Median of Two Sorted Arrays

【Problem】

There are two sorted arrays nums1 and nums2 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)).

Example 1:

nums1 = [1, 3]nums2 = [2]The median is 2.0

Example 2:

nums1 = [1, 2]nums2 = [3, 4]The median is (2 + 3)/2 = 2.5

【Thinking】

this is the fourth problem on the leetcode website. The link ishttps://leetcode.com/problems/median-of-two-sorted-arrays/description/. You can try it by yourself.

First time, I used the brute force solution that sort the two arrays into a vector, then find the middle element in the vector.  As expected, the complexity over o(log(m+n)). It is not satisfied with the problem requirement. The run time complexity should be o(log(m+n)).

Second time, I choosed to directly find the middle element whentraversal the two array.Using the two

variables---(i,j)  to record the position of biggest number between arrays. when i+j==(nums1.length+nums2.length)/2, we find the middle element. We also should consider the nums1.length+nums2.length is odd(奇数) or even number（偶数）to choose different calculate way.

【Important】

Recursive Approach

递归的方法

To solve this problem, we need to understand "What is the use of median". In statistics, the median is used for:

为了解决这个问题， 我们需要理解 “中位数的用途” 。 统计学中，中位数被用于：

Dividing a set into two equal length subsets, that one subset is always greater than the other.

将一个集合分成两个相等长度的子集，一个子集总是比另一个子集大。

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

如果我们理解了中位数在划分时的用途，我们就离答案不远了。 

First let's cut $\text{A}$ into two parts at a random position $i$:

首先我们将A以任意位置i切分成两部分:

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

Since $\text{A}$ has $m$ elements, so there are $m+1$ kinds of cutting ($i=0\sim m$).

因为A有m个元素，因此这里有m+1种切法(i=0~m)

And we know:

并且我们知道:

With the same way, cut $\text{B}$ into two parts at a random position $j$:

用同样的方法，在任意位置j上将B切成两部分:

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

Put $\text{left\_A}$ and $\text{left\_B}$ into one set, and put $\text{right\_A}$ and $\text{right\_B}$ into another set. Let's name them $\text{left\_part }$and $\text{right\_part}$:

将left_A和
\text{left\_B}left_B放入一个集合，并且将$\text{right\_A}$ 和 $\text{right\_B 放入另一个集合。我们把它们分别叫做 text{left\_part}left\_part和 text{right\_part}right\_part:}$

\text{left\_B}left_B放入一个集合，

   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:

如果我们可以确定:

1. len(left_part)=len(right_part)
2. $\max (\text{left\_part})\le \min (\text{right\_part})$

then we divide all elements in $\{\text{A},\text{B}\}$ into two parts with equal length, and one part is always greater than the other. Then

然后我们就可以将{A,B}中所有的元素分成有相同长度的两部分，并且其中的一部分总是比另一部分更大。然后

​​

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

为了确保上面的两个条件，我们只需要确保：

1. $i+j=m-i+n-j$ (or: $m-i+n-j+1$)
   if $n\ge m$, we just need to set:  i=0∼m, j=m+n+12−i i=0∼m, j=m+n+12−i

2. $\text{B}[j-1]\le \text{A}[i]$ and $\text{A}[i-1]\le \text{B}[j]$

ps.1 For simplicity, I presume $\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]$ are always valid even if $i=0$, $i=m$, $j=0$, or $j=n$. 

ps.1 简单来说，我假设 $\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]$  即使在$i=0$, $i=m$, $j=0$, or $j=n\; 都是有效的。$

I will talk about how to deal with these edge values at last.

我将会在最后讨论怎么样去处理这些边界值：

ps.2 Why $n\ge m$? Because I have to make sure $j$ is non-negative since $0\le i\le m$ and j = \frac{m + n + 1}{2} - ij=​2​​m+n+1​​−i.

为什么 $n\ge m$? 因为我必须确保 jj是一个非负数 当0≤i≤m 和 j = \frac{m + n + 1}{2} - ij=(m+n+1)/2-i

 If $n<m$, then $j$ may be negative, that will lead to wrong result.

如果 $n<m$, 那么j有可能是负数，那样会导致错误的结果。

So, all we need to do is:

所以，我们所有需要做的事情是：

Searching $i$ in $[0,m]$, to find an object $i$ such that:

找到$[0,m]中i的值，用它找到对象i，就像：$

$\text{B}[j-1]\le \text{A}[i]$ and $\text{ A}[i-1]\le \text{B}[j],$ where j = \frac{m + n + 1}{2} - ij=​2​​m+n+1​​−i

And we can do a binary search following steps described below

并且我们可以跟随下面描述的步骤写一个二分查找

1. Set $\text{imin}=0$, $\text{imax}=m$, then start searching in $[\text{imin},\text{imax}]$

      1.设置$\text{imin}=0$, \text{imax} = mimax=m，然后开始在
[\text{imin}, \text{imax}][imin,imax]范围中查找

      

       2.设置 i=(imin+imax)/2, j=(m+n+1)/2-i

       

     3.Now we have $\text{len}(\text{left}\_\text{part})=\text{len}(\text{right}\_\text{part})$. And there are only 3 situations that we may encounter:

     3.现在我们有 $\text{len}(\text{left}\_\text{part})=\text{len}(\text{right}\_\text{part})$. 并且这里我们也许会碰到的情形仅仅只有3种：

 

 一.   $\text{B}[j-1]\le \text{A}[i]$ and $\text{A}[i-1]\le \text{B}[j]$ 

      Means we have found the object $i$, so stop searching.

     意味着我们找到了对象 i, 可以停止查找。

二.   $\text{B}[j-1]>\text{A}[i]$ 
Means $\text{A}[i]$ is too small. We must adjust $i$ to get $\text{B}[j-1]\le \text{A}[i]$.

意味着 $\text{A}[i]$ 太小，我们需要调整 i 去得到 $\text{B}[j-1]\le \text{A}[i]$.

Can we increase $i$?

我们能增加 i 吗？

 Yes. Because when $i$ is increased, $j$ will be decreased.

是的。因为当 i 增加时， j 将会减少。

So $\text{B}[j-1]$ is decreased and $\text{A}[i]$ is increased, and $\text{B}[j-1]\le \text{A}[i]$ maybe satisfied.

因此$\text{B}[j-1]$ 减小 并且 $\text{A}[i]$ 增大， 这样$\text{B}[j-1]\le \text{A}[i]\; 也许会满足。$

Can we decrease ii?

$我们能够减小\; i\; 吗？$

 No! Because when ii is decreased, jj will be increased.、

$不行！因为\; 当\; i\; 减小时，\; j\; 将会被增加。$

  So \text{B}[j-1]B[j−1] is increased and \text{A}[i]A[i] is decreased, and \text{B}[j-1] \leq \text{A}[i]B[j−1]≤A[i] will  be never satisfied.

因此 B[j-1] 将会增加 并且 A[i] 将会减少， 并且 B[j−1]≤A[i] 将永远不会满足。

So we must increase i. That is, we must adjust the searching range to  [i+1, \text{imax}][i+1,imax]..
因此我们必须增加 i .那就是，我们必须调整搜索的范围到 [i+1, \text{imax}][i+1,imax].

So, set \text{imin} = i+1imin=i+1, and goto 2.

所以设置 imin= i+1, 然后跳转到第二步。

三. A[i−1]>B[j]:

Means A[i−1] is too big. And we must decrease i to get i-1   ,  A[i−1]≤B[j].

$意味着\; A[i-1]\; 太大了。并且我们必须减小\; i使它\; i-1\; ,\; A[i-1]\le B[j]$

$That\; is,\; we\; must\; adjust\; the\; searching\; range\; to\; [imin,i-1]$

$那就是，我们必须调整寻找范围到[imin,i-1]$

$So,\; set\; imax=i-1,\; and\; goto\; 2.$

$因此，设置\; imax=i-1,\; 并且转到第二步$

$When\; the\; object\; i\; is\; found,\; the\; median\; is:$

$当对象\; i\; 找到时，中位数是：$

 $i=0,i=m,j=0,j=n$ where $\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]$ may not exist. 

 i=0,i=m,j=0,j=ni=0,i=m,j=0,j=n的时候A[i−1],B[j−1],A[i],B[j] 也许不会存在。

Actually this situation is easier than you think.

事实上这个情形比你想象得要简单。

What we need to do is ensuring that $\text{max}(\text{left}\_\text{part})\le \text{min}(\text{right}\_\text{part})$. So, if $i$ and $j$ are not edges values (means 

我们需要做的就是确保 $\text{max}(\text{left}\_\text{part})\le \text{min}(\text{right}\_\text{part})$. 因此，如果i 和j 不是边缘值(意味着

$\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]$ all exist), then we must check both $\text{B}[j-1]\le \text{A}[i]$ and 

$\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]$ 都存在，我们必须检查 $\text{B}[j-1]\le \text{A}[i]和$

$\text{A}[i-1]\le \text{B}[j].\; But\; if\; some\; of\backslash text\{A\}[i-1],\backslash text\{B\}[j-1],\backslash text\{A\}[i],\backslash text\{B\}[j]\text{A}[i-1],\text{B}[j-1],\text{A}[i],\text{B}[j]don\text{'}t\; exist,$

A[i−1]≤B[j]。但是如果其中一些A[i−1],B[j−1],A[i],B[j] 不存在的话

$then\; we\; don\text{'}t\; need\; to\; check\; one\; (or\; both)\; of\; these\; two\; conditions.$

$然后我们不需要检查两个条件中的其中一个或者全部。$

$For\; example,\; ifi=0i=0,\; then\backslash text\{A\}[i-1]\text{A}[i-1]doesn\text{'}t\; exist,\; then\; we\; don\text{'}t\; need\; to\; check\backslash text\{A\}[i-1]\; \backslash leq\; \backslash text\{B\}[j]\text{A}[i-1]\le \text{B}[j].\; So,\; what\; we\; need\; to\; do\; is:$

举个例子，如果 i=0, 然后  \text{A}[i-1]A[i−1]  不存在，我们就不需要检测A[i−1]≤B[j]。 因此，我们需要做的事情就是:

Searching $i$ in $[0,m]$, to find an object $i$ such that:

找到在 [0,m] 中的 i ,找到对象 i 就像:

$(j=0$ or $i=m$ or $\text{B}[j-1]\le \text{A}[i])$ and

$(j=0$ or $i=m$ or $\text{B}[j-1]\le \text{A}[i])并且$

$(i=0$ or $j=n$ or $\text{A}[i-1]\le \text{B}[j]),$ where j = \frac{m + n + 1}{2} - ij=​2​​m+n+1​​−i

$(i=0$ or $j=n$ or $\text{A}[i-1]\le \text{B}[j])，这里$$j=(m+n+1)/$2-i

Thanks to @Quentin.chen for pointing out that: i<m⟹j>0and i>0⟹j<n. Because:

多谢@Quentin.chen 指出：i<m⟹j>0 和 i>0⟹j<n    因为：

i>0⟹j<ni>0⟹j<ni>0⟹j<n

（我感觉最后一个不等式应该是<=n+（1/2）,不过也不影响）

java 代码

class Solution {    public double findMedianSortedArrays(int[] A, int[] B) {        int m = A.length;        int n = B.length;        if (m > n) { // to ensure m<=n            int[] temp = A; A = B; B = temp;            int tmp = m; m = n; n = tmp;        }        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;        while (iMin <= iMax) {            int i = (iMin + iMax) / 2;            int j = halfLen - i;            if (i < iMax && B[j-1] > A[i]){                iMin = iMin + 1; // i is too small            }            else if (i > iMin && A[i-1] > B[j]) {                iMax = iMax - 1; // i is too big            }            else { // i is perfect                int maxLeft = 0;                if (i == 0) { maxLeft = B[j-1]; }                else if (j == 0) { maxLeft = A[i-1]; }                else { maxLeft = Math.max(A[i-1], B[j-1]); }                if ( (m + n) % 2 == 1 ) { return maxLeft; }                int minRight = 0;                if (i == m) { minRight = B[j]; }                else if (j == n) { minRight = A[i]; }                else { minRight = Math.min(B[j], A[i]); }                return (maxLeft + minRight) / 2.0;            }        }        return 0.0;    }}

python代码

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

i>0⟹j<n