I've asked a similar question before, but this time it's different.

Since our array contains only two elements, we might as well set it to 1 and -1, where 1 is on the left side of the array and -1 is on the right side of the array:

[1,...,1,1,-1,-1,...,-1]

Both 1 and -1 exist at the same time and the number of 1 and -1 is not necessarily the same. Also, the numbers of 1 and -1 are both very large.

Then, define the boundary between 1 and -1 as the index of the -1 closest to 1. For example, for the following array:

[1,1,1,-1,-1,-1,-1]

Its boundary is 3.

Now, for each number in the array, I cover it with a device that you have to unlock to see the number in it.

I want to try to unlock as few devices as possible that cover 1, because it takes much longer to see a '1' than it takes to see a '-1'. And I also want to reduce my time cost as much as possible.

How can I search to get the boundary as quickly as possible?

CodePudding user response：

The problem is very like the "egg dropping" problem, but where a wrong guess has a large fixed cost (100), and a good guess has a small cost (1).

Let E(n) be the (optimal) expected cost of finding the index of the right-most 1 in an array (or finding that the array is all -1), assuming each possible position of the boundary is equally likely. Define the index of the right-most 1 to be -1 if the array is all -1.

If you choose to look at the array element at index i, then it's -1 with probability i/(n 1), and 1 with probability (n-i 1)/(n 1).

So if you look at array element i, your expected cost for finding the boundary is (1 E(i)) * i/(n 1) (100 E(n-i-1)) * (n-i 1)/(n 1).

Thus E(n) = min((1 E(i)) * i/(n 1) (100 E(n-i-1)) * (n-i 1)/(n 1), i=0..n-1)

For each n, the i that minimizes the equation is the optimal array element to look at for an array of that length.

I don't think you can solve these equations analytically, but you can solve them with dynamic programming in O(n^2) time.

The solution is going to look like a very skewed binary search for large n. For smaller n, it'll be skewed so much that it will be a traversal from the right.

CodePudding user response：

If I am right, a strategy to minimize the expectation of the cost is to draw at a fraction of the interval that favors the -1 outcome, in inverse proportion of the cost. So instead of picking the middle index, take the right centile.

But this still corresponds to a logarithmic asymptotic complexity.

There is probably nothing that you can do regarding the worst case.