## O(n) and O(n^2) in a Dynamic Programming Problem

From: andrew cooke <andrew@...>

Date: Sat, 6 Aug 2011 14:39:51 -0400

I posted an answer at
http://stackoverflow.com/questions/6967853/dynamic-programing-can-interval-of-even-1s-and-0s-be-found-in-linear-time
that I want to copy here, beause I am worried that it will be deleted by the
moderators.

The question asks for a linear solution to a problem which generates O(N^2)
results.  That seems impossible, but if you look in more detail there's a cute
argument that hinges on the difference between "finding" and "printing" the
results:

A linear solution is possible (sorry, earlier I argued that this had to be
n^2) if you're careful to not actually print the results!

First, let's define a "score" for any set of zeros and ones as the number of
ones minus the number of zeroes.  So (0,1) has a score of 0, while (0) is -1
and (1,1) is 2.

Now, start from the right.  If the right-most digit is a 0 then it can be
combined with any group to the left that has a score of 1.  So we need to know
what groups are available to the left, indexed by score.  This suggests a
recursive procedure that accumulates groups with scores.  The sweep process is
O(n) and at each step the process has to check whether it has created a new
group and extend the table of known groups.  Checking for a new group is
constant time (lookup in a hash table).  Extending the table of known groups
is also constant time (at first I thought it wasn't, but you can maintain a
separate offset that avoids updating each entry in the table).

So we have a peculiar situation: each step of the process identifies a set of
results of size O(n), but the calculation necessary to do this is constant
time (within that step).  So the process itself is still O(n) (proportional to
the number of steps).  Of course, actually printing the results is O(n^2).

I'll write some Python code to test/demonstrate.

Here we go:

SCORE = [-1,1]

class Accumulator:

def __init__(self):
self.offset = 0
self.groups_to_right = {} # map from score to start index
self.even_groups = []
self.index = 0

def append(self, digit):
score = SCORE[digit]
# want existing groups at -score, to sum to zero
# but there's an offset to correct for, so we really want
# groups at -(score+offset)
corrected = -(score + self.offset)
if corrected in self.groups_to_right:
self.even_groups.append(
(self.index, self.groups_to_right[corrected]))
# this updates all the known groups
self.offset += score
# this adds the new one, which should be at the index so that
# index + offset = score (so index = score - offset)
groups = self.groups_to_right.get(score-self.offset, [])
groups.append(self.index)
self.groups_to_right[score-self.offset] = groups
# and move on
self.index += 1
#print self.offset
#print self.groups_to_right
#print self.even_groups
#print self.index

def dump(self):
# printing the results does take longer, of course...
for (end, starts) in self.even_groups:
for start in starts:
if start < end:
print (start, end)

@staticmethod
def run(input):
accumulator = Accumulator()
print input
for digit in input:
accumulator.append(digit)
accumulator.dump()
print

Accumulator.run([0,1,0,0,1,1,1,1,0])

And the output:

dynamic: python dynamic.py
[0, 1, 0, 0, 1, 1, 1, 1, 0]
(0, 1)
(1, 2)
(1, 4)
(3, 4)
(0, 5)
(2, 5)
(7, 8)

You might be worried that some additional processing (the filtering for start
< end) is done in the dump routine that displays the results.  But that's
because I am working around Python's lack of linked lists (I want to both
extend a list and save the previous value in constant time).

It may seem surprising that the result is of *size* O(n^2) while the process
of *finding* the results is O(n), but it's easy to see how that is possible:
at one "step" the process identifies a number of groups (of size O(n)) by
associating the current point (end in dump()) with a list of end points
(ends).

Andrew