Longest Decreasing Subsequence
by Isai Damier

/***********************************************************************
 * Author: Isai Damier
 * Title: Longest Decreasing Subsequence
 * Project: geekviewpoint
 * Package: algorithms
 *
 * Statement:
 *   Given a sequence of numbers, find a longest decreasing subsequence.
 *   
 *
 *  Time Complexity: O(n^2)
 * 
 * Sample Input: {5,0,3,2,1,8}
 * Sample Output: {5,3,2,1}
 *
 * DEFINITION OF SUBSEQUENCE:
 *   A sequence is a particular order in which related objects follow
 *   each other (e.g. DNA, Fibonacci). A sub-sequence is a sequence
 *   obtained by omitting some of the elements of a larger sequence.
 * 
 *   SEQUENCE       SUBSEQUENCE     OMISSION
 *   {3,1,2,5,4}     {1,2}            3,5,4          
 *   {3,1,2,5,4}     {3,1,4}          2,5
 * 
 *   SEQUENCE       NOT SUBSEQUENCE   REASON
 *   {3,1,2,5,4}     {4,2,5}           4 should follow 5
 * 
 * STRATEGY:
 *   Illustrating by finding
 *   a longest decreasing subsequence of {5,0,3,2,1,8}:
 *   
 *   - Start by finding all subsequences of size 1: {5},{0},{3},{2},{1},{8};
 *     each element is its own decreasing subsequence.
 *   
 *   - Since we already have the solutions for the size 1 subsequences,
 *     we can use them in solving for the size two subsequences. For
 *     instance, we already know that 5 is the smallest element of a
 *     decreasing subsequence of size 1, i.e. the subsequence {5}.
 *     Therefore, all we need to get a subsequence of size 2 is add an
 *     element smaller than 5 to {5}: {5,0}, {5,3}, {5,2}, {5,1};
 *     {3,2}, {3,1}, {2,1}.
 * 
 *   - Now we use the size 2 solutions to get the size 3 solutions:
 *     {5,3,2}, {5,3,1}, {3,2,1}
 * 
 *   - Then we use the size 3 solutions to get the size 4 solutions:
 *     {5,3,2,1}. Since there are no size 5 solutions, we are done.
 * 
 * SUMMARY:
 *   Instead of directly solving the big problem, we solved a smaller
 *   version and then 'copied and pasted' the solution of the subproblem
 *   to find the solution to the big problem. To make the 'copy and paste'
 *   part easy, we use a table (i.e. array) to track the subproblems
 *   and their solutions. This strategy as a whole is called Dynamic
 *   Programming. The tabling part is known as memoization, which means
 *   writing memo.
 * 
 *   To recognize whether you can use dynamic programming on a problem,
 *   look for the following two traits: optimal substructures and
 *   overlapping subproblems.
 * 
 *   Optimal Substructures: the ability to 'copy and paste' the solution
 *     of a subproblem plus an additional trivial amount of work so to
 *     solve a larger problem. For example, we were able to use {5,3}
 *     itself an optimal solution to the problem {5,0,3} to get {5,3,2}
 *     as an optimal solution to the problem {5,0,3,2}.
 * 
 *   Overlapping Subproblems: Okay. So in our approach the solution grew
 *     from left to right: {5} to {5,3} to {5,3,2} etc. But in reality
 *     we could have solved the problem using recursion trees so that
 *     for example {5,3} could be reached either from {5} or from {3}.
 *     That wouldn't really be a problem except we would be solving for
 *     {5,3} more than once. Anytime a recursive solution would lead to
 *     such overlaps, the bet is dynamic programming is the way to go.
 * 
 *          {5}                 {3}
 *         / | \               / | \
 *        /  |  \             /  |  \
 *       /   |   \           /   |   \
 *   {5,0} {5,3} {5,2}   {5,3} {3,2} {3,1}
 * 
 * NOTE:
 * Dynamic Programming = Optimal Substructures + Overlapping Subproblems
 * Divide and Conquer = Optimal Substructures - Overlapping Subproblems 
 *   see merge sort: http://www.geekviewpoint.com/java/sorting/mergesort
 * 
 * Alternate coding: Not really much difference here, just another code
 *   that some readers will find more intuitive:
 * 
 *   int[] m = new int[A.length];
 *   Arrays.fill(m, 1);
 * 
 *   for(int x=1; x <A.length; x++)
 *     for(int y = 0; y < x; y++) 
 *       if(m[y] >= m[x] && A[y] > A[x]) 
 *         m[x]++; 
 * 
 *   int max = m[0]; 
 *   for (int i = 0; i < m.length; i++) 
 *     if (max < m[i]) 
 *       max = m[i]; 
 * 
 *   List<Integer> result = new ArrayList<Integer>(); 
 *   for (int i = m.length-1; i >=0 ; i--) 
 *     if (max == m[i]) { 
 *       result.add(A[i]); 
 *       max--; 
 *     }
 * 
 *   Collections.reverse(result); 
 *   return result;
 *    
 **********************************************************************/ 
 package algorithm.programming.dynamic;

import java.util.ArrayList;
import java.util.List;

public class LongestDecreasingSubsequence {

  public List<Integer> LDS(Integer[] A) {
    int[] m = new int[A.length];
    //Arrays.fill(m, 1);//not important here
    for (int x = A.length - 2; x >= 0; x--) {
      for (int y = A.length - 1; y > x; y--) {
        if (m[x] <= m[y] && A[x] > A[y]) {
          m[x]=m[y]+1;//or use m[x]++
        }
      }
    }
    int max = m[0];
    for (int i = 0; i < m.length; i++) {
      if (max < m[i]) {
        max = m[i];
      }
    }
    List<Integer> result = new ArrayList<Integer>();
    for (int i = 0; i < m.length; i++) {
      if (max == m[i]) {
        result.add(A[i]);
        max--;
      }
    }
    return result;
  }
}
package algorithm.programming.dynamic;

import java.util.Arrays;
import static org.junit.Assert.*;
import java.util.List;
import org.junit.Test;

public class LongestDecreasingSubsequenceTest {

  @Test
  public void testDecreasingSubsequence() {
    Integer[] A = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    Integer[] B = {9, 8, 7, 6, 5, 4, 3, 2, 1};
    Integer[] C = {15, 37, 12, 40, 8, 1, 44, 5, 3, 45, 2, 50};
    Integer[] D = {15, 12, 8, 5, 3, 2};

    LongestDecreasingSubsequence seq = new LongestDecreasingSubsequence();
    List<Integer> actual = seq.LDS(A);
    assertEquals(1, actual.size());
    actual = seq.LDS(B);
    assertTrue(Arrays.equals(B, actual.toArray(new Integer[0])));
    actual = seq.LDS(C);
    assertTrue(Arrays.equals(D, actual.toArray(new Integer[0])));
  }
}