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//To find median, we could do the following

//1) Calculate the medians m1 and m2 of the input arrays ar1[] and ar2[] respectively.

//2) If m1 and m2 both are equal then we are done, and return m1 (or m2)

//3) If m1 is greater than m2, then median is present in one of the below two subarrays.

// a) From first element of ar1 to m1 (ar1[0...|_n/2_|])

// b) From m2 to last element of ar2 (ar2[|_n/2_|...n-1])

//4) If m2 is greater than m1, then median is present in one of the below two subarrays.

// /a) From m1 to last element of ar1 (ar1[|_n/2_|...n-1])

// b) From first element of ar2 to m2 (ar2[0...|_n/2_|])

//5) Repeat the above process until size of both the subarrays becomes 2.

//6) If size of the two arrays is 2 then use below formula to get the median.

// Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2

//Run time complexity: The overall run time complexity should be O(log (m+n)).

public static double findMedianSortedArrays(int A[], int B[]) {

int m = A.length;

int n = B.length;

if ((m + n) % 2 != 0) // odd

return (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);

else { // even

return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1)

+ findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;

}

public static int findKth(int A[], int B[], int k, int aStart, int aEnd, int bStart, int bEnd) {

int aLen = aEnd - aStart + 1;

int bLen = bEnd - bStart + 1;

// Handle special cases

if (aLen == 0) { //If A is empty, just find k by index of B

return B[bStart + k];

}

if (bLen == 0) { //If B is empty, just find k by index of A

return A[aStart + k];

}

if (k == 0) { //If k is first element, then find the smallest between 2 array

return A[aStart] < B[bStart] ? A[aStart] : B[bStart];

}

int aMid = aLen * k / (aLen + bLen); // a's middle count

int bMid = k - aMid - 1; // b's middle count

// make aMid and bMid to be array index

aMid = aMid + aStart;

bMid = bMid + bStart;

if (A[aMid] > B[bMid]) { //aEnd, bStart

k = k - (bMid - bStart + 1);

aEnd = aMid;

bStart = bMid + 1;

} else { //(A[mid] <= B[mid]) => bEnd, aStart

k = k - (aMid - aStart + 1);

bEnd = bMid;

aStart = aMid + 1;

}

return findKth(A, B, k, aStart, aEnd, bStart, bEnd);

}