
Match Editorial 
SRM 169Tuesday, October 28, 2003
Match summary
The fact that 507 coders (or should I say top coders) have registered for
this match is quite impressive. DIV I coders lived up to their high standard
and the submission and success rates where pretty solid. But I think there
was one other quite impressive thing
if not a bit suprising and frustrating....
Lets look at the following scenario:
It is probably (if memory serves me correctly) the only time that I have
seen complete silence in my room (Div II) and basically no submissions for
the longest period of time.
Here is how it went:
We all started with the 250 as usual and we all submitted the easy problem
within the normal time frame of about 5 to 10 mins. So far, so good. I check
the submission times and I see that I am ok. I was about 4th in my room at
the time when I submitted. So what do we all do next? But of course, we open
and tackle the 500 problem. So we all opened the 500 and...
And I think that any spectator watching my room would have been quite taken
aback and would have shaken his applet a bit in order to see why nothing was
happening. And nothing seemed to be happening for the longest period of
time. But in actuality we were all hard at work trying to figure out a
strategy for getting this problem into submission. Time passed, and an
occasional compilation or testing message would appear on the room history
list. Finally after what must have been like 40 minutes submissions started
to trickle in.
But alas! No time was left to even attempt the 1000 and
in my room there were no submissions for the hard problem. What is even more
interesting is the fact that most of the submissions for the DIV II 500
would fail either by a direct coder challenge or through a system test case.
There was about a 20% success rate for this problem. Any questions? Yes,
this writeup is dedicated to the coders of alternative submission
lifestyle. But do read on, as we will have a look at why there was such a
poor success rate for this problem and we will look at a strategy to
maximize chances at scoring high points especially for those who would like
to trade the green of their earthly pastures for the blue of the heavens.
The Problems
Swimmers
Used as: Division Two  Level One:
Value

250

Submission Rate

260 / 299 (86.96%)

Success Rate

196 / 260 (75.38%)

High Score

swanksax2 for 238.48 points (6 mins 18 secs)

Average Score

191.46 (for 196 correct submissions)

This is a relatively simple problem but one has to be always on the lookout
for special cases.
In a nutshell: A swimmer will plunge into a river at some point A and swim
downstream a certain given distance to point B (i.e. with the river current)
and then reverse and swim upstream back to point A (i.e. against the
current) We are given the distance that swimmer will swim before they will
reverse. We are also given the speed of the river current and the speed of
the swimmer. We are to calculate the actual time that it took the swimmer to
swim from point A to point B and back to point A. It is in fact a simple
vector addition on the xaxis. One important thing to watch out for is the
fact that the swimmer might not have enough speed to return. If the speed of
the river current is greater than or equal to the swimmer's speed then the
swimmer will never make it back. In such a case we return 1.
Simple formula:
The time to swim from point A to point B is given
by
TimeFromAtoB = ABDistance/(speedOfSwimmer +
speedOfCurrent)
Here the swimmer is taking advantage of the fact
that his/her speed gets added
to the speed of the current.
The time to swim back from point B to point A
(against the current) is given by
TimeFromBtoA = ABDistance/(speedOfSwimmer 
speedOfCurrent)
Here the swimmer has to fight the current and in
fact looses speed which
basically means that the current speed gets
subtracted from the swimmer's speed.
Thus the total time that the swimmer takes to
complete their round is going to be the addition of these two.
Contraints:
There were a number of things to watch out for though:
(1) The return value was to be rounded down (take
the floor of the result) but one
has to be careful not to do the
following:
floor(ABDistance/(speedOfSwimmer
+ speedOfCurrent)) +
floor(ABDistance/(speedOfSwimmer
 speedOfCurrent))
as this would result
occasionally in too small a value.
We need to do the
following:
floor(
(ABDistance/(speedOfSwimmer + speedOfCurrent)+
(ABDistance/(speedOfSwimmer  speedOfCurrent)
)
(2) The actual distance from A to B could be 0. This
is a special case and we
would return for such a swimmer the time of 0
regardless of what the swimmer's speed
is or the river current speed. This takes precedence
over case (3) below.
(3) If the actual speed of the swimmer is less than
(or equalto) the speed of the
river current then the swimmer would never make it
back. In such a case we return 1;
(4) Note that we can have a case where we would have
division by 0 in our
formula if the speed of the swimmer is the same as
the speed of the river current. We
need to ensure that we trap this case before it gets
to our formula. Note that test of
(equalto) in case (3) takes care of that. The code
would fail if we only checked for the lessthan condition.
Sample java code:
int[] result = new int[speeds.length];
for(int i=0; i < speeds.length; i++)
{
// case (2)
if(distances[i] <= 0)
{
result[i] = 0;
continue;
}
// case (3) note the (equalto).
if(speeds[i] <= current)
{
result[i] = 1;
}
// case (1)
else
{
double temp = ((double)distances[i]/(speeds[i]+current));
temp += ((double)distances[i]/(speeds[i]current));
result[i] = (int)(Math.floor(temp));
}
}
return result;
Where did code fail:
Most code that failed the system tests or was successfully challenged did
not ensure that
case (2) takes precedence over case (3) as well as missing the (equalto)
condition in case (3)
Twain
Used as: Division Two  Level Two:
Value

500

Submission Rate

110 / 299 (36.79%)

Success Rate

23 / 110 (20.91%)

High Score

SleepyOverlord for 273.99 points (32 mins 1 secs)

Average Score

216.66 (for 23 correct submissions)

Oh boy, oh boy! This was the killer question for div II. On the surface this
is a relatively simple problem where we are given a number of rules to
follow that transform a given input string through a succession of well
defined steps. There is a maximum of 7 transformations (and a minimum of 0
where the input string in unchanged) that can be applied to the input
string. The key to solving this problem successfully is to build the
solution in increments and test each transformation independently.Basically
if you followed the rules you would have no problem solving this question.
But then why did so many coders fail this question? And secondly why was the
fastest submission time clocked at about 30 mins?
With this particular question you have to ensure that you read each rule
carefully as mistakes in such questions are difficult to pinpoint and thus
very time
consuming to debug.
Lets look at two sample transformations necessary for this question:
Transformation (1)  year 1
 If a word starts with "x", replace the "x" with a
"z".
 Change all remaining "x"s to "ks"s.
Solution pseudocode:
find all the "x"'s in the
input string and for each "x" do the following test:
(beginning of a word
test)
if the x is the very first
character in the string (beginning of a word special
case) then
replace
this 'x' with 'z'
else if the character just
before the current 'x' is a space character ' ' then
replace
this 'x' with 'z'
(inside the word)
else
replace
the 'x' with "ks"
Faster Trick Solution:
Add to the beginning of
the input String a space: input = " "+input. This will
collapse the word test
code above from two cases to one generic case.
Replace each occurrence of
" x" with " z" (note the added space to test a beginning of a word)
Replace remaining
occurrences of "x" with "ks"
Remove the extra at the
beginning " ": input = input.substring(1, input.length());
Here is a java code snippet:
input = " " + input;
input = input.replaceAll(" x", " z");
input = input.replaceAll("x","ks" );
input = input.substring(1);
Transformation (5) year 5
 If a word starts with "sch", change the "sch" to
a "sk".
 If a "ch" is directly followed by an "r", change
the "ch" to a "k".
 After applying the above rules, change all "c"s
that are not directly followed
by an "h", to a "k".
(This includes all "c"s that are the last letter
of a word.)
Solution pseudocode:
Add to the beginning of
the input String a space: input = " "+input;
Replace each occurrence of
" sch" with " sk" (note the added space to test a
beginning of a word)
Replace each occurrence of
"chr" with " kr"
Append a space to the
input string: input = input + " " (This is a nice trick to
ensure that we can test
for the last letter without falling off the edge)
Replace all occurrences of
"c" not followed by an "h" to a "k":
(Find all 'c' and then
check if the next character to the right is an 'h'
if(not)
then
replace
the 'c' with "k"
)
Remove the helper spaces
from beginning and end.
Here is a java code snippet:
input = " " + input + " ";
input = input.replaceAll(" sch", " sk");
input = input.replaceAll("chr","kr" );
for(int i=0; i < input.length(); i++)
{
if(input.charAt(i) == 'c' && input.charAt(i+1) != 'h')
input = input.substring(0, i) + "k" +
input.substring(i+1);
}
input = input.substring(1, input.length()1);
As can be seen just going over two transformations this can be quite time
consuming especially if you do not use any tricks and
shortcuts. It is also very error prone.
Where did code fail:
Basically all over the place. It was easy to make mistakes here. In some
cases the code had some problems with correctly figuring out where the
beginning of a word was.
Here is a suggested strategy that might be useful when you hit a rather
difficult to write/understand problem:
(1) Read it carefully a
couple of times (Always a good thing to do)
(2) Formulate a strategy for
the solution before you write code
(3) Test your strategy
against a couple sample test cases (no code yet)
(4) if you find that the
question is very time consuming (i.e. for the 500 I would say if
within 25 minutes you
are not testing the code) you might consider skipping this
question for now and
peek if the next question is perhaps more to your liking.
You can always come
back (with a loss of time and thus points of course) if you find
the other question even
harder.
In SRM 169 it turns out that the 1000 point question for Div II was
actually pretty
accessible and doable. Skipping the 500 question would have been preferable
(I think) to
a lot of people.
FairWorkload
Used as: Division Two  Level Three:
Value

1000

Submission Rate

19 / 299 (6.35%)

Success Rate

11 / 19 (57.89%)

High Score

. for 914.74 points (8 mins 50 secs)

Average Score

721.75 (for 11 correct submissions)

Used as: Division One  Level Two:
Value

500

Submission Rate

147 / 184 (79.89%)

Success Rate

132 / 147 (89.80%)

High Score

antimatter for 487.35 points (4 mins 36 secs)

Average Score

348.60 (for 132 correct submissions)

This is a very nice question, which can be solved in a couple ways. But
first and foremost lets try to understand what the problem is asking us to
do.
We are given a number of filing cabinets where each cabinet contains some
number of folders. We are also given the number of available workers. What
we would like to do is to partition the cabinets amongst the N workers in a
such a way that
(1) each worker works with adjacent cabinets
(2) number of folders that each worker has to work
on (cabinets can not be shared) is spread as evenly as possible.
We are to return the largest partition in our most evenly spread
arrangement.
As an example: if we are given 9 cabinets:
10 20 30 40 50 60 70 80
90
and we have 3 workers, the best solution would be:
10 20 30 40 50  60 70  80
90
And the largest partition in this case is 170 (80+90)
Constraints:
The number of cabinets we will work with is
[2..15]
The max number of workers we will have is
[2..15]
This is a nutshell is a problem of generating all N (number of workers)
subsets of the cabinets such that the following
is always true:
(1) A subset must contain
only adjacent cabinets
(2) Cabinets can not
repeat
(3) Cabinets can not be
omitted: The total number of cabinets in all subsets gives
us the total set back
(4) Subsets cannot be
empty
Here are a couple approaches to this problem:
(1) First we can generate all the possible valid subsets as follows:
We set a globalMax to be the total number of folders.
We start with the full set of cabinets and we divide this set into two
subsets LeftSet and RightSet.
We then take the RightSet and again divide it by into two sets.
We repeat this process until we have generated the same number of sets as we
have workers.
For each subset that we generate in this manner we calcuate the
LocalMax.
We then compare this LocalMax to the GlobalMax and if GlobalMax is greater
than the LocalMax then we set
the new GlobalMax to be the value of the LocalMax.
Basically we have found at this point a solution that distributes the
folders better.
Once we go through all these sets we return the current GlobalMax.
This can be done in a recursive manner as follows (after lanenal's solution
from room 2)
int globalMax;
public int getMostWork(int[] f, int w)
{
for(int i=0; i<f.length; i++) globalMax += f[i];
genSubsets(f, 1, w, 0);
return globalMax;
}
void genSubsets(int[] f, int from, int w, int localMax)
{
int sum = 0;
// Final subset
if(w==1)
{
// get the folders for this subset
for( int i=from+1; i<=f.lengthw; i++) sum += f[i];
// is this the most folders for the current subsets?
// If yes then update the localMax
if(localMax<sum) localMax = sum;
// Is our localMax less than (better) the globalMax
if(localMax < globalMax) globalMax = localMax;
return;
}
for( int i=from+1; i<=f.lengthw; i++)
{
// the LeftSubset folders sum
sum += f[i];
// Generate the RightSubset by dividing the current
// subset in two with the pivot at i.
// Propagate localMax to the RightSubset
genSubsets(f, i, w1, sum>localMax?sum:localMax);
}
}
(2) Application of Binary Search. For a nonrecursive example of this
approach please look at Ukraine's code in room 1.
(3) We can also use Dynamic Programming. I will leave this as an exercise to
the reader.
MineField
Used as: Division One  Level One:
Value

250

Submission Rate

181 / 184 (98.37%)

Success Rate

177 / 181 (97.79%)

High Score

ZorbaTHut for 245.83 points (3 mins 42 secs)

Average Score

206.02 (for 177 correct submissions)

This is actually a very easy problem in my opinion.Basically given locations
of randomly generated mines on a playing grid
we are to return an array which for each location on the grid that is not
occupied by a mine will give the count of all the mines that
are neighboring this location.
Constraints:
9x9 grid
Mines input will be [0, 10]
Each grid location will neighbor at most 8
mines.
I will present here three steps to the solution:
(1) Create a grid char Array [10][10] initialized to
0
(The grid is extended by a 1 char buffer in every
direction)
Parse the input string and place the mines in grid.
(denoted with 'M')
(2) For each grid location in the range of
[1..9][1..9]
int count=0;
Iterate through
each neighbor of the current grid location
if(the neighbor
location contains 'M')
count++;
Set the current
grid location to count;
(3) Convert the grid representation to the output
string.
The reason why I have created the grid to be 10x10 is so that I have a
buffer (which is empty and thus will not contribute to the count) which
allows me not to have a boundary testing code when I am looking at the
neighbors:
Here is a java snippet for (2) assuming the grid is made to be [10][10]
char[][] grid = new char[10][10];
int count;
for(int row=1; row <= 9; row++)
{
for(int col=1; col <= 9; col++)
{
count=0;
if(grid[row][col] = 'M') continue;
// Test all neighbours
// (Note no need to test if outside of grid)
for(int r = row1; r < = row+1; r++)
{
for(int c = col1; c < = col+1; c++)
{
if(row==r && col==c) continue;
if(grid[r][c] == 'M')
count++;
}
}
grid[row][col] = count;
}
}
Where did code fail:
The success rate for this problem was almost 100% but a few solutions failed
mostly on incorrectly going through the neighbors. Especially one has to be
careful with the
if(row==r && col==c) continue
condition to ensure that one doesn't do this instead:
(for example)
if(row==0 && col==0) continue;
GoldMine
Used as: Division One  Level Three:
Value

900

Submission Rate

56 / 184 (30.43%)

Success Rate

39 / 56 (69.64%)

High Score

WishingBone for 757.59 points (12 mins 49 secs)

Average Score

472.93 (for 39 correct submissions)

This is a rather straight forward problem once you realize that is can be
solved in a greedy manner.
At any given point a miner has a profit value associated with him depending
on the state of each mine (i.e is the mine full? How much will the mine
yield if we add this miner to it) We need to find the mine that will yield
best profit if we add this miner to it.The basic premise has to do with
maximizing resources. Given a number of gold mines and a formula for the
amount of money that each mine would yield given a number of miners working
in the mine we can use the following observation to solve this
problem:
Take the next available miner and apply him into
each of the mines.
Calculate the profit made by applying this miner to
each mine and choose the best
profit for this miner. Apply this miner to the
goldmine that makes the most money from
having him.
We continue in this manner until all miners have
been distributed.
How do we know that this will work? Think of it this way: any given gold
mine will have profit based on the following function:
Profit(goldmine , numminers)
We could run now this function for every gold mine
for all the possible num_miners [0..6]
that could fit into that mine.
This would build for us a profit table for each
mine.
We could then sort this table based on profit and
number of miners.
Then, once we are given the number of available
miners we would simply look up the
most profitable entry in our table for the most
miners we could fit in it. Then we repeat this
process for the remaining miners (after having
removed the best entry from the profit table)
This is in essence a greedy algorithm, which is
equivalent to the algorithm given before.
Constraints:
We can have at most six miners per gold mine.
All miners must be placed.
That's all, folks. Until the next SRM. keep on coding!
By AleaActaEst
TopCoder Member