codejam 2015 qualification round: Infinite House of Pancakes - algorithm

The problem description and solutions after contest analysis
There is one case I can't figure out: If there is one plate, 9 pancakes, that is the test case
1
9
the "correct answer" is 5
But how? Here is my "faulty" thinking:
9 -> {4, 5} -> {4,3,2} -> {3,2,2,2}
So it totals 3 + 3 = 6 minutes, not 5
Anything obvious that I misunderstood?

I managed to fail on this one during the competition too because I assumed that the best way was to split things in half (to get the maximum height reduction possible).
But, in viewing your question I can see a way that would do better then halving:
9 -> {6, 3} -> {3, 3, 3}
Two swaps plus three minutes to eat: 5 minutes

This has already been answered here:
Infinite House of Pancakes
Basically this is a case where dividing unevenly will give a better solution, for example:
9
3 6
3 3 3
2 2 2
1 1 1
0 0 0
Which is better than dividing evenly

Related

Can you check for duplicates by taking the sum of the array and then the product of the array?

Let's say we have an array of size N with values from 1 to N inside it. We want to check if this array has any duplicates. My friend suggested two ways that I showed him were wrong:
Take the sum of the array and check it against the sum 1+2+3+...+N. I gave the example 1,1,4,4 which proves that this way is wrong since 1+1+4+4 = 1+2+3+4 despite there being duplicates in the array.
Next he suggested the same thing but with multiplication. i.e. check if the product of the elements in the array is equal to N!, but again this fails with an array like 2,2,3,2, where 2x2x3x2 = 1x2x3x4.
Finally, he suggested doing both checks, and if one of them fails, then there is a duplicate in the array. I can't help but feel that this is still incorrect, but I can't prove it to him by giving him an example of an array with duplicates that passes both checks. I understand that the burden of proof lies with him, not me, but I can't help but want to find an example where this doesn't work.
P.S. I understand there are many more efficient ways to solve such a problem, but we are trying to discuss this particular approach.
Is there a way to prove that doing both checks doesn't necessarily mean there are no duplicates?
Here's a counterexample: 1,3,3,3,4,6,7,8,10,10
Found by looking for a pair of composite numbers with factorizations that change the sum & count by the same amount.
I.e., 9 -> 3, 3 reduces the sum by 3 and increases the count by 1, and 10 -> 2, 5 does the same. So by converting 2,5 to 10 and 9 to 3,3, I leave both the sum and count unchanged. Also of course the product, since I'm replacing numbers with their factors & vice versa.
Here's a much longer one.
24 -> 2*3*4 increases the count by 2 and decreases the sum by 15
2*11 -> 22 decreases the count by 1 and increases the sum by 9
2*8 -> 16 decreases the count by 1 and increases the sum by 6.
We have a second 2 available because of the factorization of 24.
This gives us:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24
Has the same sum, product, and count of elements as
1,3,3,4,4,5,6,7,9,10,12,13,14,15,16,16,17,18,19,20,21,22,22,23
In general you can find these by finding all factorizations of composite numbers, seeing how they change the sum & count (as above), and choosing changes in both directions (composite <-> factors) that cancel out.
I've just wrote a simple not very effective brute-force function. And it shows that there is for example
1 2 4 4 4 5 7 9 9
sequence that has the same sum and product as
1 2 3 4 5 6 7 8 9
For n = 10 there are more such sequences:
1 2 3 4 6 6 6 7 10 10
1 2 4 4 4 5 7 9 9 10
1 3 3 3 4 6 7 8 10 10
1 3 3 4 4 4 7 9 10 10
2 2 2 3 4 6 7 9 10 10
My write-only c++ code is here: https://ideone.com/2oRCbh

Creating simple repeating number sequence in SPSS

I want to create the following sequence in SPSS syntax. I've tried LOOP and DO REPEAT, but cannot figure out how to re-create this:
1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
Your question is really not clear enough, so I'm just guessing. Please edit your question so we can know if this is the right solution (and for the benefit of future readers).
If what you want is a variable that has the values 1, 1, 1, 2, 2, 2, 3, 3, 3, etc', Here is a way to get that:
compute MyVar=trunc(($casenum-1)/3)+1.
exe.

Getting the combination of facevalues that gives the highest score in a dicegame

Working on a dicegame for school and I have trouble figuring out how to do automatic calculation of the result. (we don't have to do it automatically, so I could just let the player choose which dice to use and then just check that the user choices are valid) but now that I have started to think about it I can't stop...
the problem is as follows:
I have six dice, the dice are normal dice with the value of 1-6.
In this example I have already roled the dice and they have the following values:
[2, 2, 2, 1, 1, 1]
But I don't know how to calulate all combinations so that as many dicecombinations as possible whose value combined(addition) are 3 (in this example) are used.
The values should be added together (for example a die with value 1 and another die with the value 2 are together 3) then there are different rounds in the game where the aim is to get different values (which can be a combination(addition) of die-values for example
dicevalues: [2, 2, 2, 2, 2, 2]
could give the user a total of 12 points if 4 is the goal for the current round)
2 + 2 = 4
2 + 2 = 4
2 + 2 = 4
if the goal of the round instead where 6 then the it would be
2 + 2 + 2 = 6
2 + 2 + 2 = 6
instead which would give the player 12 points (6 + 6)
[1, 3, 6, 6, 6, 6]
with the goal of 3 would only use the dice with value 3 and discard the rest since there is no way to add them up to get three.
2 + 1 = 3
2 + 1 = 3
2 + 1 = 3
would give the user 9 points.
but if it where calculated the wrong way and the ones where used up together instead of each 1 getting apierd with a two 1 + 1 + 1 which would only give the player 3 points och the twos couldn't be used.
Another example is:
[1, 2, 3, 4, 5, 6]
and all combinations that are equal to 6 gives the user points
[6], [5, 1], [4 ,2]
user gets 18 points (3 * 6)
[1 ,2 ,3], [6]
user gets 12 points (2 * 6) (Here the user gets six points less due to adding upp 1 + 2 + 3 instead of doing like in the example above)
A dice can have a value between 1 and 6.
I haven't really done much more than think about it and I'm pretty sure that I could do it right now, but it would be a solution that would scale really bad if I for example wanted to use 8 dices instead and every time I start programming on it I start to think that have to be a better/easier way of doing it... Anyone have any suggestion on where to start? I tried searching for an answer and I'm sure it's out there but I have problem forumulating a query that gives me relevant result...
With problems that look confusing like this, it is a really good idea to start with some working and examples. We have 6 die, with range [1 to 6]. The possible combinations we could make therefore are:
target = 2
1 combination: 2
2 combination: 1+1
target = 3
1 combination: 3
2 combination: 2+1
3 combination: 1+1+1
target = 4
1 combination: 4
2 combination: 3+1
2+2
3 combination: 2+1+1
4 combination: 1+1+1+1
target = 5
1 combination: 5
2 combination: 4+1
3+2
3 combination: 2+2+1
4 combination: 2+1+1+1
5 combination: 1+1+1+1+1
See the pattern? Hint, we go backwards from target to 1 for the first number we can add, and then given this first number, and the size of the combination, there is a limit to how big subsequent numbers can be!
There is a finite list of possible combinations. You can by looking for 1 combination scores, and remove these from the die available. Then move on to look for 2 combination scores, etc.
If you want to read more about this sub-field of mathematics, the term you need to look for is "Combinatorics". Have fun!

Binary tree in concentric circles

Recently I came across a question in an interview "Print a complete binary tree in concentric circles".
1
2 3
4 5 6 7
8 9 0 1 2 3 4 5
The output should be
1 2 4 8 9 0 1 2 3 4 5 7 3
5 6
Could anyone help me out on how we can solve this problem?
Here is how you can approach the problem. Arrange the tree by levels:
1
2, 3
4, 5, 6, 7
8, 9, 0, 1, 2, 3, 4, 5
So the data you have is k levels L1, L2, ..., Lk. Now answer this questions: After we execute one step, that is when one circle is traversed, how would the tree levels would look like after the traverse elements have been removed from the levels? How should I modify the levels and which elements should I print so that it would seems like I've traversed on circle?
In your example, after the first step the levels would be modified to just:
5, 6
So what was the operation that was executed?
After you've answered the questions just apply the same procedure couple of times until you've printed all elements.

Find the number of non-decreasing and non-increasing subsequences in an array

I am attempting to complete a programming challenge from Quora on HackerRank: https://www.hackerrank.com/contests/quora-haqathon/challenges/upvotes
I have designed a solution that works with some test cases, however, for many the algorithm that I am using is incorrect.
Rather than seeking a solution, I am simply asking for an explanation to how the subsequence is created and then I will implement a solution myself.
For example, with the input:
6 6
5 5 4 1 8 7
the correct output is -5, but I fail to see how -5 is the answer. The subsequence would be [5 5 4 1 8 7] and I cannot for the life of me find a means to get -5 as the output.
Problem Statement
At Quora, we have aggregate graphs that track the number of upvotes we get each day.
As we looked at patterns across windows of certain sizes, we thought about ways to track trends such as non-decreasing and non-increasing subranges as efficiently as possible.
For this problem, you are given N days of upvote count data, and a fixed window size K. For each window of K days, from left to right, find the number of non-decreasing subranges within the window minus the number of non-increasing subranges within the window.
A window of days is defined as contiguous range of days. Thus, there are exactly N−K+1 windows where this metric needs to be computed. A non-decreasing subrange is defined as a contiguous range of indices [a,b], a<b, where each element is at least as large as the previous element. A non-increasing subrange is similarly defined, except each element is at least as large as the next. There are up to K(K−1)/2 of these respective subranges within a window, so the metric is bounded by [−K(K−1)/2,K(K−1)/2].
Constraints
1≤N≤100,000 days
1≤K≤N days
Input Format
Line 1: Two integers, N and K
Line 2: N positive integers of upvote counts, each integer less than or equal to 10^9
Output Format
Line 1..: N−K+1 integers, one integer for each window's result on each line
Sample Input
5 3
1 2 3 1 1
Sample Output
3
0
-2
Explanation
For the first window of [1, 2, 3], there are 3 non-decreasing subranges and 0 non-increasing, so the answer is 3. For the second window of [2, 3, 1], there is 1 non-decreasing subrange and 1 non-increasing, so the answer is 0. For the third window of [3, 1, 1], there is 1 non-decreasing subrange and 3 non-increasing, so the answer is -2.
Given a window size of 6, and the sequence
5 5 4 1 8 7
the non-decreasing subsequences are
5 5
1 8
and the non-increasing subsequences are
5 5
5 4
4 1
8 7
5 5 4
5 4 1
5 5 4 1
So that's +2 for the non-decreasing subsequences and -7 for the non-increasing subsequences, giving -5 as the final answer.

Resources