You have M vectors of length N. Each vector is defined by a position pos. Array on position i is pos – i if i <= pos and i – pos if i > pos. For each position, find maximum of values of all M vectors.
It turns out it’s enough to calculate minimal and maximal value of all M positions. For an index i, the answer is max(|i – minimal_position|, |i – maximal_position|), where |x| is absolute value of x.
Special case of M = 1
A good start is to see what happens if M = 1. Suppose selected position by the captain is pos. Then,
- numbering[i] = 0 if i = pos
- numbering[i] = numbering[i – 1] + 1 if i > pos
- numbering[i] = numbering[i + 1] + 1 if i < pos
where numbering is the resulting array. We can see that if i > pos, numbering[i] = i – pos and if i < pos, numbering[i] = pos – i. But this is the definition of absolute value. So numbering[i] = |i – pos| (where by |x| I denote absolute value of x).
Take general case
For general case, suppose pos, pos, …, pos[M] are positions asked by the captain during the M rounds. For each i (0 <= i < N) you need to output max(|i – pos|, |i – pos|, …, |i – pos[M]|).
So now let’s calculate for a particular i (let’s forget for a moment we need to iterate each i, we’re interested only in a single value of i). Let’s maximize |i – a|, where a can be pos, pos, …, pos[M]. A property of absolute value says that |i – a| = max(a – i, i – a). So if we maximize the expressions (a – i) and (i – a), then take their maximum value, we’re done for the given i.
In both expressions i is a constant (since we’ve decided to calculate only for a particular value of i). Since i is constant, it can’t influence maximization. So we need to maximize (a) and (–a). Now it’s obvious we need to take maximum and minimum between pos, pos, …, pos[M]. Let max_value be the maximum and min_value be the minimum. The answer for a fixed i is max(max_value – i, i – min_value).
The only remained thing is to iterate over all possible i. We calculate at the beginning max_value and min_value, then we iterate i and print max(max_value – i, i – min_value).
The algorithm solves each test in O(N + M) time complexity.