Median of Two Sorted Arrays

Binary Search; Hard;

1. Link

在两个长度相等的排序数组中找到上中位数_牛客题霸_牛客网

2. 描述: NC36 在两个长度相等的排序数组中找到上中位数

给定两个有序数组arr1和arr2，已知两个数组的长度都为N，求两个数组中所有数的上中位数。上中位数：假设递增序列长度为n，若n为奇数，则上中位数为第n/2+1个数；否则为第n/2个数[要求]时间复杂度为O(logN)O(logN)，额外空间复杂度为O(1)O(1)

示例1

输入：

[1,2,3,4],[3,4,5,6]

复制返回值：

3

复制说明：

总共有8个数，上中位数是第4小的数，所以返回3。 

示例2

输入：

[0,1,2],[3,4,5]

复制返回值：

2

复制说明：

总共有6个数，那么上中位数是第3小的数，所以返回2 

3. 思路

1. method 1: merge and sort

   1. First merge and sort two array

   2. find the middle element as median of two array

   3. Time: O(n), Space: O(n)

2. method 2: binary search

   1. Define left, right pointers of arr1 = l1, r1 and left, right pointers of arr2 = l2, r2

   2. Since two sorted arrays have same length, so first find the middle elements of two arrays (two medians of two array)

   3. if arr1[mid1] == arr2[mid2] -> two arrays have the same median and this is the global median

   4. if arr1[mid1] > arr2[mid2], then arr1[mid1] is on the right hand side of arr2[mid2] in the global array. That is, arr1[mid1] >= global median >= arr2[mid2]. So we need to search the left space of arr1[mid1] and the right space of arr2[mid2]. So let l1 = mid1, r2 = mid2

   5. if arr1[mid1] < arr2[mid2], similary, arr1[mid1] <= global median <= arr2[mid2]. So let r1 = mid1, l2 = mid2.

   6. repeat step 1~5 while l1 < r1

   7. Note:

      1. when l1=r1-1, compute middle between l1, r1 could lead to dead loop, since l1 = mid1 always. When subarray arr1[l1:r1+1] has even size, we need l1 = mid1 + 1 and l2 = mid1 + 1

   8. Time: O(logn), Space: O(1)

4. Coding

method 1:

class Solution:
    def findMedianinTwoSortedAray(self , arr1 , arr2 ):
        # write code here
        # method 1:  merge and sort to find median
        # Time: O(n ), Sppace: O(n)
        #
        # method 2: binarysearch
        #1. for each array use left, right pointers
        # 2. find median of two arrays
        # 3. compare two median
        #    1. if left median == right median: overall median = left median = right median
        #        since number of element < left median = number of element < right median
        #    2. if left median > right median
        #        it can be left median >= overal median >= right median -> search the 
        #        left space of left median and the right space of right median
        #    3. if left median < right median: reverse case 2
        #    4. we can also see num of element on the left of left median and the left of right median
        #        = num of element on the right of left median and the right of right median
        
        arr = [None]*(len(arr1) + len(arr2))
        pt1, pt2 = 0, 0
        cur = 0
        repeat = 0
        while pt1 < len(arr1) and pt2 < len(arr2):
            if arr1[pt1] < arr2[pt2]:
                arr[cur] = arr1[pt1]
                pt1 += 1
            elif arr1[pt1] >= arr2[pt2]:
                arr[cur] = arr2[pt2]
                pt2 += 1
            cur += 1
            
        while  pt1 < len(arr1):
            arr[cur] = arr1[pt1]
            pt1 += 1
            cur += 1
        while  pt2 < len(arr2):
            arr[cur] = arr2[pt2]
            pt2 += 1
            cur += 1
            
        return arr[(cur-1)//2]

method 2:

class Solution:
    def findMedianinTwoSortedAray(self , arr1 , arr2 ):
        if not arr1 or not arr2:
            return None
        if len(arr1) == 1:
            return min(arr1[0], arr2[0])
        
        n = len(arr1)
        l1, r1 = 0, n-1
        l2, r2 = 0, n-1
        while l1 < r1:
            m1 = l1 + ((r1 - l1)>>1)
            m2 = l2 + ((r2 - l2)>>1)
            flag = (r1 - l1 + 1)&1
            if arr1[m1] == arr2[m2]:
                return arr1[m1]
            elif arr1[m1] > arr2[m2]:
                #search left space of m1 and right space of m2
                r1 = m1 
                l2 = m2 if flag else m2 + 1
            else:
                l1 = m1 if flag else m1 +1
                r2 = m2 
        # return the smallest element as median
        return min(arr1[l1], arr2[l2])