The kth largest element from two sorted arrays

January 5, 2017

Introduction
Spent over 2 hours to study Leetcode 4 and Leetcode 215: the medium of two sorted array, and then Leetcode 215: the kth largest element from the array, and then, Julia spent over one hour to work on the code for the algorithm: the kth largest element from two sorted array.

Workout

Here is the C# practice: binary search, time complexity: not O(lg n + lg m), smaller one: O(lg(n+m)), n, m are the length of two sorted arrays.

Julia likes to work on the test case a little more time, 20 - 30 minutes, and then post the algorithm on stackexchange.com for a code review.

Read the article again, learn more about the analysis.

January 9, 2017
The most important feedback about the code - code review:
Just so you know, your solution appears to be O(logk∗(n+m))$O(\log k\ast (n+m))$. The reason is that ArraySplice() makes a copy of the array, which takes either O(n)$O(n)$ or O(m)$O(m)$ time. If you would just avoid doing the copy and instead pass a starting index for each array to your function, you would be down to O(logk)$O(\log k)$ time. – JS1

Actionable Items:

1. Read a few articles about the algorithm, list here:

2. Write a new version of algorithm to solve all issues, post it on code review. Learn by doing.

3. C# practice with the correction of time complexity issue - array splice. Code is here.

Leetcode 4: Median of two sorted array - a warmup practice

May 15, 2015

Review the leetcode 4: Median of two sorted array, warm up the algorithm.

Here is the last practice:

https://github.com/jianminchen/Leetcode_C-/blob/master/4MedianOfTwoSortedArrays.cs

Step 1: 
Read the last practice 10 - 20 minutes on May 15, 2016
-- write down the ideas: 
1. Brute force solution: O(nlogn)
2. Linear solution: O(n) 

3. Transform to the kth element problem first, and then: 
Try to get rid of k/2 elements once, so the algorithm will go to logk, k = (m+n)/2 

Some analysis and reasoning: 
Assuming that A and B both arrays are with length > k/2, and then, compare A[k/2-1] and B[k/2-1]. 

Read the blog again:
http://blog.csdn.net/yutianzuijin/article/details/11499917

http://blog.csdn.net/zxzxy1988/article/details/8587244

Step 2: 
Spend 30 minutes to read and then write some code. 

Related to Leetcode 215: Find kth largest element in the array. 


Question and Answer:

1. Write down what you learn through this practice.
Answer: The algorithm is very well defined, is a special case of find kth element in two sorted arrays.

2. Algorithms learning and important thing to learn:

Answer:
1. Always know how to solve the problem using naive solution, brute force one first.

O(nlogn) -> O(n) -> O(log(m+n)/2)

2. And then, discuss the improvement.

Leetcode 4: Median of two sorted arrays

Problem statement: 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)).

May 17, 2015

这道题我想了好几个小时,　读了题解,　没有读明白．直到我读了Leetcode的题解,　总算读懂了题目的分析．让我感叹计算机科学的奇妙．有大学学数学分析的感觉．

下面是题解分析:

这是一道非常经典的题。这题更通用的形式是，给定两个已经排序好的数组，找到两者所有元

素中第k 大的元素。

1. O(m + n) 的解法比较直观，直接merge 两个数组，然后求第k 大的元素。

2. O(k)时间，O(1) 空间; 但是，当k 很接近m + n 的时候，这个方法还是O(m + n) 的。

不过我们仅仅需要第k 大的元素，是不需要“排序”这么复杂的操作的。可以用一个计数器，

记录当前已经找到第m 大的元素了。同时我们使用两个指针pA 和pB，分别指向A 和B 数组的第

一个元素，使用类似于merge sort 的原理，如果数组A 当前元素小，那么pA++，同时m++；如果

数组B 当前元素小，那么pB++，同时m++。最终当m 等于k 的时候，就得到了我们的答案，O(k)

时间，O(1) 空间。但是，当k 很接近m + n 的时候，这个方法还是O(m + n) 的。

3. 更好的方案

有没有更好的方案呢？我们可以考虑从k 入手。如果我们每次都能够删除一个一定在第k 大元

素之前的元素，那么我们需要进行k 次。但是如果每次我们都删除一半呢？由于A 和B 都是有序

的，我们应该充分利用这里面的信息，类似于二分查找，也是充分利用了“有序”。

假设A 和B 的元素个数都大于k/2，我们将A 的第k/2 个元素（即A[k/2-1]）和B 的第k/2

个元素（即B[k/2-1]）进行比较，有以下三种情况（为了简化这里先假设k 为偶数，所得到的结

论对于k 是奇数也是成立的）：

• A[k/2-1] == B[k/2-1]

• A[k/2-1] > B[k/2-1]

• A[k/2-1] < B[k/2-1]

如果A[k/2-1] < B[k/2-1]，意味着A[0] 到A[k/2-1] 的肯定在A [B] 的top k 元素的范围 内，换句话说，A[k/2-1 不可能大于A [ B 的第k 大元素。留给读者证明。 因此，我们可以放心的删除A 数组的这k/2 个元素。同理，当A[k/2-1] > B[k/2-1] 时，可以删除B 数组的k/2 个元素。

当A[k/2-1] == B[k/2-1] 时，说明找到了第k 大的元素，直接返回A[k/2-1] 或B[k/2-1] 即可。

因此，我们可以写一个递归函数。那么函数什么时候应该终止呢？

当A 或B 是空时，直接返回B[k-1] 或A[k-1]；

• 当k=1 是，返回min(A[0], B[0])；

• 当A[k/2-1] == B[k/2-1] 时，返回A[k/2-1] 或B[k/2-1]

－－

Read some blogs:

http://articles.leetcode.com/2011/01/find-k-th-smallest-element-in-union-of.html

http://fisherlei.blogspot.ca/2012/12/leetcode-median-of-two-sorted-arrays.html

Practice code sharing: C# code: 

https://github.com/jianminchen/medianOfTwoSortedArrays/blob/master/Program.cs

2017 January 5, 

Rewrite C# program for each time complexity:

O(m+n)

O(k)

O(log(m + n)

Post 3 code review on stackexchange.com code review.