How to determine the longest increasing subsequence using dynamic programming?

Question

I have a set of integers. I want to find the longest increasing subsequence of that set using dynamic programming.

Answer 1

I have implemented LIS in java using Dynamic Programming and Memoization. Along with the code I have done complexity calculation i.e. why it is O(n Log(base2) n). As I feel theoretical or logical explanations are good but practical demonstration is always better for understanding.

package com.company.dynamicProgramming;

import java.util.HashMap;
import java.util.Map;

public class LongestIncreasingSequence {

    static int complexity = 0;

    public static void main(String ...args){


        int[] arr = {10, 22, 9, 33, 21, 50, 41, 60, 80};
        int n = arr.length;

        Map<Integer, Integer> memo = new HashMap<>();

        lis(arr, n, memo);

        //Display Code Begins
        int x = 0;
        System.out.format("Longest Increasing Sub-Sequence with size %S is -> ",memo.get(n));
        for(Map.Entry e : memo.entrySet()){

            if((Integer)e.getValue() > x){
                System.out.print(arr[(Integer)e.getKey()-1] + " ");
                x++;
            }
        }
        System.out.format("%nAnd Time Complexity for Array size %S is just %S ", arr.length, complexity );
        System.out.format( "%nWhich is equivalent to O(n Log n) i.e. %SLog(base2)%S is %S",arr.length,arr.length, arr.length * Math.ceil(Math.log(arr.length)/Math.log(2)));
        //Display Code Ends

    }



    static int lis(int[] arr, int n, Map<Integer, Integer> memo){

        if(n==1){
            memo.put(1, 1);
            return 1;
        }

        int lisAti;
        int lisAtn = 1;

        for(int i = 1; i < n; i++){
            complexity++;

            if(memo.get(i)!=null){
                lisAti = memo.get(i);
            }else {
                lisAti = lis(arr, i, memo);
            }

            if(arr[i-1] < arr[n-1] && lisAti +1 > lisAtn){
                lisAtn = lisAti +1;
            }
        }

        memo.put(n, lisAtn);
        return lisAtn;

    }
}

While I ran the above code -

Longest Increasing Sub-Sequence with size 6 is -> 10 22 33 50 60 80 
And Time Complexity for Array size 9 is just 36 
Which is equivalent to O(n Log n) i.e. 9Log(base2)9 is 36.0
Process finished with exit code 0