Set Request Headers in Swagger-UI

For the last 2 days, I was facing a issue with setting Global Request headers to Springfox’s Swagger-UI (version 2.8.0) for a SpringBoot Application.

The issue was more related to the new Swagger version 2.8.0 and does not any issues in prior versions. In the below code, I am only presenting the cause and the solution. Assuming the developers have prior knowledge of Swagger and its implementation. More implementation details can be found at https://springfox.github.io/springfox/docs/current/#quick-start-guides

@Bean
    public Docket apiSwaggerDocket() {
        return new Docket(DocumentationType.SWAGGER_2)
                .select()
                .apis(RequestHandlerSelectors.withClassAnnotation(Api.class))
                .paths(PathSelectors.any())
                .build()
                .pathMapping("/")
                .genericModelSubstitutes(ResponseEntity.class)
                .useDefaultResponseMessages(false)
                .forCodeGeneration(true)
                .securitySchemes(newArrayList(apiKey()))
                .apiInfo(apiInfo());

    private ApiKey apiKey() {
        return new ApiKey("access_token", "access_token", "header");
    }

Let’s only concentrate on the apiKey() method. The new ApiKey(…) constructor has different argument signatures for different versions of springfox’s swagger.

/**
* http://springfox.github.io/springfox/javadoc/current/springfox/documentation/service/ApiKey.html
* Signature of the ApiKey constructor
* return new ApiKey(name, keyName, passAs);
* 
* name - is the name of the key
* keyName - is the value of the key name
* passAs - you can pass as header or query parameter
*/
// For version 2.6.0
return new ApiKey("Authorization", "Bearer", "header");

Output at swagger-ui
name: Authorization
in: header
value: Bearer

// For version 2.7.0 - This version has reverse the constructor Argument signature
return new ApiKey("Authorization", "Bearer", "header");

Output at swagger-ui
name: Authorization
in: header
value: Bearer

// For version 2.8.0
return new ApiKey("Authorization", "Bearer", "header");

Output at swagger-ui
There is no header displayed in this version of the swagger.

For my current use case, I had to step down the swagger version to 2.7.0.

Alternatively, if one intends to use version 2.8.0, we can have globalOperationParameters to put in use, with all API’s requesting for header

@Bean
    public Docket apiSwaggerDocket() {
        return new Docket(DocumentationType.SWAGGER_2)
                .select()
                .apis(RequestHandlerSelectors.withClassAnnotation(Api.class))
                .paths(PathSelectors.any())
                .build()
                .pathMapping("/")
                .genericModelSubstitutes(ResponseEntity.class)
                .useDefaultResponseMessages(false)
                .forCodeGeneration(true)
                .apiInfo(apiInfo())
                .globalOperationParameters(
                        newArrayList(new ParameterBuilder()
                                .name("access_token")
                                .description("Access Token")
                                .modelRef(new ModelRef("string"))
                                .parameterType("header")
                                .required(true)
                                .build()));
    }

The only drawback using globalOperationParameters, the header is not sticky. Meaning, the same header value is required for each and every API in Swagger, which is not great.

Thanks for reading my post ūüôā

More Reading & Resources:

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Find the middle element of the linked list in single pass – O(n) complexity

A linked list contains Nodes that are linked to each other.

Create a Node – A node consists of data of that node and some information about the next linked node

function Node(data, nextNode){
this.data = data;
this.nextNode = nextNode;
}

Create all nodes

Note: First Node from right. This would be our last node in sequence. The reference to next node would be null since this is our last node.

var n5 = new Node("5", null); 
var n4 = new Node("4", n5);
var n3 = new Node("3", n4);
var n2 = new Node("2", n3);
var n1 = new Node("1", n2);

// So here is the representation of the linked list nodes, that we just created.
// n1 (1, n2) –> n2 (2, n3) –> n3 (3, n4) –> n4 (4, n5) –> n5 (5, null)

Here is how the linked list looks

linkedlist

The Naive Solution to this problem would be in 2 passes (2 for loops), which takes O(n^2) worst case complexity

1st pass is to loop thru to get the count of nodes in linked list and
2nd pass is to loop thru find the middle element by dividing the Count of Nodes by 2

So, for example if there are 5 Nodes.

1st pass is to count the number of nodes = 5
2nd pass is to loop thru to get the middle element by dividing 5/2 = 2.5 (so in the middle element to be 3)

There is a better and a faster way to get this in 1 Pass (1 for/while loop), with O(n) complexity

Assign 2 pointers while looping.
1st pointer moves 1 element at a time.
2nd pointer moves 2 elements at a time (twice as faster as the 1st pointer)

Some theory/story:

Assume that Person A and Person B would like to reach a destination X from same starting point.

Assume that it would actually take 10 mins to reach destination X.

Assume that Person B walks 2 times faster than Person A, then by the time Person B reaches destination X, Person A is exactly half way thru.

So, lets give a starting point to our node n1.

var slowNode = n1;
var fastNode = n1;

// Check if nextNode is not null and nextNode of the nextNode 
// (2 Nodes from the current Node) is also not null
while(fastNode.nextNode != null && fastNode.nextNode.nextNode != null) {
// slowNode should be moving to the next element of the linked list sequentially
slowNode = slowNode.nextNode;
// Now the fastNode should be twice as faster as the slowNode
fastNode = fastNode.nextNode.nextNode;
}

console.log(slowNode.data);

Relevant Code using Java

public class MiddleElementLinkedList {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		// Create Linked List Nodes
		
		LinkedListNode n5 = new LinkedListNode("5", null);
		LinkedListNode n4 = new LinkedListNode("4", n5);
		LinkedListNode n3 = new LinkedListNode("3", n4);
		LinkedListNode n2 = new LinkedListNode("2", n3);
		LinkedListNode n1 = new LinkedListNode("1", n2);
		
		// Set the starting point
		
		LinkedListNode slowNode = n1;
		LinkedListNode fastNode = n1;
		
		while(fastNode.getNextNode() != null && fastNode.getNextNode().getNextNode() != null){
			slowNode = slowNode.getNextNode();
			fastNode = fastNode.getNextNode().getNextNode();
		}
		
		System.out.println("Middle Element: " + slowNode.getData());

	}
	
	public static class LinkedListNode {
		private String data;
		/**
		 * @return the data
		 */
		public String getData() {
			return data;
		}

		/**
		 * @param data the data to set
		 */
		public void setData(String data) {
			this.data = data;
		}

		/**
		 * @return the nextNode
		 */
		public LinkedListNode getNextNode() {
			return nextNode;
		}

		/**
		 * @param nextNode the nextNode to set
		 */
		public void setNextNode(LinkedListNode nextNode) {
			if(nextNode == null)
				this.nextNode = null;
			this.nextNode = nextNode;
		}

		private LinkedListNode nextNode;
		
		LinkedListNode (String data, LinkedListNode nextNode) {
			this.data = data;
			this.nextNode = nextNode;
		}
	}

}

Here is how the linked list looks in Java

linkedlistjava

 

Happy Coding !!!

Area, Perimeter and Type of a Triangle.

Problem Statement: Given 3 sides of a triangle, find the area, perimeter and determine the type of the triangle, if isosceles, equilateral, or scalene. Also, check for Right Angled Triangle. /*Skipping Acute and Obtuse triangle types.*/

Prompt 3 user inputs for sides of the triangle and a dialog to output.

Solution:

import javax.swing.JOptionPane;

public class Triangle {

	/**
	 * This is a class to determine if the given side lengths form a triangle.
	 * It lets the user know the type of the triangle based on the given side lengths.
	 * Also, it gives the Perimeter and Area of the triangle.
	 * @param args
	 */
	public static void main(String[] args) {

//		String sideA, sideB, sideC;
		double a,b,c;
		a = Double.parseDouble(JOptionPane.showInputDialog(null,"Please input side 1 length of the triangle: ", "Triangle Side 1", JOptionPane.QUESTION_MESSAGE));
		b = Double.parseDouble(JOptionPane.showInputDialog(null,"Please input side 2 length of the triangle: ", "Triangle Side 2", JOptionPane.QUESTION_MESSAGE));
		c = Double.parseDouble(JOptionPane.showInputDialog(null,"Please input side 3 length of the triangle: ", "Triangle Side 3", JOptionPane.QUESTION_MESSAGE));
		
		// For right angled triangle, as per mathematics, square of the hypotenuse is equal to the sum of the squares of the other 2 sides. h^2 = a^2 + b^2 where h is hypotenuse (can be any side a,b,c)
		// So we first need to determine the hypotenuse
		
		double h = a > b ? (a > c ? a : c) : (b > c ? b : c);
		
		// Permiter of the triangle
		double p = a + b + c, s = p/2;
		
		//Area of triangle
		double areaOfTriangle = Math.sqrt(s * (s-a) * (s-b) * (s-c));
		
		// Check to see if the triangle is equilateral -- All sides should be equal
		if(a == b && b == c){ // no need to check a == c the value always holds true.
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is a equilateral." +
					"\nPerimeter of the triangle is " + p + 
					"\nArea of Triangle is " + areaOfTriangle);
		} 
		// Check to see if the triangle is isosceles -- At least 2 sides should be equal
		else if(a == b || b == c || c == a) {
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is a isosceles." +
					"\nPerimeter of the triangle is " + p + 
					"\nArea of Triangle is " + areaOfTriangle);
		}
		// Check to see if the triangle is right angle -- Can be verified if sum of squares of each of the sides is equal to twice the square of the hypotenuse. 
		//Ex : 3^2 + 4^2 + 5^2 = 2 * h^2 where h is the hypotenuse
		else if(Math.pow(h,2) * 2 == Math.pow(a,2) + Math.pow(b,2) + Math.pow(c,2)) {
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is a right angled triangle." +
					"\nPerimeter of the triangle is " + p + 
					"\nArea of Triangle is " + areaOfTriangle);
		}// Check to see if the triangle is scalene -- no sides should be equal
		else if(a != b && b != c && a != c) {
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is scalene." +
					"\nPerimeter of the triangle is " + p + 
					"\nArea of Triangle is " + areaOfTriangle);
		}
		// Check if the triangle is not a triangle
		else if(a > b + c || b > a + c || c > a + b ) {
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is not a triangle");
		}
		else {
			JOptionPane.showMessageDialog(null, "Based on the given sides, the triangle is a normal triangle." +
					"\nPerimeter of the triangle is " + p + 
					"\nArea of Triangle is " + areaOfTriangle);
		}
	}

triangle

 

Pascal’s Triangle with O(n^2) worst case.

public class PascalTriangle {
    
    public static void main(String[] args) {
        showPascal(9);
    }
    
    // This takes O(n^2) constraint.
    public static void showPascal(int rows){
    	// rows = Number of Pascal triangle rows to show up
    	// i = loop thru each row.
    	for(int i = 0; i<rows; i++)
    	{
    		// Start from 1. So, initialize the first number to 1
    		int number = 1;
    		// Give the spacing - Ex: If rows is 9, then the spacing  would 18 (because I specified the args as "", else if I specify args as "_", you would notice that the spacing is 17 and on 18th bit there would be a '_' ) for the first iteration of i = 0.
    		System.out.format("%"+(rows-i)*2+"s", "");
    		// j = loop thru  each column of a row by CALCULATing and display
    		for(int j=0; j<=i; j++)
    		{
    			// Give 3 more spacing and print the number on the 4th bit.
    			System.out.format("%4d", number);
    			
    			// Formula to calculate each column.
    			number = number * (i - j)/(j + 1);
    		}
    		// Print a empty line after each row.
    		System.out.println();
    	}
    }

Here is how the output look like:

pascaltriangleoutput

Happy Coding !!!

Some Points about Abstract and Interfaces.

Abstract Classes:

1). A class defined as abstract cannot be instantiated. Meaning, you cannot call just/only new AbstractClass().You need to override all the methods that are marked as abstract as well.(See AbstractMain)

2). Abstract keyword is non access modifier.

3). When a method is marked as abstract, there should not be any implementation for the method.

4). Abstract methods have to be overridden in the sub classes.

5). Abstract classes NEED NOT have abstract methods. Those methods which do not have abstract defined, should be implemented. Meaning there should be a implementation code for this method right in this abstract class method body.

6). If there is atleast one abstract method defined in the whole class, then that class should be declared as abstract.

7). Classes and Methods marked as private cannot be abstract

8). Classes and Methods marked as static cannot be abstract.

9). Classes and Methods marked as final cannot be abstract.

10). Abstract classes can have default and as well as overloaded constructors. When no constructor is defined a default constructor will be generated by compiler.

11). Constructors of abstract classes are invoked when a subclass object is created. You can call the superclass constructor by calling super().

12). Abstract classes hides the functionality of those methods that are defined as abstract. The actual functionality of the abstract methods are exposed by the implementation classes.

13). Abstract Method Overloading signatures are supported in abstract classes.

14). By making this class as abstract and implementing the methods as well, will not give any compile time error., but gives a run time error after Instantiating the AbstractSubClass.

15). ¬†@Override – Not necessary to provide this annotation, because,¬†Compiler by default takes care of overriding, when it see’s extends keyword.

Interfaces:

1). Interfaces does not have any method implementation.

2). All methods in an interface are public and/or abstract.

3). Any field that is declared, is public and static.

4). There is no multiple inheritance in Java. Which means, any Java class cannot extend more than one class at a time. Instead there is a Multiple Interface Inheritance. Which means, any Java can implement/extends more than one interface.

5). Any Java class that implements any interface, will be having a ISA relationship, between the Java Class and the interface.

6). If a method deceleration in a interface is abstract, the whole interface need not be declared as abstract, this is unlike abstract classes.

7). Java does not support multiple inheritance, because of the Diamond of Death Problem.

8). Method Overriding is allowed in Interfaces. Each method signature in an interface is unique. Meaning, you can have multiple methods of same name with different arguments.

9). Multiple Interface Inheritance example can extend any number of interfaces.

Ex: public interface MultipleInterfaceImplementation extends InterfaceExample,AnotherInterface

* here InterfaceExample and AnotherInterface are both interfaces.

Source code: https://github.com/chouhan/CoreJavaBasics

Big O, Omega and Theta Notations.

Big O, Omega and Theta Notations are used to describe not only the way an algorithm performs but the way an algorithm scales to produce a output. It measures the efficiency of an algorithm with respect to time it takes for an algorithm to run as a function of a given input. They are used to determine the Worst case complexity, Best case complexity and the Average case complexity.

Big Omega notation describes the best case of a running algorithm. In contrast, Big O notation describes the worst case of a running algorithm.

Big O Notation is denoted as O(n)  also termed as Order of n, or also termed as O of n.

Big Omega Notation is denoted as¬†ő©(n) also termed as Omega of n.

Apart from Big O (O) and Big Omega (ő©), there are Big Theta (őė), Small O (o) and Small Omega (ŌČ) notations that are used in computer science programming. You can get more information from¬†http://en.wikipedia.org/wiki/Big_O_notation#Big_Omega_notation.

O(1) – Constant Time

This means that the algorithm requires the same fixed number of steps regardless of the size of the task.

Examples:

1). Finding an element in a HashMap is usually a constant time, which is O(1). This is a constant time because, a hashing function is used to find an element, and computing a hash value does not depend on the number of elements in the HashMap.

2). Push and Pop operations for a stack containing n elements.

3). Insert and Remove the operations of a queue.

O(log n) – Logarithmic Time

This means that the algorithm requires the Logarithmic amount of time to perform the task.

Examples:

1). Binary search in a sorted list or Array list of n elements.

2). Insert and Find operations for a binary search tree with n nodes.

3). Insert and Remove operations for a heap with n nodes.

4). Fast insertion, removal and lookup time of a TreeMap (a.k.a balanced tree because a TreeMap maintains key/value objects in a sorted order by using a red black tree) is O(log n)

O(n) – Linear Time

This means that the algorithm requires a number of steps directly proportional to the size of the task.

Examples:

1). Traversal or searching of a list(a linked list or a array) with n elements. This is linear because you will have to search the entire list. This means that if a list is twice as big, searching will take twice as long.

2). Finding the maximum or minimum element in a list, or sequential search in an unsorted list of n elements.

3). Traversal of a tree with n nodes.

4). Calculating iteratively n-factorial, for example finding iteratively the nth Fibonacci number.

O(n log n) – N times Logarithmic time

This means that the algorithm requires N times the Logarithmic time of solving a algorithm.

Examples:

1).  Advanced Sorting Algorithms like quick sort and merge sort.

O(n2) – Quadratic Time

Examples:

1). Simple sorting algorithms, for example a selection sort of n elements.

2). Comparing 2 two dimensional arrays of size n by n.

3). Finding duplicates in an unsorted list of n elements.

Note: If a solution to a problem has one single iteration, in other words, if the solution is achieved by either only one for loop or one while loop or one do-while loop or a single recursive function, then that algorithm is said to perform with O(n) else if the solution is achieved by 2 nested loops, then the algorithm is said to perform with O(n2) and if it is achieved by 3 nested loops, then the algorithm is said to perform with O(n3) and so on..

O(n3) – Polynomial Time

Examples:

1). Given a expression 23n3 + 12n2 + 9, and n = large numbers, the execution time for n3 increases drastically which takes O(n3) to perform the operation.

O(an) for a > 1 – Exponential Time

Examples:

1). Recursive Fibonacci implementation

2). Problem to solve Towers of Hanoi

3). Generating all permutations of n symbols.

Here is the order of execution time, in which the way Big O notations worst case behavior is determined.

O(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) <O(an)

Constant Time < Logarithmic Time < Linear Time < N times of Logarithmic Time < Quadratic Time < Polynomial Time < Exponential Time.

If algorithm is __ then its performance is __

algorithm performance
o(n) < n
O(n) ‚ȧ n
őė(n) = n
ő©(n) ‚Č• n
ŌČ(n) > n
4 important Big O rules:

1). If you have 2 different steps in your algorithm, add up those steps

Example:

function something() {
    doStep1();	// O(a)
    doStep2();	// O(b)
}

Overall runtime: O(a+b)

2). Drop Constants

function minMax1(array) {
	min, max = null
	for each e in array
		min = MIN(e, min)	//O(n)
	for each e in array
		max = MAX(e, max)	//O(n)
}

Another example:

function minMax2(array){
	min, max = null
	for each e in array
		min = MIN(e, min)
		max = MAX(e, max)	
}

Overall runtime: O(2n)
So dropping the constants overall runtime equals to O(n)

3). Different Arrays is equivalent to (=>) Different Variables.

function getSize(arrayA, arrayB){
	int count = 0;
	for a in arrayA{
		for b in arrayB{
			if(a == b){
				count = count++;
			}
		}
	}
	return count;
}

Overall runtime is NOT O(n^2), but instead it is O(a * b)

Note that, the loops are emulated on 2 different array’s.

4). Drop non dominant terms.

function printSize(array){
	max = null;
	
	// runtime for the below loop is O(n)
	for each a in array{
		max = MAX(a, max);
	}
	
	//runtime for the below nested loop is O(n^2)
	for a in array{
		for b in array{
			print a, b;
		}
	}
}

Overall runtime: O(n + n^2)

However, since O(n) is too small and can be neglected.
In other words,

O(n^2) <= O(n + n^2) <= O(n^2 + n^2)
O(n^2) <= O(n + n^2) <= O(2 * n^2)

Now, dropping the constants based on Rule 2,

O(n^2) <= O(n + n^2) <= O(n^2)

if left and right are equivalent, then center is too..

i.e O(n + n^2) is equivalent to O(n^2)

Hence, overall runtime is O(n^2)


Sources:

http://www.apcomputerscience.com/apcs/misc/BIG-O.pdf
http://bigocheatsheet.com/

This post was more or less taken from the afore mentioned sites, with little or more changes based on my knowledge and observation. I will be adding more examples to one or more above mentioned time analysis Big O notations as and when I come across in the near future of programming career.

Multiplication Table..

package solutions.java;

/**
 * Created by rchouhan on 11/1/14.
 *
 * For Zero iterations
 * O(N) < Olog(N)
 *
 * replace N with 0; and log(0) = 1
 *
 * For N > 0
 *
 * O(N) < Olog(N) < O(Nlog(N)) < O(N + N) (also O(2N)) (or also O(N^2)) < O(N^N)
 */


public class MultiplicationTable {

    public static void main(String[] args){
        multiplicationTable(3);
        multiplicationTableUntil(3);
    }

    // Time Complexity is O(N)
    static void multiplicationTable(int i){

        for(int j=0; j<=10; j++){
            System.out.println(i +" X " + j +" = " + i*j);
        }
    }

    // Time Complexity is O(N^2)
    static void multiplicationTableUntil(int input){
        for(int i=1; i<= input; i++){
            System.out.println("\nMultiplication Table for " + i + "\n");
            for(int j=0; j<=10; j++){
                System.out.println(i +" X " + j +" = " + i*j);
            }
         }
    }
}