**Objective**: Given two sorted arrays of size n. Write an algorithm to find the median of combined array (merger of both the given arrays, size = 2n).

**What is Median?**

The **median** is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. For a data set, it may be thought of as the “middle” value. For example, in the data set {1, 3, 3, 6, 7, 8, 9}, the median is 6, the fourth largest, and also the fourth smallest, number in the sample. For a continuous probability distribution, the median is the value such that a number is equally likely to fall above or below it.

If *n* is odd then Median (*M*) = value of ((*n* + 1)/2)th item term.

If *n* is even then Median (*M*) = value of [(*n*/2)th item term + (*n*/2 + 1)th item term]/2

**Reference: **https://en.wikipedia.org/wiki/Median

**Example**

A = 2, 6, 9
B = 1, 5, 7
Median = (5+6)/2 = 5.5
Explanation: combined array: 1, 2, 5, 6, 7, 9

**Approach**:

**Naïve**: Create third array, which will be the merger of both the arrays based upon the formula we have mentioned above calculate the median.

Time Complexity: O(n)

**Binary Approach**: Compare the medians of both arrays

- Say arrays are array1[] and array2[].
- Calculate the median of both the arrays, say m1 and m2 for array1[] and array2[].
- If m1 == m2 then return m1 or m2 as final result.
- If m1 > m2 then median will be present in either of the sub arrays.
- Array1[] – from first element to m1.
- Array2[] – from m2 to last element.

- If m2 > m1 then median will be present in either of the sub arrays.
- Array1[] – from m1 to last element.
- Array2[] – from first element to m2.

- Repeat the steps from 1 to 5 recursively until 2 elements are left in both the arrays.
- Then apply the formula to get the median
- Median = (max(array1[0],array2[0]) + min(array1[1],array2[1]))/2

Let take an example and walk through-

**int [] a = {2,6,9,10,11};
int [] b = {1,5,7,12,15};**
m1 = 9 , m2 = 7
We have m1 > m2
Array1[] - from first element to m1, Array2[] – from m2 to last element.
**int [] a = {2,6,9};
int [] b = {7,12,15};**
We have m1 < m2
Array1[] - from m1 to last element, Array2[] – from first element to m2.
**int [] a = {6,9};
int [] b = {7,12};**
Now we have 2 elements left in both the arrays so now apply the formula
**Median** = (max(array[0],array[0]) + min(array[1],array[1]))/2
=(max(6,7) + min(9,12))/2
= (7+9)/2
=8

Time Complexity: O(logn)

**Code:**

**Output:**

Median of combined sorted array is: 8.0

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.

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