You are given an array of positive numbers. You can can convert any element (i) to any value in range [i/2, i-1] or [i+1, 2*i] in one step. Mind minimum steps needed to make the all elements equal.
1 <= n <= 1e5
Each element in range [1,1e5]
№ | Пользователь | Рейтинг |
---|---|---|
1 | jiangly | 3898 |
2 | tourist | 3840 |
3 | orzdevinwang | 3706 |
4 | ksun48 | 3691 |
5 | jqdai0815 | 3682 |
6 | ecnerwala | 3525 |
7 | gamegame | 3477 |
8 | Benq | 3468 |
9 | Ormlis | 3381 |
10 | maroonrk | 3379 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | Dominater069 | 161 |
4 | Um_nik | 159 |
4 | atcoder_official | 159 |
6 | djm03178 | 157 |
7 | adamant | 153 |
8 | luogu_official | 151 |
9 | awoo | 149 |
10 | TheScrasse | 146 |
You are given an array of positive numbers. You can can convert any element (i) to any value in range [i/2, i-1] or [i+1, 2*i] in one step. Mind minimum steps needed to make the all elements equal.
1 <= n <= 1e5
Each element in range [1,1e5]
Название |
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I think I found a solution.
First off, let's try to fix the element we will convert all elements into and minimize the step count across all elements. Can we calculate the minimum steps in $$$O(\log n)$$$? Turns out we can.
Let $$$x$$$ be the final element. First, we need a frequency array. Now we will use this to find all elements in a certain interval. It is clear we can find the number of elements in the interval $$$[x,2x]$$$ in $$$O(1)$$$ by turning the frequency array into a prefix sum array. We can do the same for $$$[x,\frac {x}{2}]$$$. For both of these we add $$$1$$$ to the total step count. For the interval $$$[2x,4x]$$$ though the step count will be 2, same goes for $$$[\frac {x}{2},\frac {x}{4}]$$$. The amount of such segments will clearly be $$$O(\log n)$$$.
Since there's only $$$10^5$$$ possible final elements, the total time complexity will be $$$O(n \log n)$$$
I'm sorry if I got this wrong
Let target element be 10
so according to you any element in range
[10,20] will take 1 move(s) [20,40] will take 2 move(s) [40,80] will take 3 move(s)
So let's take the element 50 which is to be converted to 10 : Then 50 -> 25 -> 12 -> 10 (3 moves)
Yeah, this approach seems right. You can also think about it as having only log(n) checkpoints that matter. So instead of 50->25->12-10, you can do 50->40->20->10 since the checkpoints are powers of 2 times your target (10), and there are log(n) ranges that matter.