ZIO 2013: Dolls - algorithm

Question

Your shop sells several different types of dolls. Each doll has a suggested price, and no
two types of doll have the same price. You would like to fix an actual selling price for
each doll so that dolls of different types are as different in price as possible. Due to
some government regulations, you can only modify the suggested price within a fixed
band of ±K—in other words, if the suggested price is p, you can pick any selling price
in the range {p− K, p− K + 1, . . . , p+ K −1, p+ K}. Of course, the selling price must
always be non-negative.
For instance, suppose there are four types of dolls with suggested prices 130, 210, 70
and 90 and you are allowed to modify prices within a band of 20. Then, you can adjust
the prices to 150, 210, 50 and 100, respectively, so that the minimum difference in price
between any two types of dolls is 50. (For the second doll, you could have picked any
price between 200 and 230.) You can check that this is the largest separation that you
can achieve given the constraint.
In each of the cases below, you are given a sequence of prices and the value of K. You
have to determine the maximum separation that you can achieve between all pairs in
the sequence if you are allowed to modify each price by upto ±K.
(a) K = 13. Sequence: 144, 152, 214, 72, 256, 3, 39, 117, 238, 280.
(b) K = 10. Sequence: 10, 48, 57, 32, 61, 74, 33, 45, 99, 81, 19, 24, 101.
(c) K = 20. Sequence: 10, 19, 154, 67, 83, 39, 54, 110, 124, 99, 139, 170
So basically, I just need to find the value of maximum separation without coding. I tried to devise an algorithm, but failed miserably, so I just started brute forcing it, by basically increasing/decreasing each of the prices by a certain value, but the bruteforcing applied here is just too tough due to the value of K. (It would have been simple for any K<6).
Can someone define a function or recurrence relation to calculate it? The solutions are up online, but they only give the answer as an integer and don't explain how to reach the solution. I am a beginner in programming, so try explaining using pseudocode/ little bit of C++, please. Thank you.
Source: http://www.iarcs.org.in/inoi/2013/zio2013/zio2013-qpaper.pdf
Solution: http://www.iarcs.org.in/inoi/2013/zio2013/zio2013-solutions.pdf

Answer 1

Here is a O(nlogn) algorithm.
To illustrate I will use the second example: 10, 48, 57, 32, 61, 74, 33, 45, 99, 81, 19, 24, 101 with K=10
Sort the list (10, 19, 24, 32, 33, 45, 48, 57, 61, 74, 81, 99, 101)
Use bisection to find the minimum separation x
For a trial value of x, assign the final values greedily placing them as small as possible while satisfying the conditions (non-negative, within K of original value, at least x greater than previous).
So let us start with x=10.
We will move as follows:
10->0 (can't go negative so this is smallest allowed)
19->10 (can't go within K=10 of the previous value)
24->20
32->30
33->40
45->50
48 becomes impossible. We can only assign values between 38 and 58, but none of these are more than 10 away from the previous 50.
We conclude that x=10 is too high a separation and we need to move lower.
You might try x=7 and find it is possible, x=9 find it is impossible, then try x=8:
10->0
19->9 (can only move to values 9->29)
24->17
32->25
33->33
45->41
48->49
57->56
61->64
74->72
81->80
99->89
101->97
And so we have found that x=8 is possible, x=9 is impossible and therefore x=8 is the maximum possible separation.