WAP to Find Median of Two Sorted Array in O(logn) Time

There are 2 sorted arrays A and B of size n each. Write an algorithm to find the median of the array obtained after merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n))


Median: In probability theory and statistics, a median is described as the number separating the higher half of a sample from the lower half.
The median of a finite list of numbers can be found by arranging all the numbers from lowest value to highest value and picking the middle one.

For getting the median of input array { 12, 11, 15, 10, 20 }, first sort the array. We get { 10, 11, 12, 15, 20 } after sorting. Median is the middle element of the sorted array which is 12.

Algo 1 merge both array & return avg of a[n/2]+a[n/2-1]/2;but merging requires extra space of size O(2n)

Algo 2 Linear Time O(n)

Use merge procedure of merge sort. Keep track of count while comparing elements of two arrays. If count becomes n(For 2n elements), we have reached the median. Take the average of the elements at indexes n-1 and n in the merged array. See the below implementation.

if (i==n) base case when all elements of 1st array is less then 1st element of 2nd Array
then m1=m2;
m2=a[0];//obvious
break

if(j==n)
base case when all elements of 2nd array is less then 1st element of 1st Array
then
m1=m2;
m2=a1[0]; //obvious
& break



#include

/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian(int ar1[], int ar2[], int n)
{
int i = 0; /* Current index of i/p array ar1[] */
int j = 0; /* Current index of i/p array ar2[] */
int count;
int m1 = -1, m2 = -1;

/* Since there are 2n elements, median will be average
of elements at index n-1 and n in the array obtained after
merging ar1 and ar2 */
for(count = 0; count <= n; count++)
{
/*Below is to handle case where all elements of ar1[] are
smaller than smallest(or first) element of ar2[]*/
if(i == n)
{
m1 = m2;
m2 = ar2[0];
break;
}

/*Below is to handle case where all elements of ar2[] are
smaller than smallest(or first) element of ar1[]*/
else if(j == n)
{
m1 = m2;
m2 = ar1[0];
break;
}

if(ar1[i] < ar2[j])
{
m1 = m2; /* Store the prev median */
m2 = ar1[i];
i++;
}
else
{
m1 = m2; /* Store the prev median */
m2 = ar2[j];
j++;
}
}

return (m1 + m2)/2;
}

/* Driver program to test above function */
int main()
{
int ar1[] = {1, 12, 15, 26, 38};
int ar2[] = {2, 13, 17, 30, 45};

printf("%d", getMedian(ar1, ar2, 5)) ;

getchar();
return 0;
}

Time Complexity O(n)
Space Complexity O(1)
Run Here https://ideone.com/er0L7

3rd Algo O(logn)

This method works by first getting medians of the two sorted arrays and then comparing them.

Let ar1 and ar2 be the input arrays.

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




Complete Reference From http://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-046JFall-2005/30C68118-E436-4FE3-8C79-6BAFBB07D935/0/ps9sol.pdf & geeksForgeeks