To solve this problem, we need to understand "What is the use of median". In statistics, the median is used for `dividing a set into two equal length subsets, that one subset is always greater than the other`

. If we understand the use of median for dividing, we are very close to the answer.

First let's cut **A** into two parts at a random position **i**:

```
left_A | right_A
A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
```

Since **A** has **m** elements, so there are **m+1** kinds of cutting( **i = 0 ~ m** ). And we know: **len(left_A) = i, len(right_A) = m - i** . Note: when **i = 0** , **left_A** is empty, and when **i = m** , **right_A** is empty.

With the same way, cut **B** into two parts at a random position **j**:

```
left_B | right_B
B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
```

Put **left_A** and **left_B** into one set, and put **right_A** and **right_B** into another set. Let's name them **left_part** and **right_part** :

```
left_part | right_part
A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
```

If we can ensure:

```
1) len(left_part) == len(right_part)
2) max(left_part) <= min(right_part)
```

then we divide all elements in **{A, B}** into two parts with equal length, and one part is always greater than the other. Then **median = (max(left_part) + min(right_part))/2**.

To ensure these two conditions, we just need to ensure:

```
(1) i + j == m - i + n - j (or: m - i + n - j + 1)
if n >= m, we just need to set: i = 0 ~ m, j = (m + n + 1)/2 - i
(2) B[j-1] <= A[i] and A[i-1] <= B[j]
```

ps.1 For simplicity, I presume **A[i-1],B[j-1],A[i],B[j]** are always valid even if **i=0/i=m/j=0/j=n** . I will talk about how to deal with these edge values at last.

ps.2 Why n >= m? Because I have to make sure j is non-nagative since 0 <= i <= m and j = (m + n + 1)/2 - i. If n < m , then j may be nagative, that will lead to wrong result.

So, all we need to do is:

```
Searching i in [0, m], to find an object `i` that:
B[j-1] <= A[i] and A[i-1] <= B[j], ( where j = (m + n + 1)/2 - i )
```

And we can do a binary search following steps described below:

```
<1> Set imin = 0, imax = m, then start searching in [imin, imax]
<2> Set i = (imin + imax)/2, j = (m + n + 1)/2 - i
<3> Now we have len(left_part)==len(right_part). And there are only 3 situations
that we may encounter:
<a> B[j-1] <= A[i] and A[i-1] <= B[j]
Means we have found the object `i`, so stop searching.
<b> B[j-1] > A[i]
Means A[i] is too small. We must `ajust` i to get `B[j-1] <= A[i]`.
Can we `increase` i?
Yes. Because when i is increased, j will be decreased.
So B[j-1] is decreased and A[i] is increased, and `B[j-1] <= A[i]` may
be satisfied.
Can we `decrease` i?
`No!` Because when i is decreased, j will be increased.
So B[j-1] is increased and A[i] is decreased, and B[j-1] <= A[i] will
be never satisfied.
So we must `increase` i. That is, we must ajust the searching range to
[i+1, imax]. So, set imin = i+1, and goto <2>.
<c> A[i-1] > B[j]
Means A[i-1] is too big. And we must `decrease` i to get `A[i-1]<=B[j]`.
That is, we must ajust the searching range to [imin, i-1].
So, set imax = i-1, and goto <2>.
```

When the object **i** is found, the median is:

```
max(A[i-1], B[j-1]) (when m + n is odd)
or (max(A[i-1], B[j-1]) + min(A[i], B[j]))/2 (when m + n is even)
```

Now let's consider the edges values **i=0,i=m,j=0,j=n** where **A[i-1],B[j-1],A[i],B[j]** may not exist. Actually this situation is easier than you think.

What we need to do is ensuring that `max(left_part) <= min(right_part)`

. So, if **i** and **j** are not edges values(means **A[i-1],B[j-1],A[i],B[j]** all exist), then we must check both **B[j-1] <= A[i]** and **A[i-1] <= B[j]**. But if some of **A[i-1],B[j-1],A[i],B[j]** don't exist, then we don't need to check one(or both) of these two conditions. For example, if **i=0**, then **A[i-1]** doesn't exist, then we don't need to check **A[i-1] <= B[j]**. So, what we need to do is:

```
Searching i in [0, m], to find an object `i` that:
(j == 0 or i == m or B[j-1] <= A[i]) and
(i == 0 or j == n or A[i-1] <= B[j])
where j = (m + n + 1)/2 - i
```

And in a searching loop, we will encounter only three situations:

```
<a> (j == 0 or i == m or B[j-1] <= A[i]) and
(i == 0 or j = n or A[i-1] <= B[j])
Means i is perfect, we can stop searching.
<b> j > 0 and i < m and B[j - 1] > A[i]
Means i is too small, we must increase it.
<c> i > 0 and j < n and A[i - 1] > B[j]
Means i is too big, we must decrease it.
```

Thank @Quentin.chen , him pointed out that: `i < m ==> j > 0`

and `i > 0 ==> j < n`

. Because:

```
m <= n, i < m ==> j = (m+n+1)/2 - i > (m+n+1)/2 - m >= (2*m+1)/2 - m >= 0
m <= n, i > 0 ==> j = (m+n+1)/2 - i < (m+n+1)/2 <= (2*n+1)/2 <= n
```

So in situation <b> and <c>, we don't need to check whether `j > 0`

and whether `j < n`

.

Below is the accepted code:

```
def median(A, B):
m, n = len(A), len(B)
if m > n:
A, B, m, n = B, A, n, m
if n == 0:
raise ValueError
imin, imax, half_len = 0, m, (m + n + 1) / 2
while imin <= imax:
i = (imin + imax) / 2
j = half_len - i
if i < m and B[j-1] > A[i]:
# i is too small, must increase it
imin = i + 1
elif i > 0 and A[i-1] > B[j]:
# i is too big, must decrease it
imax = i - 1
else:
# i is perfect
if i == 0: max_of_left = B[j-1]
elif j == 0: max_of_left = A[i-1]
else: max_of_left = max(A[i-1], B[j-1])
if (m + n) % 2 == 1:
return max_of_left
if i == m: min_of_right = B[j]
elif j == n: min_of_right = A[i]
else: min_of_right = min(A[i], B[j])
return (max_of_left + min_of_right) / 2.0
```