LeetCode 4 Median of Two Sorted Arrays (C++, Python)

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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)).
Solution C++


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class Solution {
private:
    double intFindMedianSortedArrays(vector<int>& nums1, int low1, int high1, vector<int>& nums2, int low2, int high2, int k) {
        if (high1 - low1 > high2 - low2) {
            return intFindMedianSortedArrays(nums2, low2, high2, nums1, low1, high1, k);
        }
        if (low1 > high1) {
            return nums2[low2 + k - 1];
        }
        if (k == 1) {
            return min(nums1[low1], nums2[low2]);    
        }
        int na = min(k / 2, high1 - low1 + 1);
        int nb = k - na;
        if (nums1[low1 + na - 1] > nums2[low2 + nb - 1]) {
            return intFindMedianSortedArrays(nums1, low1, high1, nums2, low2 + nb, high2, k - nb);
        } else if (nums1[low1 + na - 1] < nums2[low2 + nb - 1]) {
            return intFindMedianSortedArrays(nums1, low1 + na, high1, nums2, low2, high2, k - na);
        } else {
            return nums1[low1 + na - 1];
        }
    }
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        const int M = nums1.size();
        const int N = nums2.size();
        int total = M + N;
        if (total % 2) {
            return intFindMedianSortedArrays(nums1, 0, M - 1, nums2, 0, N - 1, total / 2 + 1);
        } else {
            return (intFindMedianSortedArrays(nums1, 0, M - 1, nums2, 0, N - 1, total / 2) +
                intFindMedianSortedArrays(nums1, 0, M - 1, nums2, 0, N - 1, total / 2 + 1)) / 2.0;    
        }
    }
};
Solution Python


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class Solution:
    def findKth(self, nums1, low1, high1, nums2, low2, high2, k):
        if high1 - low1 > high2 - low2:
            return self.findKth(nums2, low2, high2, nums1, low1, high1, k)
        if low1 > high1:
            return nums2[low2 + k - 1]
        if k == 1:
            return min(nums1[low1], nums2[low2])
        pa = min(k / 2, high1 - low1 + 1)
        pb = k - pa
        if nums1[low1 + pa - 1] < nums2[low2 + pb - 1]:
            return self.findKth(nums1, low1 + pa, high1, nums2, low2, high2, k - pa)
        elif nums1[low1 + pa - 1] > nums2[low2 + pb - 1]:
            return self.findKth(nums1, low1, high1, nums2, low2 + pb, high2, k - pb)
        else:
            return nums1[low1 + pa - 1]
    # @param {integer[]} nums1
    # @param {integer[]} nums2
    # @return {float}
    def findMedianSortedArrays(self, nums1, nums2):
        M = len(nums1)
        N = len(nums2)
        if (M + N) % 2:
            return self.findKth(nums1, 0, M - 1, nums2, 0, N - 1, (M + N) / 2 + 1)
        else:
            return (self.findKth(nums1, 0, M - 1, nums2, 0, N - 1, (M + N) / 2) + self.findKth(nums1, 0, M - 1, nums2, 0, N - 1, (M + N) / 2 + 1)) / 2.0

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