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
Heap is a specialized treebased data structure that satisfies the heap property. The time complexity of its operations are important (e.g., findmin, deletemin, insert, etc). In Java, PriorityQueue is important to know.
Classic problems:
1) Binary Tree Preorder Traversal
2) Binary Tree Inorder Traversal [Palantir]
3) Binary Tree Postorder Traversal
4) Binary Tree Level Order Traversal
4) Binary Tree Level Order Traversal II
5) Validate Binary Search Tree
6) Flatten Binary Tree to Linked List
7) Path Sum
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 *
13) Balanced Binary Tree
14) Symmetric Tree
15) Binary Search Tree Iterator
16) Binary Tree Right Side View
4. Graph
Graph related questions mainly focus on depth first search and breath first search. Depth first search is straightforward, you can just loop through neighbors starting from the root node.
Below is a simple implementation of a graph and breath first search. The key is using a queue to store nodes.
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: 4
Classic Problems:
1) Clone Graph
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 
Insertion sort  n^2  n^2  
Quick sort  n log(n)  n^2  
Merge sort  n log(n)  n log(n)  depends 
* 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)
Here are some implementations/demos, and in addition, you may want to check out how Java developers sort in practice.
1) Mergesort
2) Quicksort
3) InsertionSort.
4) Maximum Gap (Bucket Sort)
6. Recursion vs. Iteration
Recursion should be a builtin 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 n1.
To get n, there are only two ways, one 1stair from n1 or 2stairs from n2. If f(n) is the number of ways to climb to n, then f(n) = f(n1) + f(n2)
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(n1) + f(n2); return x; } 
The time complexity of the recursive method is exponential to n. There are a lot of redundant computations.
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(n1) + f3(n2);//store results so only calculate once! return A[n]; } 
Classic problems:
1) Edit Distance
2) Longest Palindromic Substring
3) Word Break
3) Word Break II
4) Maximum Subarray
4) Maximum Product Subarray
5) Palindrome Partitioning
5) Palindrome Partitioning II
6) Candy [Google]
7) Jump Game
8) Best Time to Buy and Sell Stock III (DP)
8) Best Time to Buy and Sell Stock IV (DP)
9) Dungeon Game (DP)
8. Bit Manipulation
Bit operators:
OR ()  AND (&)  XOR (^)  Left Shift (<<)  Right Shift (>>)  Not (~) 
10=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<<i); if(result == 0){ return false; }else{ return true; } } 
For example, get second bit of number 10.
1<<1= 10
1010&10=10
10 is not 0, so return true;
Classic Problems:
1) Single Number
1) Single Number II
2) Maximum Binary Gap
3) Number of 1 Bits
4) Reverse Bits
5) Repeated DNA Sequences
6) Bitwise AND of Numbers Range
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 * ... * 365n/365 * ... * 36549/365. 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 *= (365.0i)/365.0; } double pro = Math.round((1x) * 100); return pro/100; } 
calculateProbability(50) = 0.97
10. Combinations and Permutations
The difference between combination and permutation is whether order matters.
Example 1:
Given 5 numbers  1, 2, 3, 4 and 5, print out different sequence of the 5 numbers. 4 can not be the third one, 3 and 5 can not be adjacent. How many different combinations?
Example 2:
Given 5 banaba, 4 pear, and 3 apple, assuming one kind of fruit are the same, how many different combinations?
Class Problems:
1) Permutations
2) Permutations II
3) Permutation Sequence
4) Generate Parentheses
11. Math
Solving math problems usually require us to get some observations and form rules:
1) Reverse Integer
2) Palindrome Number
3) Pow(x,n)
4) Subsets
5) Subsets II
6) Fraction to Recurring Decimal [Google]
7) Excel Sheet Column Number
8) Excel Sheet Column Title
9) Factorial Trailing Zeroes
10) Happy Number
11) Count Primes
Additional Resources
1. Share your code to Github/BitBucket

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