# Sharing my 108ms Python Solution

• ``````class Solution(object):
def findMedianSortedArrays(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: float
"""
def median(a):
if len(a) % 2 != 0:
return a[len(a)//2]
else:
return ((a[len(a)/2]+a[(len(a)/2)-1])/2.)
temp = nums1+nums2
temp.sort()
return (median(temp))``````

• I also used this way, but the complexity of sort() is O(nlog(n))
There is space to improve.
I guess bi-search is the direction.

• How can I improve on the sort step?

• I don't know either :) There should be others who already posted their O(n) solution.

• Shocked...but obviously this code is O(nlogn), as @dewei said, why it can pass all of the test cases?

• Well, by the definition of a median of a sorted array, even numbered arrays wok have a median of the middle 2 elements, and odd numbered arrays will have a median of the middle element. So by manipulating the indicies of a sorted array, we can ignore all other elements and reach the median quickly. My only downfal is the sort function, which I Appreciate Any help with

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