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ANUARM - Editorial


Author: Anudeep Nekkanti
Tester: Constantine Sokol
Editorialist: Florin Chirica






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[1], pos[2], ..., 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[1]|, |i – pos[2]|, ...., |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[1], pos[2], ..., 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[1], pos[2], ..., 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).

Time Complexity

The algorithm solves each test in O(N + M) time complexity.


Author's solution
Tester's solution

This question is marked "community wiki".

asked 20 Oct '14, 00:34

elfus's gravatar image

3★elfus ♦♦
accept rate: 0%

edited 20 Oct '14, 03:29

dpraveen's gravatar image

4★dpraveen ♦♦

Getting access denied errors in XML when I open the Author's and Tester's solutions.

(20 Oct '14, 01:00) chetan_kakkar3★

What has the tester done in so much code?

(20 Oct '14, 10:26) paramjitrohit2★

where is the practice link? Btw nice question. :)


answered 20 Oct '14, 05:34

tictactoecoder's gravatar image

accept rate: 7%

(25 Oct '14, 15:45) vidudaya3★

What is wrong with the below approach? For every M in the input calculate the maximum value of the left most guy(index 0)[leftMax] and right most guy(index n-1)[rightMax] At the end the maximum value assigned to guy at position i is maximum(|leftMax-i|,|rightMax-(n-1-i)|)?


answered 31 Dec '14, 06:31

anudeep_reddy's gravatar image

accept rate: 0%

You are not printing a space after each number

(31 Dec '14, 23:34) aalhadkulkarni2★

Please help me find the reason for WA link:


answered 05 Mar '15, 17:28

joel1794's gravatar image

accept rate: 0%

Please help me find the reason for wrong answer.


answered 04 Jun, 23:41

utkarsha_dr20's gravatar image

accept rate: 0%

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Asked: 20 Oct '14, 00:34

Seen: 3,768 times

Last updated: 04 Jun, 23:41