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01/4.py

+'''
+	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
+'''
+
+class Solution(object):
+    def findMedianSortedArrays(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: float
+        """
+        
+        if len(nums1) > len(nums2):
+        	nums1, nums2 = nums2, nums1
+
+        x, y = len(nums1), len(nums2)
+        low , high = 0, x
+
+        while  low <= high:
+        	partitionx = (low+high)/2
+        	partitiony = (x+y+1)/2 - partitionx
+        	if partitionx == 0:
+        		maxLeftX = float('-inf')
+        	else:
+	        	maxLeftX = nums1[partitionx-1]
+
+	        if partitionx == x:
+	        	minRightX = float('inf')
+	        else:
+	        	minRightX = nums1[partitionx]
+
+	        if partitiony == 0:
+	        	maxLeftY = float('-inf')
+	        else:
+	        	maxLeftY = nums2[partitiony-1]
+
+	        if partitiony == y:
+	        	minRightY = float('inf')
+	        else:
+	        	minRightY = nums2[partitiony]
+
+	        if maxLeftX <= minRightY and maxLeftY <= minRightX:
+	        	if((x+y)%2 == 0):
+	        		return (max(maxLeftX, maxLeftY) + min(minRightX, minRightY))/2.0
+	        	else:
+	        		return max(maxLeftY, maxLeftX)
+	        elif maxLeftX > minRightY:
+	        	high = partitionx - 1
+	        else:
+	        	low = partitionx + 1
+
+
+print Solution().findMedianSortedArrays([1,2], [3, 4])
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