Daily Coding Problem 260 : Reconstruct a jumbled array - Intuition?

Question

I'm going through the question below.

The sequence [0, 1, ..., N] has been jumbled, and the only clue you have for its order is an array representing whether each number is larger or smaller than the last. Given this information, reconstruct an array that is consistent with it.

For example, given [None, +, +, -, +], you could return [1, 2, 3, 0, 4].

I went through the solution on this post but still unable to understand it as to why this solution works. I don't think I would be able to come up with the solution if I had this in front of me during an interview. Can anyone explain the intuition behind it? Thanks in advance!

Answer 1

This answer tries to give a general strategy to find an algorithm to tackle this type of problems. It is not trying to prove why the given solution is correct, but lying out a route towards such a solution.

A tried and tested way to tackle this kind of problem (actually a wide range of problems), is to start with small examples and work your way up. This works for puzzles, but even so for problems encountered in reality.

First, note that the question is formulated deliberately to not point you in the right direction too easily. It makes you think there is some magic involved. How can you reconstruct a list of N numbers given only the list of plusses and minuses?

Well, you can't. For 10 numbers, there are 10! = 3628800 possible permutations. And there are only 2? = 512 possible lists of signs. It's a very huge difference. Most original lists will be completely different after reconstruction.

Here's an overview of how to approach the problem:

• Start with very simple examples
• Try to work your way up, adding a bit of complexity
• If you see something that seems a dead end, try increasing complexity in another way; don't spend too much time with situations where you don't see progress
• While exploring alternatives, revisit old dead ends, as you might have gained new insights
• Try whether recursion could work:
  • given a solution for N, can we easily construct a solution for N+1?
  • or even better: given a solution for N, can we easily construct a solution for 2N?
• Given a recursive solution, can it be converted to an iterative solution?
• Does the algorithm do some repetitive work that can be postponed to the end?
• ....

So, let's start simple (writing 0 for the None at the start):

• very short lists are easy to guess:
  • '0++' ? 0 1 2 ? clearly only one solution
  • '0--' ? 2 1 0 ? only one solution
  • '0-+' ? 1 0 2 or 2 0 1 ? hey, there is no unique outcome, though the question only asks for one of the possible outcomes
• lists with only plusses:
  • '0++++++' ? 0 1 2 3 4 5 6 ? only possibility
• lists with only minuses:
  • '0-------'? 7 6 5 4 3 2 1 0 ? only possibility
• lists with one minus, the rest plusses:
  • '0-++++' ? 1 0 2 3 4 5 or 5 0 1 2 3 4 or ...
  • '0+-+++' ? 0 2 1 3 4 5 or 5 0 1 2 3 4 or ...
  • ? no very obvious pattern seem to emerge
• maybe some recursion could help?
  • given a solution for N, appending one sign more?
  • appending a plus is easy: just repeat the solution and append the largest plus 1
  • appending a minus, after some thought: increase all the numbers by 1 and append a zero
  • ? hey, we have a working solution, but maybe not the most efficient one
    • the algorithm just appends to an existing list, no need to really write it recursively (although the idea is expressed recursively)
    • appending a plus can be improved, by storing the largest number in a variable so it doesn't need to be searched at every step; no further improvements seem necessary
    • appending a minus is more troublesome: the list needs to be traversed with each append
      • what if instead of appending a zero, we append -1, and do the adding at the end?
      • this clearly works when there is only one minus
      • when two minus signs are encountered, the first time append -1, the second time -2
      • ? hey, this works for any number of minuses encountered, just store its counter in a variable and sum with it at the end of the algorithm

This is in bird's eye view one possible route towards coming up with a solution. Many routes lead to Rome. Introducing negative numbers might seem tricky, but it is a logical conclusion after contemplating the recursive algorithm for a while.

Answer 2

It works because all changes are sequential, either adding one or subtracting one, starting both the increasing and the decreasing sequences from the same place. That guarantees we have a sequential list overall. For example, given the arbitrary

[None, +, -, +, +, -]

turned vertically for convenience, we can see

None   0
+      1
-     -1
+      2
+      3
-     -2

Now just shift them up by two (to account for -2):

2 3 1 4 5 0
 + - + + -

Answer 3

Let's look at first to a solution which (I think) is easier to understand, formalize and demonstrate for correctness (but I will only explain it and not demonstrate in a formal way):

We name A[0..N] our input array (where A[k] is None if k = 0 and is + or - otherwise) and B[0..N] our output array (where B[k] is in the range [0, N] and all values are unique)

At first we see that our problem (find B such that B[k] > B[k-1] if A[k] == + and B[k] < B[k-1] if A[k] == -) is only a special case of another problem:

Find B such that B[k] == max(B[0..k]) if A[k] == + and B[k] == min(B[0..k]) if A[k] == -.
Which generalize from "A value must larger or smaller than the last" to "A value must be larger or smaller than everyone before it"

So a solution to this problem is a solution to the original one as well.

Now how do we approach this problem?

A greedy solution will be sufficient, indeed is easy to demonstrate that the value associated with the last + will be the biggest number in absolute (which is N), the one associated with the second last + will be the second biggest number in absolute (which is N-1) ecc...
And in the same time the value associated with the last - will be the smallest number in absolute (which is 0), the one associated with the second last - will be the second smallest (which is 1) ecc...

So we can start filling B from right to left remembering how many + we have seen (let's call this value X), how many - we have seen (let's call this value Y) and looking at what is the current symbol, if it is a + in B we put N-X and we increase X by 1 and if it is a - in B we put 0+Y and we increase Y by 1.
In the end we'll need to fill B[0] with the only remaining value which is equal to Y+1 and to N-X-1.

An interesting property of this solution is that if we look to only the values associated with a - they will be all the values from 0 to Y (where in this case Y is the total number of -) sorted in reverse order; if we look to only the values associated with a + they will be all the values from N-X to N (where in this case X is the total number of +) sorted and if we look at B[0] it will always be Y+1 and N-X-1 (which are equal).
So the - will have all the values strictly smaller than B[0] and reverse sorted and the + will have all the values strictly bigger than B[0] and sorted.

This property is the key to understand why the solution proposed here works:
It consider B[0] equals to 0 and than it fills B following the property, this isn't a solution because the values are not in the range [0, N], but it is possible with a simple translation to move the range and arriving to [0, N]

Answer 4

The idea is to produce a permutation of [0,1...N] which will follow the pattern of [+,-...]. There are many permutations which will be applicable, it isn't a single one. For instance, look the the example provided:

[None, +, +, -, +], you could return [1, 2, 3, 0, 4].

But you also could have returned other solutions, just as valid: [2,3,4,0,1], [0,3,4,1,2] are also solutions. The only concern is that you need to have the first number having at least two numbers above it for positions [1],[2], and leave one number in the end which is lower then the one before and after it.

So the question isn't finding the one and only pattern which is scrambled, but to produce any permutation which will work with these rules.

This algorithm answers two questions for the next member of the list: get a number who’s both higher/lower from previous - and get a number who hasn’t been used yet. It takes a starting point number and essentially create two lists: an ascending list for the ‘+’ and a descending list for the ‘-‘. This way we guarantee that the next member is higher/lower than the previous one (because it’s in fact higher/lower than all previous members, a stricter condition than the one required) and for the same reason we know this number wasn’t used before.

So the intuition of the referenced algorithm is to start with a referenced number and work your way through. Let's assume we start from 0. The first place we put 0+1, which is 1. we keep 0 as our lowest, 1 as the highest.

l[0] h[1] list[1]

the next symbol is '+' so we take the highest number and raise it by one to 2, and update both the list with a new member and the highest number.

l[0] h[2] list [1,2]

The next symbol is '+' again, and so:

l[0] h[3] list [1,2,3]

The next symbol is '-' and so we have to put in our 0. Note that if the next symbol will be - we will have to stop, since we have no lower to produce.

l[0] h[3] list [1,2,3,0]

Luckily for us, we've chosen well and the last symbol is '+', so we can put our 4 and call is a day.

l[0] h[4] list [1,2,3,0,4]

This is not necessarily the smartest solution, as it can never know if the original number will solve the sequence, and always progresses by 1. That means that for some patterns [+,-...] it will not be able to find a solution. But for the pattern provided it works well with 0 as the initial starting point. If we chose the number 1 is would also work and produce [2,3,4,0,1], but for 2 and above it will fail. It will never produce the solution [0,3,4,1,2].

I hope this helps understanding the approach.

Answer 5

This is not an explanation for the question put forward by OP. Just want to share a possible approach. Given: N = 7 Index: 0 1 2 3 4 5 6 7 Pattern: X + - + - + - + //X = None

Go from 0 to N
[1] fill all '-' starting from right going left. Index: 0 1 2 3 4 5 6 7 Pattern: X + - + - + - + //X = None Answer: 2 1 0

[2] fill all the vacant places i.e [X & +] starting from left going right. Index: 0 1 2 3 4 5 6 7 Pattern: X + - + - + - + //X = None Answer: 3 4 5 6 7

Final: Pattern: X + - + - + - + //X = None Answer: 3 4 2 5 1 6 0 7