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)).
Note : Since size of the set for which we are looking for median is even (2n), we need take average of middle two numbers and return floor of the average.
Method 1 (Simply count while Merging)
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 n1 and n in the merged array. See the below implementation.
 C

// A Simple Merge based O(n) solution to find median of
// two sorted arrays
#include <stdio.h>
/* 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 n1 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};
int
n1 =
sizeof
(ar1)/
sizeof
(ar1[0]);
int
n2 =
sizeof
(ar2)/
sizeof
(ar2[0]);
if
(n1 == n2)
printf
(
"Median is %d"
, getMedian(ar1, ar2, n1));
else
printf
(
"Doesn't work for arrays of unequal size"
);
getchar
();
return
0;
}
Output
Median is 16
Time Complexity: O(n)
Method 2 (By comparing the medians of two arrays)
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_...n1]) 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_...n1]) 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
Implementation:

// A divide and conquer based efficient solution to find median
// of two sorted arrays of same size.
#include<bits/stdc++.h>
using
namespace
std;
int
median(
int
[],
int
);
/* to get median of a sorted array */
/* 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)
{
/* return 1 for invalid input */
if
(n <= 0)
return
1;
if
(n == 1)
return
(ar1[0] + ar2[0])/2;
if
(n == 2)
return
(max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;
int
m1 = median(ar1, n);
/* get the median of the first array */
int
m2 = median(ar2, n);
/* get the median of the second array */
/* If medians are equal then return either m1 or m2 */
if
(m1 == m2)
return
m1;
/* if m1 < m2 then median must exist in ar1[m1....] and
ar2[....m2] */
if
(m1 < m2)
{
if
(n % 2 == 0)
return
getMedian(ar1 + n/2  1, ar2, n  n/2 +1);
return
getMedian(ar1 + n/2, ar2, n  n/2);
}
/* if m1 > m2 then median must exist in ar1[....m1] and
ar2[m2...] */
if
(n % 2 == 0)
return
getMedian(ar2 + n/2  1, ar1, n  n/2 + 1);
return
getMedian(ar2 + n/2, ar1, n  n/2);
}
/* Function to get median of a sorted array */
int
median(
int
arr[],
int
n)
{
if
(n%2 == 0)
return
(arr[n/2] + arr[n/21])/2;
else
return
arr[n/2];
}
/* Driver program to test above function */
int
main()
{
int
ar1[] = {1, 2, 3, 6};
int
ar2[] = {4, 6, 8, 10};
int
n1 =
sizeof
(ar1)/
sizeof
(ar1[0]);
int
n2 =
sizeof
(ar2)/
sizeof
(ar2[0]);
if
(n1 == n2)
printf
(
"Median is %d"
, getMedian(ar1, ar2, n1));
else
printf
(
"Doesn't work for arrays of unequal size"
);
return
0;
}
Output :
Median is 5
Time Complexity: O(logn)
Algorithmic Paradigm: Divide and Conquer
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/medianoftwosortedarrays/
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