Median of two sorted array

Data StructureDivide and Conquer AlgorithmsAlgorithms

Medians are the middle numbers, in other words, the median value is the middle observation in an ordered list. It corresponds to the cumulative percentage of 50%.

The size of two arrays must be same, we will find the median of two separate arrays at first, then compare the separate medians to get an actual median of two lists.

Input and Output

Input:
Two sorted array are given.
Array 1: {1, 2, 3, 6, 7}
Array 2: {4, 6, 8, 10, 11}
Output:
The median from two array. Here the median value is 6.
Merge the given lists into one. {1, 2, 3, 4, 6, 6, 7, 8, 10, 11}
From the merged list find the average of two middle elements. here (6+6)/2 = 6.

Algorithm

median(list, n)

Input: List of data, and the number of data.

Output: Median of the given list.

Begin
   if the list has even number of data, then
      return (list[n/2] + list[n/2-1])/2
   else
      return list[n/2]
End

findMedian(list1, list2, n)

Input − Two sorted lists, and the number of lists.

Output − The median from two sorted lists.

Begin
   if n <= 0, then
      it is invalid, and return invalid number
   if n = 1, then
      return (list1[0] + list2[0])/2
   if n = 2, then
      return ((max of list1[0], list2[0]) + (min of list1[1], list2[1]))/2
   med1 := median(list1, n)
   med2 := median(list2, n)

   if med1 = med2, then
      return med1
   if med1 < med2, then
      if item has even number of data, then
         subList := data from list2, from 0 to n/2 – 1 data
         return findMedian(subList, list1, n – (n/2) + 1)
      subList := data from list2, from 0 to n/2 data
      return findMedian(subList, list2, n – (n/2))
End

Example

#include<iostream>
using namespace std;

int median(int list[], int n) {
   if (n%2 == 0)     //when array containts even number of data
      return (list[n/2] + list[n/2-1])/2;
   else        //for odd number of data
      return list[n/2];
}

intfindMedian(int list1[], int list2[], int n) {
   if (n <= 0)
      return -1;      //invalid length of lists
   if (n == 1)
      return (list1[0] + list2[0])/2;    //for single element simply get average from two array
   if (n == 2)
      return (max(list1[0], list2[0]) + min(list1[1], list2[1])) / 2;

   int med1 = median(list1, n);     //Find median from first array
   int med2 = median(list2, n);     //Find median from second array

   if (med1 == med2)    //when both medians are same, they are the final median
       return med1;
   if (med1 < med2) {
       if (n % 2 == 0)
          return findMedian(list1 + n/2 - 1, list2, n - n/2 +1);
       return findMedian(list1 + n/2, list2, n - n/2);
   }

   if (n % 2 == 0)    //when med1 > med2
      return findMedian(list2 + n/2 - 1, list1, n - n/2 + 1);
   return findMedian(list2 + n/2, list1, n - n/2);
}

int main() {
   int list1[] = {1, 2, 3, 6, 7};
   int list2[] = {4, 6, 8, 10, 11};

   int n1 = 5;
   int n2 = 5;

   if (n1 == n2)
      cout<< "Median is "<<findMedian(list1, list2, n1);
   else
      cout<< "Doesn't work for lists of unequal size";
}

Output

Median is 6
raja
Published on 10-Jul-2018 07:34:59
Advertisements