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Possible solutions to this question ?
By guesshere, history, 15 months ago, In English

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]

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15 months ago, # |
← Rev. 3   Vote: I like it +13 Vote: I do not like it

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

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    15 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    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)

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      15 months ago, # ^ |
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      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.

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