Hard Amazon OA Problem | Which greedy strategy is correct ?

Правка en2, от LovesProgramming, 2024-11-11 11:08:00

Algorithm to solve this problem — If all weights were distinct solution is pretty simple — find 'r' the last and only occurrence of smallest number in the weight array — now travel efficiently through the second smallest number and push it to smallest position > r -> now do the same algorithm for the third number — this time 'r' will be the last and only occurrence of the second smallest number

Problem — This algorithm does not work if there are duplicate numbers in the weight array

-> [2 1 1 1 1 2 3 3 3 3 2]

-> [5 1 1 1 1 2 1 1 1 1 100]

It will take 3 moves on w[0] so it gets to a free position on the co-ordinate axis.

But if you make single move on w[5] towards the right — then w[0] can take place of w[5] in just total two moves.

So how to deal with which strategy to apply for duplicated numbers ?

I think it is impossible to solve this problem for duplicate numbers as it is NP-hard problem as per this blog — https://mirror.codeforces.com/blog/entry/103255

Теги greedy, amazon, oa

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en3 Английский LovesProgramming 2024-11-11 13:05:03 53 Tiny change: '![ ](/pred' -> 'This problem was asked in Amazon OA few days back\n\n![ ](/pred'
en2 Английский LovesProgramming 2024-11-11 11:08:00 164
en1 Английский LovesProgramming 2024-11-11 08:13:15 1080 Initial revision (published)