4-median-of-two-sorted-arrays

Try it on leetcode

Description

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

Solution(Python)

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        return self.OptmialbinarySeach(nums1, nums2)

    # Time Complexity: O(m+n)
    # Space Complexity: O(m+n)
    def bruteforce(self, nums1: List[int], nums2: List[int]) -> float:
        n = len(nums1)
        m = len(nums2)

        nums1 = nums1[:] + [0] * m

        i = n - 1
        j = m - 1
        k = m + n - 1

        while i >= 0 and j >= 0:
            if nums1[i] > nums2[j]:
                nums1[k] = nums1[i]
                i -= 1
                k -= 1

            else:
                nums1[k] = nums2[j]
                j -= 1
                k -= 1
        while j >= 0:
            nums1[k] = nums2[j]
            j -= 1
            k -= 1

        l = m + n
        mid = l // 2
        print(nums1)
        if l % 2 == 0:
            return (nums1[mid] + nums1[mid - 1]) / 2
        else:
            return nums1[mid]

    # Time Complexity: O(log(m+n))
    def binarySeach(self, A, B):
        l = len(A) + len(B)
        if l % 2 == 1:
            return self.kth(A, B, l // 2)
        else:
            return (self.kth(A, B, l // 2) + self.kth(A, B, l // 2 - 1)) / 2.0

    def kth(self, a, b, k):
        if not a:
            return b[k]
        if not b:
            return a[k]
        ia, ib = len(a) // 2, len(b) // 2
        ma, mb = a[ia], b[ib]

        # when k is bigger than the sum of a and b's median indices
        if ia + ib < k:
            # if a's median is bigger than b's, b's first half doesn't include k
            if ma > mb:
                return self.kth(a, b[ib + 1:], k - ib - 1)
            else:
                return self.kth(a[ia + 1:], b, k - ia - 1)
        # when k is smaller than the sum of a and b's indices
        else:
            # if a's median is bigger than b's, a's second half doesn't include k
            if ma > mb:
                return self.kth(a[:ia], b, k)
            else:
                return self.kth(a, b[:ib], k)

    # Time Complexity: O(min(log(m),log(n))))
    def OptmialbinarySeach(self, A, B):
        l = len(A) + len(B)
        if l % 2 == 1:
            return self.kth(A, B, l // 2)
        else:
            return (self.kth(A, B, l // 2) + self.kth(A, B, l // 2 - 1)) / 2.0

    def kth(self, a, b, k):
        if not a:
            return b[k]
        if not b:
            return a[k]
        ia, ib = len(a) // 2, len(b) // 2
        ma, mb = a[ia], b[ib]

        # when k is bigger than the sum of a and b's median indices
        if ia + ib < k:
            # if a's median is bigger than b's, b's first half doesn't include k
            if ma > mb:
                return self.kth(a, b[ib + 1:], k - ib - 1)
            else:
                return self.kth(a[ia + 1:], b, k - ia - 1)
        # when k is smaller than the sum of a and b's indices
        else:
            # if a's median is bigger than b's, a's second half doesn't include k
            if ma > mb:
                return self.kth(a[:ia], b, k)
            else:
                return self.kth(a, b[:ib], k)