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

The approach discussed in this post is similar to method 2 of equal 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[0] and B[0].

**Case 2:** N = 1, M is odd

Let B[5] = {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[0] is smaller than 10, the median is average of 10 and 12.

…**2.2 ** If A[0] lies between 10 and 12, the median is average of A[0] and 12.

…**2.3** If A[0] lies between 12 and 15, the median is average of 12 and A[0].

…**2.4 ** If A[0] 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[4] = {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[0] is smaller than 10, the median is 10.

…**3.2 ** If A[0] lies between 10 and 12, the median is A[0].

…**3.3** If A[0] 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] = {5, 10, 12, 15, 20}

The median is given by median of following three elements: B[M/2], max(A[0], B[M/2 – 1]), min(A[1], B[M/2 + 1]).

**Case 6:** N = 2, M is even

Let B[4] = {5, 10, 12, 15}

The median is given by median of following four elements: B[M/2], B[M/2 – 1], max(A[0], B[M/2 – 2]), min(A[1], 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 C implementation of the above approach.

// A C++ program to find median of two sorted arrays of // unequal sizes #include <bits/stdc++.h> 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 (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[0], B[0]); // 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[0], 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[0] ); } // 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[0], A[1], B[0], B[1]); // 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[0], B[M/2 - 1]), min(A[1], 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[0], B[M/2 - 2] ), min( A[1], 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[0]); int M = sizeof(B) / sizeof(B[0]); printf("%f", findMedian( A, N, B, M ) ); return 0; }

Output:

10

Time Complexity: O(LogM + LogN)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.