如何找到两个排序数组之间的中位数？

Question

I'm working on a competitive programming problem where we're trying to find the median of two sorted arrays. The optimal algorithm is to perform a binary search and identify splitting points, i and j , between the two arrays.

I'm having trouble deriving the solution myself. I don't understand the initial logic. I will follow how I think of the problem so far.

The concept of the median is to partition the given array into two sets. Consider a hypothetical left array and a hypothetical right array after merging the two given arrays. Both these arrays are of the same length.

We know that the median given both those hypothetical arrays works out to be [max(left) + min(right)]/2 . This makes sense so far. But the issue here is now knowing how to construct the left and right arrays.

We can choose a splitting point on ArrayA as i and a splitting point on ArrayB as j . Note that len(ArrayB[:j] + ArrayB[:i]) == len(ArrayB[j:] +ArrayB[i:]) .

Now we just need to find the cutting points. We could try all splitting points i , j such that they satisfy the median condition. However this works out to be O(m*n) where M is size of ArrayB and where N is size of ArrayA .

I'm not sure how to get where I am to the binary search solution using my train of thought. If someone could give me pointers - that would be awesome.

Answer 1

Here is my approach that I managed to come up with.

First of all we know that the resulting array will contain N+M elements, meaning that the left part will contain (N+M)/2 elements, and the right part will contain (N+M)/2 elements as well. Let's denote the resulting array as Ans , and denote the size of one of its parts as PartSize .

Perform a binary search operation on array A . The range of such binary search will be [ 0 , N ]. This binary search operation will help you determine the number of elements from array A that will form the left part of the resulting array.

Now, suppose we are testing the value i . If i elements from array A are supposed to be included in the left part of the resulting array, this means that j = PartSize - i elements must be included from array B in the first part as well. We have the following possibilities:

• j > M this is an invalid state. In this case it means we still need to choose more elements from array A , so our new binary search range becomes [ i + 1 , N ].

• j <= M & A[i+1] < B[j] This is a tricky case. Think about it. If the next element in array A is smaller than the element j in array B , this means that element A[i+1] is supposed to be in the left part rather than element B[j] . In this case our new binary search range becomes [ i+1 , N ].

• j <= M & A[i] > B[j+1] This is close to the previous case. If the next element in array B is smaller than the element i in array A , the means that element B[j+1] is supposed to be in the left part rather than element A[i] . In this case our new binary search range becomes [ 0 , i-1 ].

• j <= M & A[i+1] >= B[j] & A[i] <= B[j+1] this is the optimal case, and you have finally found your answer.

After the binary search operation is finished, and you managed to calculate both i and j , you can now easily find the value of the median. You need to handle a few cases here depending on whether N+M is odd or even.

Hope it helps!