median of two sorted arrays

Question

My question is with reference to Method 2 of this link. Here two equal length sorted arrays are given and we have to find the median of the two arrays merged.

Algorithm:

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.
     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

Example:

   ar1[] = {1, 12, 15, 26, 38}
   ar2[] = {2, 13, 17, 30, 45}

For above two arrays m1 = 15 and m2 = 17

For the above ar1[] and ar2[], m1 is smaller than m2. So median is present in one of the following two subarrays.

   [15, 26, 38] and [2, 13, 17]
Let us repeat the process for above two subarrays:

    m1 = 26 m2 = 13.
m1 is greater than m2. So the subarrays become

  [15, 26] and [13, 17]
Now size is 2, so median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
                       = (max(15, 13) + min(26, 17))/2 
                       = (15 + 17)/2
                       = 16

I understand how they exclude halves of the arrays and say that the median element would be in particular halves of the arrays i.e. steps 1, 2, 3, 4, 5.

But what I can't fathom, how can they say that the median of the merged arrays would be the median of the merged arrays resulting after pruning the halves of the arrays i.e. the median of merge array of {1, 12, 15, 26, 38} and {2, 13, 17, 30, 45} would be the median of the merge array of {2,13,17} and {15, 26, 38}.

Please explain. Thanks in advance.

Answer 1

For variable length you just have to check the special cases when either of the array is having only 1 element in each level of recursion . If one of them such , don't divide further just pass it as it is till the other one also becomes of length 2. While giving final answer handle the case when one of them is having only 1 element.

    //Median of two sorted arrays
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
    public static void main (String[] args) throws java.lang.Exception {
        int[] A = {1, 3, 11};
        int[] B = {2, 4, 12, 14, 15};
        System.out.println("Ans. "+findMedian(A, B));
        //System.out.println(median(A));
    }

    private static int findMedian(int[] A, int[] B) {
        System.out.println(Arrays.toString(A)+" --- "+ Arrays.toString(B));
        int sA = A.length;
        int sB = B.length;

        if(sA <= 2 && sB <= 2) {
            if(sA <= 1 && sA <= 1) {
                return (A[0]+B[0])/2; 
            } else if(sA <= 1) {
                return (max(A[0], B[0]) + min(A[0], B[1])) / 2;
            } else if(sB <= 1) {
                return (max(A[0], B[0]) + min(A[1], B[0]) ) / 2;
            } else {
                System.out.println("xxx");
                return (max(A[0], B[0]) + min(A[1],B[1])) / 2;
            }
        }

        int mA = median(A);
        int mB = median(B);

        if(mA == mB) {
            return mA;
        } else if(mA < mB) {
            if(sA <= 2) {
                return findMedian(A, Arrays.copyOfRange(B, 0, sB/2+1));     
            } else if(sB <= 2) {
                return findMedian(Arrays.copyOfRange(A, sA/2, sA), B); 
            } else {
                return findMedian(Arrays.copyOfRange(A, sA/2, sA)
                          ,Arrays.copyOfRange(B, 0, sB/2+1)); 
            }
        } else {
            if(sA <= 2) {
                return findMedian(A, Arrays.copyOfRange(B, sB/2, sB));  
            } else if(sB <= 2) {
                return findMedian(Arrays.copyOfRange(A, 0, sA/2+1),B); 
            } else {
                return findMedian(Arrays.copyOfRange(A, 0, sA/2+1)
                          ,Arrays.copyOfRange(B, sB/2, sB)); 
            }
        }
    }

    private static int median(int[] A) {
        int size = A.length;
        if(size == 0 ){
            return 0;
        } else if(size == 1) {
            return A[0];
        }

        if(size%2 == 0 ) {
            return (A[size/2 -1 ] + A[size/2  ])/2;
        }else {
            return A[size/2];
        }
    }

    private static int max(int a, int b) {
        return a > b ? a : b;
    }

    private static int min(int a, int b) {
        return a < b ? a : b;
    }
}