There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

The trivial way, O(m + n):
Merge both arrays and the k-th smallest element could be accessed directly. Merging would require extra space of O(m+n). The linear run time is pretty good, but could we improve it even further?

A better way, O(k):
There is an improvement from the above method, thanks to readers who suggested this. (See comments below by Martin for an implementation). Using two pointers, you can traverse both arrays without actually merging them, thus without the extra space. Both pointers are initialized to point to head of A and B respectively, and the pointer that has the larger smaller (thanks to a reader for this correction) of the two is incremented one step. The k-th smallest is obtained by traversing a total of k steps. This algorithm is very similar to finding intersection of two sorted arrays.

The best solution, but non-trivial, O(lg m + lg n):
Although the above solution is an improvement both in run time and space complexity, it only works well for small values of k, and thus is still in linear run time. Could we improve the run time further?

The above logarithmic complexity gives us one important hint. Binary search is a great example of achieving logarithmic complexity by halving its search space in each iteration. Therefore, to achieve the complexity ofO(lg m  + lg n), we must halved the search space of A and B in each iteration.

We try to approach this tricky problem by comparing middle elements of A and B, which we identify as A_i and B_j. If A_i is between B_j and B_j-1, we have just found the i + j< + 1 smallest element. Why? Therefore, if we choose i and j such that i + j = k - 1, we are able to find the k-th smallest element. This is an important invariant that we must maintain for the correctness of this algorithm.

Summarizing the above,

If one of the above conditions are satisfied, we are done. If not, we will use i and j as the pivot index to subdivide the arrays. But how? Which portion should we discard? How about A_i and B_j itself?

We make an observation that when A_i < B_j, then it must be true that A_i < B_j-1. On the other hand, if B_j < A_i, then B_j < A_i-1. Why?

Using the above relationship, it becomes clear that when A_i < B_j, A_i and its lower portion could never be the k-th smallest element. So do B_j and its upper portion. Therefore, we could conveniently discard A_i with its lower portion and B_j with its upper portion.

If you are still not convince why the above argument is true, try drawing blocks representing elements in A and B. Try visualize inserting blocks of A up to A_i in front of B_j-1. You could easily see that no elements in the inserted blocks would ever be the k-th smallest. For the latter, you might want to keep the invariant i + j = k - 1 in mind to reason why B_j and its upper portion could never be the k-th smallest.

On the other hand, the case for A_i > B_j is just the other way around. Easy.

Below is the code and I have inserted lots of assertion (highly recommended programming style by the way) to help you understand the code. Note that the below code is an example of tail recursion, so you could technically convert it to an iterative method in a straightforward manner. However, I would leave it as it is, since this is how I derive the solution and it seemed more natural to be expressed in a recursive manner.

Another side note is regarding the choices of i and j. The below code would subdivide both arrays using its array sizes as weights. The reason is it might be able to guess the k-th element quicker (as long as the A and B is not differed in an extreme way; ie, all elements in A are smaller than B). If you are wondering, yes, you could choose i to be A‘s middle. In theory, you could choose any values for i and j as long as the invariant i+j = k-1 is satisfied.

 1 //please refer to this link http://blog.csdn.net/jiyanfeng1/article/details/8619985 2  3 class Solution { 4 public: 5     int find_kth_small(int A[], int m, int B[], int n, int k) 6     { 7  8         int i = (int)(((double) m)/(m+n)*(k-1)); 9         int j = k-1-i;10         // invariant: i + j = k-1  11         // Note: A[-1] = -INF and A[m] = +INF to maintain invariant  12         13         int Ai_1 = ((i == 0) ? INT_MIN :A[i-1]);14         int Bj_1 = ((j == 0) ? INT_MIN :B[j-1]);15         int Ai = ((i == m) ? INT_MAX :A[i]);16         int Bj = ((j == n) ? INT_MAX :B[j]);17         18         if(Ai == Bj)19             return Ai;20             21         if(Bj_1 < Ai && Ai < Bj)22             return Ai;23         if(Ai_1 < Bj && Bj < Ai)24             return Bj;25 26         // if none of the cases above, then it is either:    27         if(Ai < Bj)28         // exclude Ai and below portion  29         // exclude Bj and above portion 30             return find_kth_small(A+i+1, m-i-1, B, j, k-i-1);31         else /* Bj < Ai */32         // exclude Ai and above portion  33         // exclude Bj and below portion 34             return find_kth_small(A, i, B+j+1, n-j-1, k-j-1);35             //return find_kth_small(B+j+1, n-j-1, A, i, k-j-1);36     }37     double findMedianSortedArrays(int A[], int m, int B[], int n) {38         39         int cnt = m + n;40         41         if(cnt& 1 == 1)42         {43             return find_kth_small(A, m, B, n, cnt/2 +1 );44         }45         else 46             return (double)(find_kth_small(A, m, B, n, cnt/2 +1) + find_kth_small(A, m, B, n, cnt/2 ))/2.0;47     }48 };