Current chat capability is restricted to GPT-3.5 (or lower) rather than GPT-4
sample question:
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
The overall run time complexity should be O(log (m+n)).
Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:
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
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106
sample output from Copilot Chat:
sample output from GPT-3.5:
To find the median of two sorted arrays, we need to merge the two arrays and then find the median of the merged array. However, we can optimize this process to achieve a time complexity of O(log(m+n)).
The idea is to use binary search to find the partition points in both arrays such that the elements to the left of the partition points are smaller than the elements to the right. Once we have found the partition points, we can merge the two arrays and find the median.
Here's the algorithm:
- Initialize two variables,
left and right, to 0 and m (the length of nums1).
- While
left is less than or equal to right:
- Calculate the partition points
partition1 and partition2 as follows:
partition1 = (left + right) // 2
partition2 = (m + n + 1) // 2 - partition1
- Calculate the left and right elements of the partitions as follows:
left1 = nums1[partition1 - 1] if partition1 > 0 else float('-inf')
left2 = nums2[partition2 - 1] if partition2 > 0 else float('-inf')
right1 = nums1[partition1] if partition1 < m else float('inf')
right2 = nums2[partition2] if partition2 < n else float('inf')
- If
left1 is less than or equal to right2 and left2 is less than or equal to right1, we have found the partition points. If the total length of the arrays is odd, the median is the maximum of left1 and left2. If the total length is even, the median is the average of the maximum of left1 and left2 and the minimum of right1 and right2.
- If
left1 is greater than right2, move the right pointer to partition1 - 1.
- If
left2 is greater than right1, move the left pointer to partition1 + 1.
- If we haven't found the partition points, return -1 (this should never happen).
Here's the Python code:
def findMedianSortedArrays(nums1, nums2):
m, n = len(nums1), len(nums2)
if m > n:
nums1, nums2, m, n = nums2, nums1, n, m
left, right = 0, m
while left <= right:
partition1 = (left + right) // 2
partition2 = (m + n + 1) // 2 - partition1
left1 = nums1[partition1 - 1] if partition1 > 0 else float('-inf')
left2 = nums2[partition2 - 1] if partition2 > 0 else float('-inf')
right1 = nums1[partition1] if partition1 < m else float('inf')
right2 = nums2[partition2] if partition2 < n else float('inf')
if left1 <= right2 and left2 <= right1:
if (m + n) % 2 == 0:
return (max(left1, left2) + min(right1, right2)) / 2
else:
return max(left1, left2)
elif left1 > right2:
right = partition1 - 1
else:
left = partition1 + 1
return -1
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