# Median of two sorted arrays of different sizes

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

Method 1: (Linear and Simpler Approach)
Here we need to find the median of the two sorted arrays of different sizes so we keep two variables to point to the arrays and one used to count the no of elements read. We used a simple Merge based O(n) solution just we are not merging the array instead we are keeping track of the last element read till we reach the median
There are two cases :
Case 1: m+n is odd
Then we will find a clear median at (m+n)/2 index in the array obtained after merging both the arrays so we just traverse both the arrays and keep the last value in m1 after the loop, m1 will contain the value of the median
Case 2: m+n is even
Median will be average of elements at index ((m+n)/2 – 1) and (m+n)/2 in the array obtained after merging both the arrays so we need to keep track of not only the last element but also the second last element (m2 is used for this) so we traverse both the arrays and keep the last value in m1 and second last value in m2 after the loop, (m1+m2)/2 will contain the value of the median.

Below is the implementation of above approach:

## 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[].  ` `Assumption in this function:  ` `Both ar1[] and ar2[] are sorted arrays */` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n, ``int` `m)  ` `{  ` `    ``int` `i = 0; ``/* Current index of input array ar1[] */` `    ``int` `j = 0; ``/* Current index of input array ar2[] */` `    ``int` `count;  ` `    ``int` `m1 = -1, m2 = -1;  ` ` `  `    ``// Since there are (n+m) elements,  ` `    ``// There are following two cases  ` `    ``// if n+m is odd then the middle  ` `    ``//index is median i.e. (m+n)/2  ` `    ``if``((m + n) % 2 == 1)  ` `    ``{  ` `        ``for` `(count = 0; count <= (n + m)/2; count++) ` `        ``{  ` `            ``if``(i != n && j != m) ` `            ``{  ` `                ``m1 = (ar1[i] > ar2[j]) ? ar2[j++] : ar1[i++];  ` `            ``}  ` `            ``else` `if``(i < n) ` `            ``{  ` `                ``m1 = ar1[i++];  ` `            ``}  ` `            ``// for case when j ar2[j]) ? ar2[j++] : ar1[i++];  ` `            ``}  ` `            ``else` `if``(i < n) ` `            ``{  ` `                ``m1 = ar1[i++];  ` `            ``}  ` `            ``// for case when j

## C

 `// A Simple Merge based O(n) solution to find  ` `// median of two sorted arrays  ` `#include   ` ` `  `/* This function returns median of ar1[] and ar2[].  ` `Assumption in this function:  ` `Both ar1[] and ar2[] are sorted arrays */` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n, ``int` `m)  ` `{  ` `    ``int` `i = 0; ``/* Current index of input array ar1[] */` `    ``int` `j = 0; ``/* Current index of input array ar2[] */` `    ``int` `count;  ` `    ``int` `m1 = -1, m2 = -1;  ` ` `  `    ``// Since there are (n+m) elements,  ` `    ``// There are following two cases ` `    ``// if n+m is odd then the middle  ` `    ``//index is median i.e. (m+n)/2 ` `    ``if``((m + n) % 2 == 1) { ` `        ``for` `(count = 0; count <= (n + m)/2; count++) { ` `            ``if``(i != n && j != m){ ` `            ``m1 = (ar1[i] > ar2[j]) ? ar2[j++] : ar1[i++]; ` `            ``} ` `            ``else` `if``(i < n){ ` `            ``m1 = ar1[i++]; ` `            ``} ` `            ``// for case when j ar2[j]) ? ar2[j++] : ar1[i++]; ` `            ``} ` `            ``else` `if``(i < n){ ` `            ``m1 = ar1[i++]; ` `            ``} ` `            ``// for case when j

Output:

` 10 `

Method 2: (Effective but a little Complex Approach)
The approach discussed in this post is similar to method 2 of equal the size post. The basic idea is same, we find the median of two arrays and compare the medians to discard almost half of the elements in both arrays. Since the number of elements may differ here, there are many base cases that need to be handled separately. Before we proceed to complete solution, let us first talk about all base cases.

Let the two arrays be A[N] and B[M]. In the following explanation, it is assumed that N is smaller than or equal to M.

Base cases:
The smaller array has only one element
Case 0: N = 0, M = 2
Case 1: N = 1, M = 1.
Case 2: N = 1, M is odd
Case 3: N = 1, M is even
The smaller array has only two elements
Case 4: N = 2, M = 2
Case 5: N = 2, M is odd
Case 6: N = 2, M is even

Case 0: There are no elements in first array, return median of second array. If second array is also empty, return -1.

Case 1: There is only one element in both arrays, so output the average of A and B.

Case 2: N = 1, M is odd
Let B = {5, 10, 12, 15, 20}
First find the middle element of B[], which is 12 for above array. There are following 4 sub-cases.
2.1 If A is smaller than 10, the median is average of 10 and 12.
2.2 If A lies between 10 and 12, the median is average of A and 12.
2.3 If A lies between 12 and 15, the median is average of 12 and A.
2.4 If A is greater than 15, the median is average of 12 and 15.
In all the sub-cases, we find that 12 is fixed. So, we need to find the median of B[ M / 2 – 1 ], B[ M / 2 + 1], A[ 0 ] and take its average with B[ M / 2 ].

Case 3: N = 1, M is even
Let B = {5, 10, 12, 15}
First find the middle items in B[], which are 10 and 12 in above example. There are following 3 sub-cases.
3.1 If A is smaller than 10, the median is 10.
3.2 If A lies between 10 and 12, the median is A.
3.3 If A is greater than 12, the median is 12.
So, in this case, find the median of three elements B[ M / 2 – 1 ], B[ M / 2] and A[ 0 ].

Case 4: N = 2, M = 2
There are four elements in total. So we find the median of 4 elements.

Case 5: N = 2, M is odd
Let B = {5, 10, 12, 15, 20}
The median is given by median of following three elements: B[M/2], max(A, B[M/2 – 1]), min(A, B[M/2 + 1]).

Case 6: N = 2, M is even
Let B = {5, 10, 12, 15}
The median is given by median of following four elements: B[M/2], B[M/2 – 1], max(A, B[M/2 – 2]), min(A, B[M/2 + 1])

Remaining Cases:
Once we have handled the above base cases, following is the remaining process.
1) Find the middle item of A[] and middle item of B[].
…..1.1) If the middle item of A[] is greater than middle item of B[], ignore the last half of A[], let length of ignored part is idx. Also, cut down B[] by idx from the start.
…..1.2) else, ignore the first half of A[], let length of ignored part is idx. Also, cut down B[] by idx from the last.

Following is implementation of the above approach.

## 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 ` `float` `MO2(``int` `a, ``int` `b) ` `{ ``return` `( a + b ) / 2.0; } ` ` `  `// A utility function to find median of three integers ` `float` `MO3(``int` `a, ``int` `b, ``int` `c) ` `{ ` `    ``return` `a + b + c - max(a, max(b, c)) ` `                     ``- min(a, min(b, c)); ` `} ` ` `  `// A utility function to find median of four integers ` `float` `MO4(``int` `a, ``int` `b, ``int` `c, ``int` `d) ` `{ ` `    ``int` `Max = max( a, max( b, max( c, d ) ) ); ` `    ``int` `Min = min( a, min( b, min( c, d ) ) ); ` `    ``return` `( a + b + c + d - Max - Min ) / 2.0; ` `} ` ` `  `// Utility function to find median of single array ` `float` `medianSingle(``int` `arr[], ``int` `n) ` `{ ` `   ``if` `(n == 0) ` `      ``return` `-1; ` `   ``if` `(n%2 == 0) ` `        ``return` `(``double``)(arr[n/2] + arr[n/2-1])/2; ` `   ``return` `arr[n/2]; ` `} ` ` `  `// This function assumes that N is smaller than or equal to M ` `// This function returns -1 if both arrays are empty ` `float` `findMedianUtil( ``int` `A[], ``int` `N, ``int` `B[], ``int` `M ) ` `{ ` `    ``// If smaller array is empty, return median from second array ` `    ``if` `(N == 0) ` `      ``return` `medianSingle(B, M); ` ` `  `    ``// If the smaller array has only one element ` `    ``if` `(N == 1) ` `    ``{ ` `        ``// Case 1: If the larger array also has one element, ` `        ``// simply call MO2() ` `        ``if` `(M == 1) ` `            ``return` `MO2(A, B); ` ` `  `        ``// Case 2: If the larger array has odd number of elements, ` `        ``// then consider the middle 3 elements of larger array and ` `        ``// the only element of smaller array. Take few examples ` `        ``// like following ` `        ``// A = {9}, B[] = {5, 8, 10, 20, 30} and ` `        ``// A[] = {1}, B[] = {5, 8, 10, 20, 30} ` `        ``if` `(M & 1) ` `            ``return` `MO2( B[M/2], MO3(A, B[M/2 - 1], B[M/2 + 1]) ); ` ` `  `        ``// Case 3: If the larger array has even number of element, ` `        ``// then median will be one of the following 3 elements ` `        ``// ... The middle two elements of larger array ` `        ``// ... The only element of smaller array ` `        ``return` `MO3( B[M/2], B[M/2 - 1], A ); ` `    ``} ` ` `  `    ``// If the smaller array has two elements ` `    ``else` `if` `(N == 2) ` `    ``{ ` `        ``// Case 4: If the larger array also has two elements, ` `        ``// simply call MO4() ` `        ``if` `(M == 2) ` `            ``return` `MO4(A, A, B, B); ` ` `  `        ``// Case 5: If the larger array has odd number of elements, ` `        ``// then median will be one of the following 3 elements ` `        ``// 1. Middle element of larger array ` `        ``// 2. Max of first element of smaller array and element ` `        ``//    just before the middle in bigger array ` `        ``// 3. Min of second element of smaller array and element ` `        ``//    just after the middle in bigger array ` `        ``if` `(M & 1) ` `            ``return` `MO3 ( B[M/2], ` `                         ``max(A, B[M/2 - 1]), ` `                         ``min(A, B[M/2 + 1]) ` `                       ``); ` ` `  `        ``// Case 6: If the larger array has even number of elements, ` `        ``// then median will be one of the following 4 elements ` `        ``// 1) & 2) The middle two elements of larger array ` `        ``// 3) Max of first element of smaller array and element ` `        ``//    just before the first middle element in bigger array ` `        ``// 4. Min of second element of smaller array and element ` `        ``//    just after the second middle in bigger array ` `        ``return` `MO4 ( B[M/2], ` `                     ``B[M/2 - 1], ` `                     ``max( A, B[M/2 - 2] ), ` `                     ``min( A, B[M/2 + 1] ) ` `                   ``); ` `    ``} ` ` `  `    ``int` `idxA = ( N - 1 ) / 2; ` `    ``int` `idxB = ( M - 1 ) / 2; ` ` `  `     ``/* if A[idxA] <= B[idxB], then median must exist in ` `        ``A[idxA....] and B[....idxB] */` `    ``if` `(A[idxA] <= B[idxB] ) ` `      ``return` `findMedianUtil(A + idxA, N/2 + 1, B, M - idxA ); ` ` `  `    ``/* if A[idxA] > B[idxB], then median must exist in ` `       ``A[...idxA] and B[idxB....] */` `    ``return` `findMedianUtil(A, N/2 + 1, B + idxA, M - idxA ); ` `} ` ` `  `// A wrapper function around findMedianUtil(). This function ` `// makes sure that smaller array is passed as first argument ` `// to findMedianUtil ` `float` `findMedian( ``int` `A[], ``int` `N, ``int` `B[], ``int` `M ) ` `{ ` `    ``if` `(N > M) ` `       ``return` `findMedianUtil( B, M, A, N ); ` ` `  `    ``return` `findMedianUtil( A, N, B, M ); ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``int` `A[] = {900}; ` `    ``int` `B[] = {5, 8, 10, 20}; ` ` `  `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` `    ``int` `M = ``sizeof``(B) / ``sizeof``(B); ` ` `  `    ``printf``(``"%f"``, findMedian( A, N, B, M ) ); ` `    ``return` `0; ` `} `

## PHP

 ` \$B[\$idxB],  ` `    ``then median must exist in ` `    ``\$A[...\$idxA] and \$B[\$idxB....] */` `    ``return` `findMedianUtil(``\$A``, ``\$N``/2 + 1,  ` `                          ``\$B` `+ ``\$idxA``, ``\$M` `- ``\$idxA` `); ` `} ` ` `  `// A wrapper function around ` `// findMedianUtil(). This  ` `// function makes sure that  ` `// smaller array is passed as  ` `// first argument to findMedianUtil ` `function` `findMedian(&``\$A``, ``\$N``,  ` `                    ``&``\$B``, ``\$M` `) ` `{ ` `    ``if` `(``\$N` `> ``\$M``) ` `    ``return` `findMedianUtil(``\$B``, ``\$M``,  ` `                          ``\$A``, ``\$N` `); ` ` `  `    ``return` `findMedianUtil(``\$A``, ``\$N``,  ` `                          ``\$B``, ``\$M` `); ` `} ` ` `  `// Driver Code ` `\$A` `= ``array``(900); ` `\$B` `= ``array``(5, 8, 10, 20); ` ` `  `\$N` `= sizeof(``\$A``); ` `\$M` `= sizeof(``\$B``); ` ` `  `echo` `findMedian( ``\$A``, ``\$N``, ``\$B``, ``\$M` `); ` ` `  `// This code is contributed ` `// by ChitraNayal ` `?> `

Output:

` 10 `

Time Complexity: O(LogM + LogN)