# Find median of two sorted arrays of same size

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

1. Say arrays are array1[] and array2[].
2. Calculate the median of both the arrays, say m1 and m2 for array1[] and array2[].
3. If m1 == m2 then return m1 or m2 as final result.
4. If m1 > m2 then median will be present in either of the sub arrays.
1. Array1[] – from first element to m1.
2. Array2[] – from m2 to last element.
5. If m2 > m1 then median will be present in either of the sub arrays.
1. Array1[] – from m1 to last element.
2. Array2[] – from first element to m2.
6. Repeat the steps from 1 to 5 recursively until 2 elements are left in both the arrays.
7. Then apply the formula to get the median
1. Median = (max(array1,array2) + min(array1,array2))/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,array) + min(array,array))/2
=(max(6,7) + min(9,12))/2
= (7+9)/2
=8
```

Time Complexity: O(logn)

Code:

 public class MedianBinary { public float find(int [] a, int start_a, int end_a, int [] b, int start_b, int end_b){ if(end_a–start_a+1==2 && end_b–start_b+1==2){ float x = Math.max(a[start_a],b[start_b]); float y = Math.min(a[end_a],b[end_b]); return (x+y)/2; } float median_a = getMedian(a, start_a, end_a); float median_b = getMedian(b, start_b, end_b); int mid_a = (start_a+end_a)/2; int mid_b = (start_b+end_b)/2; if(median_a>median_b){ return find(a,start_a,mid_a,b,mid_b,end_b); }else{ return find(a,mid_a,end_a,b,start_b,mid_b); } } public float getMedian(int [] x, int start, int end){ int size = end–start+1; double median; if(size%2==0){ float m = x[start+(size/2)]; float n = x[start+(size–1)/2]; return (m+n)/2; }else{ return x[start+(size–1)/2]; } } public static void main(String[] args) { MedianBinary m = new MedianBinary(); int [] a = {2,6,9,10,11}; int [] b = {1,5,7,12,15}; float x = m.find(a,0,a.length–1,b,0,b.length–1); System.out.println("Median of combined sorted array is: " + x); } }

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MedianBinary.java
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Output:

```Median of combined sorted array is: 8.0
```