feat(work): implement binary search solution for median of two sorted arrays

work/1/Hard/Array/4.median-of-two-sorted-arrays.py

@@ -10,7 +10,7 @@ return the median of the two sorted arrays.
 
 The overall run time complexity should be O(log (m+n)).
 
- 
+
 
 Example 1:
 
@@ -24,23 +24,47 @@ Input: nums1 = [1,2], nums2 = [3,4]
 Output: 2.50000
 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
 
- 
+
 
 Constraints:
 
-	• nums1.length == m
+        • nums1.length == m
 
-	• nums2.length == n
+        • nums2.length == n
 
-	• 0 <= m <= 1000
+        • 0 <= m <= 1000
 
-	• 0 <= n <= 1000
+        • 0 <= n <= 1000
 
-	• 1 <= m + n <= 2000
+        • 1 <= m + n <= 2000
 
-	• -10^6 <= nums1[i], nums2[i] <= 10^6
+        • -10^6 <= nums1[i], nums2[i] <= 10^6
 """
 
+
 class Solution:
     def findMedianSortedArrays(self, nums1: list[int], nums2: list[int]) -> float:
-        
+        if len(nums1) > len(nums2):
+            return self.findMedianSortedArrays(nums2, nums1)
+
+        m, n = len(nums1), len(nums2)
+        left, right = 0, m
+
+        while left <= right:
+            partA = (left + right) // 2
+            partB = (m + n + 1) // 2 - partA
+
+            maxLeftA = float("-inf") if partA == 0 else nums1[partA - 1]
+            minRightA = float("inf") if partA == m else nums1[partA]
+            maxLeftB = float("-inf") if partB == 0 else nums2[partB - 1]
+            minRightB = float("inf") if partB == n else nums2[partB]
+
+            if maxLeftA <= minRightB and maxLeftB <= minRightA:
+                if (m + n) % 2 == 0:
+                    return (max(maxLeftA, maxLeftB) + min(minRightA, minRightB)) / 2
+                else:
+                    return max(maxLeftA, maxLeftB)
+            elif maxLeftA > minRightB:
+                right = partA - 1
+            else:
+                left = partA + 1