// 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 );
}
float findMedianUtil( int A[], int N, int B[], int 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 )//leetcode acepted !
{
if (N == 0)
return getMedian(B,M);
if(M == 0)
return getMedian(A,N);
if ( N > M )
return findMedianUtil( B, M, A, N );
return findMedianUtil( A, N, B, M );
}
