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Median of two sorted arrays of different sizes | Set 1 (Linear)
• Last Updated : 14 May, 2021

This is an extension of the median of two sorted arrays of equal size problem. Here we handle arrays of unequal size also.

Examples:

```Input: a[] = {1, 12, 15, 26, 38}
b[] = {2, 13, 24}
Output: 14
Explanation:
After merging arrays the result is
1, 2, 12, 13, 15, 24, 26, 38
median = (13 + 15)/2 = 14.

Input: a[] = {1, 3, 5, 7}
b[] = {2, 4, 6, 8}
Output: 5
Explanation :
After merging arrays the result is
1, 2, 3, 4, 5, 6, 7, 8, 9
median = 5 ```

Approach:

1. The approach discussed in this post is similar to method 1 of equal size median. Use merge procedure of merge sort.
2. Keep a variable to track of count while comparing elements of two arrays.
3. Perform merge of two sorted arrays. Increase the value of count as a new element is inserted.
4. The most efficient way of merging two arrays is to compare the top of both arrays and pop the smaller value and increase the count.
5. Continue comparing the top/first elements of both arrays until there are no elements left in any arrays
6. Now to find the median there are two cases to be handled.
• If the sum of length of array sizes is even then store the elements when the count is of value ((n1 + n2)/2)-1 and (n1 + n2)/2 in the merged array add the elements and divide by 2 return the value as median.
• If the value(sum of sizes) is odd then return the (n1 + n2)/2 th element(when value of count is (n1 + n2)/2) as median.

## C++

 `// A C++ program to find median of two sorted arrays of``// unequal sizes``#include ``using` `namespace` `std;` `// A utility function to find median of two integers``/* This function returns median of a[] and b[].``Assumptions in this function: Both a[] and b[]``are sorted arrays  */``float` `findmedian(``int` `a[], ``int` `n1, ``int` `b[], ``int` `n2)``{``    ``int` `i = 0; ``/* Current index of``             ``i/p array a[] */``    ``int` `j = 0; ``/* Current index of``                  ``i/p array b[] */``    ``int` `k;``    ``int` `m1 = -1, m2 = -1;``    ``for` `(k = 0; k <= (n1 + n2) / 2; k++) {` `        ``if` `(i < n1 && j < n2) {``            ``if` `(a[i] < b[j]) {``                ``m2 = m1;``                ``m1 = a[i];``                ``i++;``            ``}``            ``else` `{``                ``m2 = m1;``                ``m1 = b[j];``                ``j++;``            ``}``        ``}` `        ``/* Below is to handle the case where``           ``all elements of a[] are``           ``smaller than smallest(or first)``           ``element of b[] or a[] is empty*/``        ``else` `if` `(i == n1) {``            ``m2 = m1;``            ``m1 = b[j];``            ``j++;``        ``}` `        ``/* Below is to handle case where``           ``all elements of b[] are``           ``smaller than smallest(or first)``           ``element of a[] or b[] is empty*/``        ``else` `if` `(j == n2) {``            ``m2 = m1;``            ``m1 = a[i];``            ``i++;``        ``}``    ``}` `    ``/*Below is to handle the case where``     ``sum of number of elements``     ``of the arrays is even */``    ``if` `((n1 + n2) % 2 == 0)``        ``return` `(m1 + m2) * 1.0 / 2;` `    ``/* Below is to handle the case where``       ``sum of number of elements``       ``of the arrays is odd */``    ``return` `m1;``}` `// Driver program to test above functions``int` `main()``{``    ``int` `a[] = { 1, 12, 15, 26, 38 };``    ``int` `b[] = { 2, 13, 24 };` `    ``int` `n1 = ``sizeof``(a) / ``sizeof``(a);``    ``int` `n2 = ``sizeof``(b) / ``sizeof``(b);` `    ``printf``(``"%f"``, findmedian(a, n1, b, n2));` `    ``return` `0;``}`

## Java

 `// A Java program to find median of two sorted arrays of``// unequal sizes` `import` `java.io.*;` `class` `GFG {` `    ``// A utility function to find median of two integers``    ``/* This function returns median of a[] and b[].``Assumptions in this function: Both a[] and b[]``are sorted arrays */``    ``static` `float` `findmedian(``int` `a[], ``int` `n1, ``int` `b[], ``int` `n2)``    ``{``        ``int` `i = ``0``; ``/* Current index of``            ``i/p array a[] */``        ``int` `j = ``0``; ``/* Current index of``                ``i/p array b[] */``        ``int` `k;``        ``int` `m1 = -``1``, m2 = -``1``;``        ``for` `(k = ``0``; k <= (n1 + n2) / ``2``; k++) {` `            ``if` `(i < n1 && j < n2) {``                ``if` `(a[i] < b[j]) {``                    ``m2 = m1;``                    ``m1 = a[i];``                    ``i++;``                ``}``                ``else` `{``                    ``m2 = m1;``                    ``m1 = b[j];``                    ``j++;``                ``}``            ``}` `            ``/* Below is to handle the case where``        ``all elements of a[] are``        ``smaller than smallest(or first)``        ``element of b[] or a[] is empty*/``            ``else` `if` `(i == n1) {``                ``m2 = m1;``                ``m1 = b[j];``                ``j++;``            ``}` `            ``/* Below is to handle case where``        ``all elements of b[] are``        ``smaller than smallest(or first)``        ``element of a[] or b[] is empty*/``            ``else` `if` `(j == n2) {``                ``m2 = m1;``                ``m1 = a[i];``                ``i++;``            ``}``        ``}` `        ``/*Below is to handle the case where``    ``sum of number of elements``    ``of the arrays is even */``        ``if` `((n1 + n2) % ``2` `== ``0``) {``            ``return` `(m1 + m2) * (``float``)``1.0` `/ ``2``;``        ``}``        ``/* Below is to handle the case where``    ``sum of number of elements``    ``of the arrays is odd */``        ``return` `m1;``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``1``, ``12``, ``15``, ``26``, ``38` `};``        ``int` `b[] = { ``2``, ``13``, ``24` `};` `        ``int` `n1 = a.length;``        ``int` `n2 = b.length;` `        ``System.out.println(findmedian(a, n1, b, n2));``    ``}``}` `// This code has been contributed by inder_verma.`

## Python3

 `# Python3 program to find median of``# two sorted arrays of unequal sizes` `# A utility function to find median``# of two integers``''' This function returns median of``a[] and b[]. Assumptions in this``function: Both a[] and b[] are sorted``arrays '''` `def` `findmedian(a, n1, b, n2):` `    ``i ``=` `0` `# Current index of i / p array a[]``    ``j ``=` `0` `# Current index of i / p array b[]` `    ``m1 ``=` `-``1``    ``m2 ``=` `-``1``    ``for` `k ``in` `range``(((n1 ``+` `n2) ``/``/` `2``) ``+` `1``) :` `        ``if` `(i < n1 ``and` `j < n2) :``            ``if` `(a[i] < b[j]) :``                ``m2 ``=` `m1``                ``m1 ``=` `a[i]``                ``i ``+``=` `1``            ` `            ``else` `:``                ``m2 ``=` `m1``                ``m1 ``=` `b[j]``                ``j ``+``=` `1` `        ``# Below is to handle the case where``        ``# all elements of a[] are``        ``# smaller than smallest(or first)``        ``# element of b[] or a[] is empty``            ` `        ``elif``(i ``=``=` `n1) :``            ``m2 ``=` `m1``            ``m1 ``=` `b[j]``            ``j ``+``=` `1``            ` `        ``# Below is to handle case where``        ``# all elements of b[] are``        ``# smaller than smallest(or first)``        ``# element of a[] or b[] is empty``        ``elif` `(j ``=``=` `n2) :``            ``m2 ``=` `m1``            ``m1 ``=` `a[i]``            ``i ``+``=` `1` `    ``'''Below is to handle the case``    ``where sum of number of elements``    ``of the arrays is even '''``    ``if` `((n1 ``+` `n2) ``%` `2` `=``=` `0``):``        ``return` `(m1 ``+` `m2) ``*` `1.0` `/` `2` `    ``''' Below is to handle the case``    ``where sum of number of elements``    ``of the arrays is odd '''``    ``return` `m1` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``a ``=` `[ ``1``, ``12``, ``15``, ``26``, ``38` `]``    ``b ``=` `[ ``2``, ``13``, ``24` `]` `    ``n1 ``=` `len``(a)``    ``n2 ``=` `len``(b)` `    ``print``(findmedian(a, n1, b, n2))` `# This code is contributed``# by ChitraNayal`

## C#

 `// A C# program to find median``// of two sorted arrays of``// unequal sizes``using` `System;` `class` `GFG {` `    ``// A utility function to find``    ``// median of two integers``    ``/* This function returns``   ``median of a[] and b[].``   ``Assumptions in this``   ``function: Both a[] and b[]``   ``are sorted arrays */``    ``static` `float` `findmedian(``int``[] a, ``int` `n1,``                            ``int``[] b, ``int` `n2)``    ``{``        ``int` `i = 0; ``/* Current index of``              ``i/p array a[] */``        ``int` `j = 0; ``/* Current index of``              ``i/p array b[] */``        ``int` `k;``        ``int` `m1 = -1, m2 = -1;``        ``for` `(k = 0; k <= (n1 + n2) / 2; k++) {``            ``if` `(i < n1 && j < n2) {``                ``if` `(a[i] < b[j]) {``                    ``m2 = m1;``                    ``m1 = a[i];``                    ``i++;``                ``}``                ``else` `{``                    ``m2 = m1;``                    ``m1 = b[j];``                    ``j++;``                ``}``            ``}` `            ``/* Below is to handle the case where``    ``all elements of a[] are``    ``smaller than smallest(or first)``    ``element of b[] or a[] is empty*/``            ``else` `if` `(i == n1) {``                ``m2 = m1;``                ``m1 = b[j];``                ``j++;``            ``}` `            ``/* Below is to handle case where``    ``all elements of b[] are``    ``smaller than smallest(or first)``    ``element of a[] or b[] is empty*/``            ``else` `if` `(j == n2) {``                ``m2 = m1;``                ``m1 = a[i];``                ``i++;``            ``}``        ``}` `        ``/*Below is to handle the case where``sum of number of elements``of the arrays is even */``        ``if` `((n1 + n2) % 2 == 0) {``            ``return` `(m1 + m2) * (``float``)1.0 / 2;``        ``}``        ``/* Below is to handle the case``where sum of number of elements``of the arrays is odd */``        ``return` `m1;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] a = { 1, 12, 15, 26, 38 };``        ``int``[] b = { 2, 13, 24 };` `        ``int` `n1 = a.Length;``        ``int` `n2 = b.Length;` `        ``Console.WriteLine(findmedian(a, n1, b, n2));``    ``}``}` `// This code is contributed``// by Subhadeep Gupta`

## PHP

 ``

## Javascript

 ``
Output:
`14.000000`

Complexity Analysis:

• Time Complexity: O(n), Both the lists need to be traversed so the time complexity is O(n).
• Auxiliary Space: O(1), no extra space is required.

More Efficient (Log time complexity methods)

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