Design Patterns & Principles

Design patterns and principles
1.	Design Principle vs. Design Pattern
2.	What algorithm is used in modern day elevators?
3.	Singleton Design Pattern - A Creational Design Patterns
4.	Factory Method Pattern (Virtual Constructor)
5.	Prototype Design Pattern
6.	Builder Design Pattern in Java
7.	Abstract Factory
8	Proxy Pattern
9	Flyweight Pattern - A Structural design pattern
10	Decorator design pattern
11	Adapter Pattern
12	Bridge Pattern
13	Composite Pattern
14	Facade pattern
15	Template Method Pattern
16	Strategy Pattern (Policy Pattern) - A behavioral design pattern
17	Iterator pattern or Cursor
18	Mediator pattern
19	Memento Design Pattern in Java
20	Observer pattern or Publish-subscribe pattern
21	State
22	Pizza Factory design in Java

Longest Repeating Subsequence

Find length of the longest repeating subsequence such that the two subsequences don’t have same string character at same position, i.e., any k’th character in the two subsequences shouldn’t have the same index in the original string.

This problem can be solved using the Longest Common Subsequence problem. LRS(str, str) where str is the input string with the restriction that when both the characters are same, they shouldn’t be on the same index in the two strings.

package amazon.dp;

public classLongestRptSubseq {

     private static int getLRS(String string) {

          

           char[] charArray = string.toCharArray();

           int len = charArray.length;

          

           int[][] dp = new int[len+1][len+1];

          

           for(int i=1;i<=len;i++) {

               

                for (int j=1;j<=len;j++) {

                     // If characters match and indexes are not same

                     if (charArray[i-1] == charArray[j-1] && i!=j) {

                           dp[i][j] =  1 + dp[i-1][j-1];

                          

                     } else {

                           // If characters do not match

                           dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j]);

                     }

                }

           }

          

           return dp[len-1][len-1];

     }

     public static voidmain(String[] args) {

           String string = "amazonazom";

           int len = getLRS(string);

           System.out.println("Longest Repeating Subsequence "+len);

     }

}

Count number of ways to cover a distance

Count total number of ways to cover the distance with 1, 2 and 3 steps.

Recursion solution time complexity is exponential i.e. O(3n).

Since same sub problems are solved again, this problem has overlapping sub problems property. So min square sum problem has both properties of a dynamic programming problem.

public class MaxStepsCount {

     /** Dynamic Programming. */

     private static int getMaxWaysDP(int distance) {

           int[] count = new int[distance+1];

           count[0] = 1;

           count[1] = 1;

           count[2] = 2;

           /** Memorize the Sub-problem in bottom up manner*/

           for (int i=3; i<=distance; i++) {

                count[i] = count[i-1] + count[i-2] + count[i-3];               

           }

           return count[distance];

     }

     /** Recursion Approach. */

     private static int getMaxWaysRecur(intdistance) {

           if(distance<0) {

                return 0;

           } else if(distance==0) {

                return 1;

           }

           return getMaxWaysRecur(distance-1)+getMaxWaysRecur(distance-2)

                     +getMaxWaysRecur(distance-3);

     }

     public static void main(String[] args) {

           // Steps pf 1, 2 and 3.

           int distance = 10;

           /** Recursion Approach. */

           int ways = getMaxWaysRecur(distance);

           System.out.println(ways);

           /** Dynamic Programming. */

           ways = getMaxWaysDP(distance);

           System.out.println(ways);

     }

}