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]     [3,1,5]          2,4
#
#   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. list) 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. Any time 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/python/sorting/mergesort
#
# Alternate coding: Not really much difference here, just another code
#   that some readers will find more intuitive:
#
#      m = [1] * len( A )
#
#      for x in range(len(A)):
#        for y in range(x):
#         if m[y] >= m[x] and A[y] > A[x]:
#           m[x]+=1
#
#      max_value = max(m)
#
#      result = []
#      for i in range(m-1,-1,-1):
#        if max == m[i]:
#          result.append(A[i])
#          max-=1
#
#      result.reverse()
#      return result
#=======================================================================
 
 def LDS( A ):
  m = [0] * len( A ) # starting with m = [1] * len( A ) is not necessary
  for x in range( len( A ) - 2, -1, -1 ):
    for y in range( len( A ) - 1, x, -1 ):
      if m[x] <= m[y] and A[x] > A[y]:
        m[x] = m[y] + 1 # or use m[x]+=1

  #===================================================================
  # Use the following snippet or the one line below to get max_value
  # max_value=m[0]
  # for i in range(m):
  #  if max_value < m[i]:
  #    max_value = m[i]
  #===================================================================
  max_value = max( m )

  result = []
  for i in range( len( m ) ):
    if max_value == m[i]:
      result.append( A[i] )
      max_value -= 1

  return result
import unittest
from dynamic_programming import LongestDecreasingSubsequence as subsequence

class Test( unittest.TestCase ):

  def testDecreasingSubsequence( self ):
    A = [1, 2, 3, 4, 5, 6, 7, 8, 9]
    B = [9, 8, 7, 6, 5, 4, 3, 2, 1]
    C = [15, 37, 12, 40, 8, 1, 44, 5, 3, 45, 2, 50]
    D = [15, 12, 8, 5, 3, 2]

    actual = subsequence.LDS( A )
    self.assertEquals( 1, len( actual ) )
    actual = subsequence.LDS( B )
    self.assertEquals( B, actual )
    actual = subsequence.LDS( C )
    self.assertEquals( D, actual )