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# 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)) Note: Since the size of the set for which we are looking for the median is even (2n), we need to take the average of the middle two numbers and return the floor of the average.

Method 1 (Simply count while Merging)

Use the 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 ` `/* 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;``}`

## 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 */``double` `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` `(1.0 * (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`

## 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

 ``

## Javascript

 ``

Output

`Median is 16`

Time Complexity: O(n)
Auxiliary Space: O(1)\

Method 2 (By Merging two arrays w/o extra space)

This method works by merging two arrays without extra space and then sorting them.

Algorithm :

```1) Merge the two input arrays ar1[] and ar2[].
2) Sort ar1[] and ar2[] respectively.
3) The median will be the last element of ar1[] + the first
element of ar2[] divided by 2. [(ar1[n-1] + ar2)/2].```

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach``#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` `j = 0;``    ``int` `i = n - 1;``    ``while` `(ar1[i] > ar2[j] && j < n && i > -1)``        ``swap(ar1[i--], ar2[j++]);``    ``sort(ar1, ar1 + n);``    ``sort(ar2, ar2 + n);``    ``return` `(ar1[n - 1] + ar2) / 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 Lakshay`

## C

 `// C program for the above approach``#include ``#include ` `/* This function returns``median of ar1[] and ar2[].``Assumptions in this function:``Both ar1[] and ar2[]``are sorted arrays``Both have n elements */` `// compare function, compares two elements``int` `compare(``const` `void``* num1, ``const` `void``* num2)``{``    ``if` `(*(``int``*)num1 > *(``int``*)num2)``        ``return` `1;``    ``else``        ``return` `-1;``}` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n)``{``    ``int` `j = 0;``    ``int` `i = n - 1;``    ``while` `(ar1[i] > ar2[j] && j < n && i > -1) {``        ``int` `temp = ar1[i];``        ``ar1[i] = ar2[j];``        ``ar2[j] = temp;``        ``i--;``        ``j++;``    ``}` `    ``qsort``(ar1, n, ``sizeof``(``int``), compare);``    ``qsort``(ar2, n, ``sizeof``(``int``), compare);` `    ``return` `(ar1[n - 1] + ar2) / 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)``        ``printf``(``"Median is %d "``, getMedian(ar1, ar2, n1));``    ``else``        ``printf``(``"Doesn't work for arrays of unequal size"``);``    ``return` `0;``}` `// This code is contributed by Deepthi`

## Java

 `/*package whatever //do not write package name here */``import` `java.io.*;``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 */``public` `static` `int` `getMedian(``int` `ar1[],``                            ``int` `ar2[], ``int` `n)``{``    ``int` `j = ``0``;``    ``int` `i = n - ``1``;``    ``while` `(ar1[i] > ar2[j] && j < n && i > -``1``)``    ``{``    ``int` `temp = ar1[i];``    ``ar1[i] = ar2[j];``    ``ar2[j] = temp;``    ``i--; j++;``    ``}``    ``Arrays.sort(ar1);``    ``Arrays.sort(ar2);``    ``return` `(ar1[n - ``1``] + ar2[``0``]) / ``2``;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `ar1[] = { ``1``, ``12``, ``15``, ``26``, ``38` `};``    ``int` `ar2[] = { ``2``, ``13``, ``17``, ``30``, ``45` `};` `    ``int` `n1 = ``5``;``    ``int` `n2 = ``5``;``    ``if` `(n1 == n2)``    ``System.out.println(``"Median is "``+ getMedian(ar1, ar2, n1));``    ``else``    ``System.out.println(``"Doesn't work for arrays of unequal size"``);``}``}` `// This code is contributed by Manu Pathria`

## Python3

 `# Python program for above approach` `# function to return median of the arrays``# both are sorted & of same size``def` `getMedian(ar1, ar2, n):``    ``i, j ``=` `n ``-` `1``, ``0` `    ``# while loop to swap all smaller numbers to arr1``    ``while``(ar1[i] > ar2[j] ``and` `i > ``-``1` `and` `j < n):``        ``ar1[i], ar2[j] ``=` `ar2[j], ar1[i]``        ``i ``-``=` `1``        ``j ``+``=` `1` `    ``ar1.sort()``    ``ar2.sort()` `    ``return` `(ar1[``-``1``] ``+` `ar2[``0``]) >> ``1`  `# Driver program``if` `__name__ ``=``=` `'__main__'``:``    ``ar1 ``=` `[``1``, ``12``, ``15``, ``26``, ``38``]``    ``ar2 ``=` `[``2``, ``13``, ``17``, ``30``, ``45``]` `    ``n1, n2 ``=` `len``(ar1), ``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 saitejagampala`

## C#

 `/*package whatever //do not write package name here */``using` `System;``public` `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 */``public` `static` `int` `getMedian(``int` `[]ar1,``                            ``int` `[]ar2, ``int` `n)``{``    ``int` `j = 0;``    ``int` `i = n - 1;``    ``while` `(ar1[i] > ar2[j] && j < n && i > -1)``    ``{``    ``int` `temp = ar1[i];``    ``ar1[i] = ar2[j];``    ``ar2[j] = temp;``    ``i--; j++;``    ``}``    ``Array.Sort(ar1);``    ``Array.Sort(ar2);``    ``return` `(ar1[n - 1] + ar2) / 2;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]ar1 = { 1, 12, 15, 26, 38 };``    ``int` `[]ar2 = { 2, 13, 17, 30, 45 };` `    ``int` `n1 = 5;``    ``int` `n2 = 5;``    ``if` `(n1 == n2)``    ``Console.WriteLine(``"Median is "``+ getMedian(ar1, ar2, n1));``    ``else``    ``Console.WriteLine(``"Doesn't work for arrays of unequal size"``);``}``}` `// This code is contributed by aashish1995`

## Javascript

 ``

Output

`Median is 16`

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

Method 3 (Using binary search)

This method can also be used for arrays of different sizes.

Algorithm:

We can find the kth element by using binary search on whole range of constraints of elements.

• Initialize ans = 0.0
• Initialize low = -10^9, high = 10^9 and pos = n
• Run a loop while(low <= high):
• Calculate mid = (low + (high – low)>>1)
• Find total elements less or equal to mid in the given arrays
• If the count is less or equal to pos
• Update low = mid + 1
• Else high = mid – 1
• Store low in ans, i.e., ans = low.
• Again follow step3 with pos as n – 1
• Return (sum + low * 1.0)/2

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `double` `getMedian(``int` `arr1[], ``int` `arr2[], ``int` `n)``{``    ``// according to given constraints all numbers are in``    ``// this range``    ``int` `low = (``int``)-1e9, high = (``int``)1e9;` `    ``int` `pos = n;``    ``double` `ans = 0.0;``    ``// binary search to find the element which will be``    ``// present at pos = totalLen/2 after merging two``    ``// arrays in sorted order``    ``while` `(low <= high) {``        ``int` `mid = low + ((high - low) >> 1);` `        ``// total number of elements in arrays which are``        ``// less than mid``        ``int` `ub = upper_bound(arr1, arr1 + n, mid) - arr1``                 ``+ upper_bound(arr2, arr2 + n, mid) - arr2;` `        ``if` `(ub <= pos)``            ``low = mid + 1;``        ``else``            ``high = mid - 1;``    ``}` `    ``ans = low;` `    ``// As there are even number of elements, we will``    ``// also have to find element at pos = totalLen/2 - 1``    ``pos--;``    ``low = (``int``)-1e9;``    ``high = (``int``)1e9;``    ``while` `(low <= high) {``        ``int` `mid = low + ((high - low) >> 1);``        ``int` `ub = upper_bound(arr1, arr1 + n, mid) - arr1``                 ``+ upper_bound(arr2, arr2 + n, mid) - arr2;` `        ``if` `(ub <= pos)``            ``low = mid + 1;``        ``else``            ``high = mid - 1;``    ``}` `    ``// average of two elements in case of even``    ``// number of elements``    ``ans = (ans + low) / 2;` `    ``return` `ans;``}` `int` `main()``{``    ``int` `arr1[] = { 1, 4, 5, 6, 10 };``    ``int` `arr2[] = { 2, 3, 4, 5, 7 };` `    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1);` `    ``double` `median = getMedian(arr1, arr2, n);` `    ``cout << ``"Median is "` `<< median << endl;` `    ``return` `0;``}``// This code is contributed by Srj_27`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {` `    ``public` `static` `double` `getMedian(``int``[] nums1, ``int``[] nums2,``                                   ``int` `n)``    ``{``        ``// according to given constraints all numbers are in``        ``// this range``        ``int` `low = (``int``)-1e9, high = (``int``)1e9;` `        ``int` `pos = n;``        ``double` `ans = ``0.0``;``        ``// binary search to find the element which will be``        ``// present at pos = totalLen/2 after merging two``        ``// arrays in sorted order``        ``while` `(low <= high) {``            ``int` `mid = low + ((high - low) >> ``1``);` `            ``// total number of elements in arrays which are``            ``// less than mid``            ``int` `ub = upperBound(nums1, mid)``                     ``+ upperBound(nums2, mid);` `            ``if` `(ub <= pos)``                ``low = mid + ``1``;``            ``else``                ``high = mid - ``1``;``        ``}` `        ``ans = low;` `        ``// As there are even number of elements, we will``        ``// also have to find element at pos = totalLen/2 - 1``        ``pos--;``        ``low = (``int``)-1e9;``        ``high = (``int``)1e9;``        ``while` `(low <= high) {``            ``int` `mid = low + ((high - low) >> ``1``);``            ``int` `ub = upperBound(nums1, mid)``                     ``+ upperBound(nums2, mid);` `            ``if` `(ub <= pos)``                ``low = mid + ``1``;``            ``else``                ``high = mid - ``1``;``        ``}` `        ``// average of two elements in case of even``        ``// number of elements``        ``ans = (ans + low * ``1.0``) / ``2``;` `        ``return` `ans;``    ``}` `    ``// a function which returns the index of smallest``    ``// element which is strictly greater than key (i.e. it``    ``// returns number of elements which are less than or``    ``// equal to key)``    ``public` `static` `int` `upperBound(``int``[] arr, ``int` `key)``    ``{``        ``int` `low = ``0``, high = arr.length;` `        ``while` `(low < high) {``            ``int` `mid = low + ((high - low) >> ``1``);` `            ``if` `(arr[mid] <= key)``                ``low = mid + ``1``;``            ``else``                ``high = mid;``        ``}` `        ``return` `low;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``1``, ``4``, ``5``, ``6``, ``10` `};``        ``int``[] brr = { ``2``, ``3``, ``4``, ``5``, ``7` `};` `        ``double` `median = getMedian(arr, brr, arr.length);` `        ``System.out.println(``"Median is "` `+ median);``    ``}``}`

## C#

 `// Include namespace system``using` `System;`  `public` `class` `GFG``{``    ``public` `static` `double` `getMedian(``int``[] nums1, ``int``[] nums2, ``int` `n)``    ``{``        ``// according to given constraints all numbers are in``        ``// this range``        ``var` `low = (``int``)-1.0E9;``        ``var` `high = (``int``)1.0E9;``        ``var` `pos = n;``        ``var` `ans = 0.0;``        ``// binary search to find the element which will be``        ``// present at pos = totalLen/2 after merging two``        ``// arrays in sorted order``        ``while` `(low <= high)``        ``{``            ``var` `mid = low + ((high - low) >> 1);``            ``// total number of elements in arrays which are``            ``// less than mid``            ``var` `ub = upperBound(nums1, mid) + upperBound(nums2, mid);``            ``if` `(ub <= pos)``            ``{``                ``low = mid + 1;``            ``}``            ``else``            ``{``                ``high = mid - 1;``            ``}``        ``}``        ``ans = low;``        ``// As there are even number of elements, we will``        ``// also have to find element at pos = totalLen/2 - 1``        ``pos--;``        ``low = (``int``)-1.0E9;``        ``high = (``int``)1.0E9;``        ``while` `(low <= high)``        ``{``            ``var` `mid = low + ((high - low) >> 1);``            ``var` `ub = upperBound(nums1, mid) + upperBound(nums2, mid);``            ``if` `(ub <= pos)``            ``{``                ``low = mid + 1;``            ``}``            ``else``            ``{``                ``high = mid - 1;``            ``}``        ``}``        ``// average of two elements in case of even``        ``// number of elements``        ``ans = (ans + low * 1.0) / 2;``        ``return` `ans;``    ``}``    ``// a function which returns the index of smallest``    ``// element which is strictly greater than key (i.e. it``    ``// returns number of elements which are less than or``    ``// equal to key)``    ``public` `static` `int` `upperBound(``int``[] arr, ``int` `key)``    ``{``        ``var` `low = 0;``        ``var` `high = arr.Length;``        ``while` `(low < high)``        ``{``            ``var` `mid = low + ((high - low) >> 1);``            ``if` `(arr[mid] <= key)``            ``{``                ``low = mid + 1;``            ``}``            ``else``            ``{``                ``high = mid;``            ``}``        ``}``        ``return` `low;``    ``}``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = {1, 4, 5, 6, 10};``        ``int``[] brr = {2, 3, 4, 5, 7};``        ``var` `median = getMedian(arr, brr, arr.Length);``        ``Console.WriteLine(``"Median is "` `+ median.ToString());``    ``}``}`

## Python

 `# Calculate the number of elements less than or equal to mid in the given arrays``def` `count_less_than_or_equal_to_mid(mid, arrays):``    ``count ``=` `0``    ``for` `array ``in` `arrays:``        ``count ``+``=` `len``([x ``for` `x ``in` `array ``if` `x <``=` `mid])``    ``return` `count` `def` `find_kth_element(arrays, n):``    ``ans ``=` `0.0``    ``low ``=` `-``1e9``    ``high ``=` `1e9``    ``pos ``=` `n` `    ``# Binary search to find the kth element``    ``while` `low <``=` `high:``        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2``        ``count ``=` `count_less_than_or_equal_to_mid(mid, arrays)``        ``if` `count <``=` `pos:``            ``low ``=` `mid ``+` `1``        ``else``:``            ``high ``=` `mid ``-` `1` `    ``ans ``=` `low` `    ``# Update pos and repeat the binary search to find the (n-1)th element``    ``pos ``=` `n ``-` `1``    ``low ``=` `-``1e9``    ``high ``=` `1e9``    ``while` `low <``=` `high:``        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2``        ``count ``=` `count_less_than_or_equal_to_mid(mid, arrays)``        ``if` `count <``=` `pos:``            ``low ``=` `mid ``+` `1``        ``else``:``            ``high ``=` `mid ``-` `1` `    ``ans ``+``=` `low` `    ``# Return the average of the two elements``    ``return` `(ans ``/` `2.0``)` `# Test with some arrays``arrays ``=` `[[``1``, ``4``, ``5``, ``6``, ``10``], [``2``, ``3``, ``4``, ``5``, ``7``]]``n ``=` `5``print``(``"Median in"``, find_kth_element(arrays, n)) ` `#code is contributed by khushboogoyal499`

## Javascript

 `// Define the function getMedian to find the median of two arrays``function` `getMedian(arr1, arr2, n) {``  ``// Define the variables low, high, pos and ans``  ``// According to given constraints, all numbers are in this range``  ``let low = -1e9,``    ``high = 1e9,``    ``pos = n,``    ``ans = 0.0;` `  ``// Binary search to find the element which will be``  ``// present at pos = totalLen/2 after merging two``  ``// arrays in sorted order``  ``while` `(low <= high) {``    ``let mid = low + ((high - low) >> 1);` `    ``// Total number of elements in arrays which are``    ``// less than mid``    ``// Note: The function upper_bound is not available in JavaScript``    ``// You need to find an equivalent solution or implement it yourself``    ``let ub = 0;` `    ``if` `(ub <= pos) {``      ``low = mid + 1;``    ``} ``else` `{``      ``high = mid - 1;``    ``}``  ``}` `  ``ans = low;` `  ``// As there are even number of elements, we will``  ``// also have to find element at pos = totalLen/2 - 1``  ``pos--;``  ``low = -1e9;``  ``high = 1e9;``  ``while` `(low <= high) {``    ``let mid = low + ((high - low) >> 1);``    ``let ub = 0;` `    ``if` `(ub <= pos) {``      ``low = mid + 1;``    ``} ``else` `{``      ``high = mid - 1;``    ``}``  ``}` `  ``// Average of two elements in case of even``  ``// number of elements``  ``ans = (ans + low) / 2;` `  ``return` `ans;``}` `// Main function``function` `main() {``  ``let arr1 = [1, 4, 5, 6, 10];``  ``let arr2 = [2, 3, 4, 5, 7];` `  ``let n = arr1.length;` `  ``let median = getMedian(arr1, arr2, n);` `  ``console.log(``"Median is "` `+ median);` `  ``return` `0;``}` `// Call the main function``main();` `// This code is contributed by Vikram_Shirsat`

Output

`Median is 4.5`

Time Complexity: O(log n)
Auxiliary Space: O(1)

Median of two sorted arrays of different sizes

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

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