Leetcode: Median of Two Sorted Arrays

Median of Two Sorted Arrays

 

There are two sorted arrays A and B 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)).

Idea: The trivial solution is to consolidate these two sorted arrays and then find the median, then the time complexity is O(m+n) and we also need the extra space. Is there any way to utilize the properties of these two arrays: "sorted"? Tje answer is YES. A helper function will be created: findKthFromTwoSortedArrays(int A[], int starta, int enda, int B[], int startb, int endb, int k) ---- to find the k-th element from two sorted arrays. 

class Solution {public:    double findMedianSortedArrays(int A[], int m, int B[], int n) {        int mid = m+n;        if( mid%2 == 0) {            return (findKthFromTwoSortedArrays(A, 0, m, B, 0, n, mid/2) +                     findKthFromTwoSortedArrays(A, 0, m, B, 0, n, mid/2+1))/2.0;        }        else {            return findKthFromTwoSortedArrays(A, 0, m, B, 0, n, mid/2+1);        }    }        double findKthFromTwoSortedArrays(int A[], int starta, int enda,                                       int B[], int startb, int endb, int k) {        if(starta == enda) {            return B[startb+k-1];        }                if(startb == endb) {            return A[starta+k-1];        }        if(k == 1) return std::min(A[starta], B[startb]);                int mid = k/2;        int min_len = std::min(enda-starta, endb-startb);        mid = std::min(mid, min_len);        if(A[starta+mid-1] > B[startb+mid-1]) {            return findKthFromTwoSortedArrays(A, starta, enda,                                               B, startb+mid, endb, k-mid);        }        else {            return findKthFromTwoSortedArrays(A, starta+mid, enda,                                               B, startb, endb, k-mid);        }    }};