Median of Two Sorted Arrays

This solution O(lg(m + n)) is not good for interview. Try kth smallest element solution extension.

public class MedianTwoSortedArrays {
	private double findMedian(int A[], int B[], int left, int right) {
		int m = A.length, n = B.length, mid = (m + n) / 2;
		if (left > right) {
			return findMedian(B, A, Math.max(0, mid - m), Math.min(n - 1, mid));
		}

		int i = (left + right) / 2, j = mid - i - 1;
		if (j >= 0 && A[i] < B[j]) // A[i] < median < B[j]
			return findMedian(A, B, i + 1, right);
		if (j < n - 1 && A[i] > B[j + 1]) // A[i] > median > B[j + 1]
			return findMedian(A, B, left, i - 1);
		// found median
		// m+n is odd
		if (((m + n) & 0x1) > 0 || (i <= 0 && (j < 0 || j >= n)))
			return A[i];
		// m+n is even
		if (j < 0 || j >= n)
			return (A[i] + A[i - 1]) / 2.0;
		if (i <= 0)
			return (A[i] + B[j]) / 2.0;
		return (A[i] + Math.max(B[j], A[i - 1])) / 2.0;
	}

	public double findMedianSortedArrays(int A[], int B[]) {
		// Start typing your Java solution below
		// DO NOT write main() function
		int m = A.length, n = B.length;
		if (m < n)// median in A
			return findMedian(A, B, 0, m - 1);
		else// median in B
			return findMedian(B, A, 0, n - 1);
	}
}

O(n), through kth smallest function.

 

public class Solution {
    public double findMedianSortedArrays(int A[], int B[]){
		// Start typing your Java solution below
		// DO NOT write main() function
		int m = A.length, n = B.length;
		if ((m + n) % 2 == 0)// median in A
			return (findKthSmallest(A, m , B, n, (m + n) / 2 + 1) + findKthSmallest(A, m , B, n, (m + n) / 2)) / 2.0;
		else// median in B
			return findKthSmallest(A, m , B, n, (m + n) / 2 + 1);
	}
	
	public double findKthSmallest(int A[], int m, int B[], int n, int k){
		int ia = 0, ib = 0, res = 0;
		while(--k >= 0 && ia <m && ib < n){
			if(A[ia] < B[ib])
				res = A[ia++];
			else
				res = B[ib++];
		}
		while(ia < m && k-- >=0)
			res = A[ia++];
		while(ib < n && k-- >=0)
			res = B[ib++];
		return res;
	}
}