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 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
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.