# Median of two sorted arrays of same size

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)). ## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 n-1 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 ` `using` `namespace` `std; ` ` `  `/* 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; ` `            ``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; ` `            ``break``; ` `        ``} ` `        ``/* equals sign because if two  ` `           ``arrays have some common elements */` `        ``if` `(ar1[i] <= ar2[j]) ` `        ``{ ` `            ``/* Store the prev median */` `            ``m1 = m2;  ` `            ``m2 = ar1[i]; ` `            ``i++; ` `        ``} ` `        ``else` `        ``{ ` `            ``/* Store the prev median */` `            ``m1 = m2;  ` `            ``m2 = ar2[j]; ` `            ``j++; ` `        ``} ` `    ``} ` ` `  `    ``return` `(m1 + m2)/2; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `ar1[] = {1, 12, 15, 26, 38}; ` `    ``int` `ar2[] = {2, 13, 17, 30, 45}; ` ` `  `    ``int` `n1 = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n2 = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``if` `(n1 == n2) ` `        ``cout << ``"Median is "`  `             ``<< getMedian(ar1, ar2, n1) ; ` `    ``else` `        ``cout << ``"Doesn't work for arrays"`  `             ``<< ``" of unequal size"` `; ` `    ``getchar``(); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed  ` `// by Shivi_Aggarwal `

## C

 `// A Simple Merge based O(n) solution to find median of ` `// two sorted arrays ` `#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; ` `            ``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; ` `            ``break``; ` `        ``} ` `         ``/* equals sign because if two  ` `            ``arrays have some common elements */` `        ``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); ` `    ``int` `n2 = ``sizeof``(ar2)/``sizeof``(ar2); ` `    ``if` `(n1 == n2) ` `        ``printf``(``"Median is %d"``, getMedian(ar1, ar2, n1)); ` `    ``else` `        ``printf``(``"Doesn't work for arrays of unequal size"``); ` `    ``getchar``(); ` `    ``return` `0; ` `} `

## Java

 `// A Simple Merge based O(n) solution  ` `// to find median of two sorted arrays ` ` `  `class` `Main ` `{ ` `    ``// function to calculate median ` `    ``static` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n) ` `    ``{    ` `        ``int` `i = ``0``;   ` `        ``int` `j = ``0``;  ` `        ``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``; ` `            ``} ` `            ``/* equals sign because if two  ` `               ``arrays have some common elements */` `            ``if` `(ar1[i] <= ar2[j]) ` `            ``{    ` `                ``/* Store the prev median */` `                ``m1 = m2;   ` `                ``m2 = ar1[i]; ` `                ``i++; ` `            ``} ` `            ``else` `            ``{ ` `                ``/* Store the prev median */` `                ``m1 = m2;   ` `                ``m2 = ar2[j]; ` `                ``j++; ` `            ``} ` `        ``} ` `      `  `        ``return` `(m1 + m2)/``2``; ` `    ``} ` `      `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `ar1[] = {``1``, ``12``, ``15``, ``26``, ``38``}; ` `        ``int` `ar2[] = {``2``, ``13``, ``17``, ``30``, ``45``}; ` `      `  `        ``int` `n1 = ar1.length; ` `        ``int` `n2 = ar2.length; ` `        ``if` `(n1 == n2) ` `            ``System.out.println(``"Median is "` `+ ` `                        ``getMedian(ar1, ar2, n1)); ` `        ``else` `            ``System.out.println(``"arrays are of unequal size"``); ` `    ``}     ` `} `

## Python3

 `# A Simple Merge based O(n) Python 3 solution  ` `# to find median of two sorted lists ` ` `  `# This function returns median of ar1[] and ar2[]. ` `# Assumptions in this function: ` `# Both ar1[] and ar2[] are sorted arrays ` `# Both have n elements ` `def` `getMedian( ar1, ar2 , n): ` `    ``i ``=` `0` `# Current index of i/p list ar1[] ` `     `  `    ``j ``=` `0` `# Current index of i/p list ar2[] ` `     `  `    ``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 ` `    ``count ``=` `0` `    ``while` `count < n ``+` `1``: ` `        ``count ``+``=` `1` `         `  `        ``# 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[] ` `        ``elif` `j ``=``=` `n: ` `            ``m1 ``=` `m2 ` `            ``m2 ``=` `ar1[``0``] ` `            ``break` `        ``# equals sign because if two  ` `        ``# arrays have some common elements  ` `        ``if` `ar1[i] <``=` `ar2[j]: ` `            ``m1 ``=` `m2 ``# Store the prev median ` `            ``m2 ``=` `ar1[i] ` `            ``i ``+``=` `1` `        ``else``: ` `            ``m1 ``=` `m2 ``# Store the prev median ` `            ``m2 ``=` `ar2[j] ` `            ``j ``+``=` `1` `    ``return` `(m1 ``+` `m2)``/``2` ` `  `# Driver code to test above function ` `ar1 ``=` `[``1``, ``12``, ``15``, ``26``, ``38``] ` `ar2 ``=` `[``2``, ``13``, ``17``, ``30``, ``45``] ` `n1 ``=` `len``(ar1) ` `n2 ``=` `len``(ar2) ` `if` `n1 ``=``=` `n2: ` `    ``print``(``"Median is "``, getMedian(ar1, ar2, n1)) ` `else``: ` `    ``print``(``"Doesn't work for arrays of unequal size"``) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// A Simple Merge based O(n) solution  ` `// to find median of two sorted arrays ` `using` `System; ` `class` `GFG ` `{ ` `    ``// function to calculate median ` `    ``static` `int` `getMedian(``int` `[]ar1,  ` `                         ``int` `[]ar2,  ` `                         ``int` `n) ` `    ``{  ` `        ``int` `i = 0;  ` `        ``int` `j = 0;  ` `        ``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; ` `                ``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; ` `                ``break``; ` `            ``} ` `            ``/* equals sign because if two  ` `            ``arrays have some common elements */` `            ``if` `(ar1[i] <= ar2[j]) ` `            ``{  ` `                ``// Store the prev median  ` `                ``m1 = m2;  ` `                ``m2 = ar1[i]; ` `                ``i++; ` `            ``} ` `            ``else` `            ``{ ` `                ``// Store the prev median  ` `                ``m1 = m2;  ` `                ``m2 = ar2[j]; ` `                ``j++; ` `            ``} ` `        ``} ` `     `  `        ``return` `(m1 + m2)/2; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `[]ar1 = {1, 12, 15, 26, 38}; ` `        ``int` `[]ar2 = {2, 13, 17, 30, 45}; ` `     `  `        ``int` `n1 = ar1.Length; ` `        ``int` `n2 = ar2.Length; ` `        ``if` `(n1 == n2) ` `            ``Console.Write(``"Median is "` `+ ` `                        ``getMedian(ar1, ar2, n1)); ` `        ``else` `            ``Console.Write(``"arrays are of unequal size"``); ` `    ``}  ` `} `

## PHP

 ` `

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_|...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, ar2) + min(ar1, ar2))/2
```

Examples :

```   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, ar2) + min(ar1, ar2))/2
= (max(15, 13) + min(26, 17))/2
= (15 + 17)/2
= 16
```

Implementation :

## C++

 `// A divide and conquer based  ` `// efficient solution to find  ` `// median of two sorted arrays  ` `// of same size. ` `#include ` `using` `namespace` `std; ` ` `  `/* to get median of a ` `   ``sorted array */` `int` `median(``int` `[], ``int``);  ` ` `  `/* 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 +  ` `                ``ar2) / 2; ` `    ``if` `(n == 2) ` `        ``return` `(max(ar1, ar2) +  ` `                ``min(ar1, ar2)) / 2; ` ` `  `    ``/* get the median of  ` `       ``the first array */` `    ``int` `m1 = median(ar1, n);  ` `     `  `    ``/* get the median of  ` `       ``the second array */` `    ``int` `m2 = median(ar2, n);  ` ` `  `    ``/* 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 / 2 - 1]) / 2; ` `    ``else` `        ``return` `arr[n / 2]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `ar1[] = {1, 2, 3, 6}; ` `    ``int` `ar2[] = {4, 6, 8, 10}; ` `    ``int` `n1 = ``sizeof``(ar1) /  ` `             ``sizeof``(ar1); ` `    ``int` `n2 = ``sizeof``(ar2) /  ` `             ``sizeof``(ar2); ` `    ``if` `(n1 == n2) ` `        ``cout << ``"Median is "` `             ``<< getMedian(ar1, ar2, n1); ` `    ``else` `        ``cout << ``"Doesn't work for arrays "`  `             ``<< ``"of unequal size"``; ` `    ``return` `0; ` `} ` ` `  `// This code is contributed ` `// by Shivi_Aggarwal `

## C

 `// A divide and conquer based efficient solution to find median ` `// of two sorted arrays of same size. ` `#include ` `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 + ar2)/2; ` `    ``if` `(n == 2) ` `        ``return` `(max(ar1, ar2) + min(ar1, ar2)) / 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/2-1])/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); ` `    ``int` `n2 = ``sizeof``(ar2)/``sizeof``(ar2); ` `    ``if` `(n1 == n2) ` `        ``printf``(``"Median is %d"``, getMedian(ar1, ar2, n1)); ` `    ``else` `        ``printf``(``"Doesn't work for arrays of unequal size"``); ` `    ``return` `0; ` `} `

## Java

 `// A Java program to divide and conquer based ` `// efficient solution to find ` `// median of two sorted arrays ` `// of same size. ` `import` `java.util.*; ` `class` `GfG { ` ` `  `    ``/* This function returns median ` `    ``of ar1[] and ar2[]. ` `    ``Assumptions in this function: ` `        ``Both ar1[] and ar2[] are ` `        ``sorted arrays ` `        ``Both have n elements */` ` `  `    ``static` `int` `getMedian( ` `        ``int``[] a, ``int``[] b, ``int` `startA, ` `        ``int` `startB, ``int` `endA, ``int` `endB) ` `    ``{ ` `        ``if` `(endA - startA == ``1``) { ` `            ``return` `( ` `                       ``Math.max(a[startA], ` `                                ``b[startB]) ` `                       ``+ Math.min(a[endA], b[endB])) ` `                ``/ ``2``; ` `        ``} ` `        ``/* get the median of ` `    ``the first array */` `        ``int` `m1 = median(a, startA, endA); ` ` `  `        ``/* get the median of ` `    ``the second array */` `        ``int` `m2 = median(b, startB, endB); ` ` `  `        ``/* 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] */` `        ``else` `if` `(m1 < m2) { ` `            ``return` `getMedian( ` `                ``a, b, (endA + startA + ``1``) / ``2``, ` `                ``startB, endA, ` `                ``(endB + startB + ``1``) / ``2``); ` `        ``} ` ` `  `        ``/* if m1 > m2 then median must ` `    ``exist in ar1[....m1] and ` `    ``ar2[m2...] */` `        ``else` `{ ` `            ``return` `getMedian( ` `                ``a, b, startA, ` `                ``(endB + startB + ``1``) / ``2``, ` `                ``(endA + startA + ``1``) / ``2``, endB); ` `        ``} ` `    ``} ` ` `  `    ``/* Function to get median ` `    ``of a sorted array */` `    ``static` `int` `median( ` `        ``int``[] arr, ``int` `start, ``int` `end) ` `    ``{ ` `        ``int` `n = end - start + ``1``; ` `        ``if` `(n % ``2` `== ``0``) { ` `            ``return` `( ` `                       ``arr[start + (n / ``2``)] ` `                       ``+ arr[start + (n / ``2` `- ``1``)]) ` `                ``/ ``2``; ` `        ``} ` `        ``else` `{ ` `            ``return` `arr[start + n / ``2``]; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `ar1[] = { ``1``, ``2``, ``3``, ``6` `}; ` `        ``int` `ar2[] = { ``4``, ``6``, ``8``, ``10` `}; ` `        ``int` `n1 = ar1.length; ` `        ``int` `n2 = ar2.length; ` `        ``if` `(n1 != n2) { ` `            ``System.out.println( ` `                ``"Doesn't work for arrays "` `                ``+ ``"of unequal size"``); ` `        ``} ` `        ``else` `if` `(n1 == ``0``) { ` `            ``System.out.println(``"Arrays are empty."``); ` `        ``} ` `        ``else` `if` `(n1 == ``1``) { ` `            ``System.out.println((ar1[``0``] + ar2[``0``]) / ``2``); ` `        ``} ` `        ``else` `{ ` `            ``System.out.println( ` `                ``"Median is "` `                ``+ getMedian( ` `                      ``ar1, ar2, ``0``, ``0``, ` `                      ``ar1.length - ``1``, ar2.length - ``1``)); ` `        ``} ` `    ``} ` `} `

## Python

 `# using divide and conquer we divide ` `# the 2 arrays accordingly recursively ` `# till we get two elements in each  ` `# array, hence then we calculate median ` ` `  `#condition len(arr1)=len(arr2)=n ` `def` `getMedian(arr1, arr2, n):  ` `     `  `    ``# there is no element in any array ` `    ``if` `n ``=``=` `0``:  ` `        ``return` `-``1` `         `  `    ``# 1 element in each => median of  ` `    ``# sorted arr made of two arrays will     ` `    ``elif` `n ``=``=` `1``:  ` `        ``# be sum of both elements by 2 ` `        ``return` `(arr1[``0``]``+``arr2[``1``])``/``2` `         `  `    ``# Eg. [1,4] , [6,10] => [1, 4, 6, 10] ` `    ``# median = (6+4)/2     ` `    ``elif` `n ``=``=` `2``:  ` `        ``# which implies median = (max(arr1, ` `        ``# arr2)+min(arr1,arr2))/2 ` `        ``return` `(``max``(arr1[``0``], arr2[``0``]) ``+`  `                ``min``(arr1[``1``], arr2[``1``])) ``/` `2` `     `  `    ``else``: ` `        ``#calculating medians      ` `        ``m1 ``=` `median(arr1, n) ` `        ``m2 ``=` `median(arr2, n) ` `         `  `        ``# then the elements at median  ` `        ``# position must be between the  ` `        ``# greater median and the first  ` `        ``# element of respective array and  ` `        ``# between the other median and  ` `        ``# the last element in its respective array. ` `        ``if` `m1 > m2: ` `             `  `            ``if` `n ``%` `2` `=``=` `0``: ` `                ``return` `getMedian(arr1[:``int``(n ``/` `2``) ``+` `1``], ` `                        ``arr2[``int``(n ``/` `2``) ``-` `1``:], ``int``(n ``/` `2``) ``+` `1``) ` `            ``else``: ` `                ``return` `getMedian(arr1[:``int``(n ``/` `2``) ``+` `1``],  ` `                        ``arr2[``int``(n ``/` `2``):], ``int``(n ``/` `2``) ``+` `1``) ` `         `  `        ``else``: ` `            ``if` `n ``%` `2` `=``=` `0``: ` `                ``return` `getMedian(arr1[``int``(n ``/` `2` `-` `1``):], ` `                        ``arr2[:``int``(n ``/` `2` `+` `1``)], ``int``(n ``/` `2``) ``+` `1``) ` `            ``else``: ` `                ``return` `getMedian(arr1[``int``(n ``/` `2``):],  ` `                        ``arr2[``0``:``int``(n ``/` `2``) ``+` `1``], ``int``(n ``/` `2``) ``+` `1``) ` ` `  ` ``# function to find median of array ` `def` `median(arr, n): ` `    ``if` `n ``%` `2` `=``=` `0``: ` `        ``return` `(arr[``int``(n ``/` `2``)] ``+` `                ``arr[``int``(n ``/` `2``) ``-` `1``]) ``/` `2` `    ``else``: ` `        ``return` `arr[``int``(n``/``2``)] ` ` `  `     `  `# Driver code ` `arr1 ``=` `[``1``, ``2``, ``3``, ``6``] ` `arr2 ``=` `[``4``, ``6``, ``8``, ``10``] ` `n ``=` `len``(arr1) ` `print``(``int``(getMedian(arr1,arr2,n))) ` ` `  `# This code is contributed by ` `# baby_gog9800 `

Output :

`Median is 5`

Time Complexity : O(logn)

Median of two sorted arrays of different sizes

References:
http://en.wikipedia.org/wiki/Median

http://ocw.alfaisal.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-046JFall-2005/30C68118-E436-4FE3-8C79-6BAFBB07D935/0/ps9sol.pdf ds3etph5wn

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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