find median of two sorted arrays

find_median.py

# From: https://www.geeksforgeeks.org/median-two-sorted-arrays-different-sizes-ologminn-m/
# def to find median of two sorted arrays 
def findMedianSortedArrays(A, len_A, B, len_B):
    median = 0
    index_in_A = 0
    index_in_B = 0
    min_index = 0 
    max_index = len_A
    half_length = (len_A + len_B + 1) // 2

    while (min_index <= max_index) : 
      
        index_in_A = (min_index + max_index) // 2
        index_in_B = half_length - index_in_A
       
        # if index_in_A = len_A, it means that  
        # Elements from A[] in the 
        # second half is an empty  
        # set. and if index_in_B = 0, it  
        # means that Elements from  
        # B[] in the first half is  
        # an empty set. so it is  
        # necessary to check that,  
        # because we compare elements  
        # from these two groups.  
        # Searching on right 
        if (index_in_A < len_A and index_in_B > 0 and B[index_in_B - 1] > A[index_in_A]) : 
            min_index = index_in_A + 1
                   
        # if index_in_A = 0, it means that  
        # Elements from A[] in the 
        # first half is an empty  
        # set and if index_in_B = len_B, it means 
        # that Elements from B[] in  
        # the second half is an empty  
        # set. so it is necessary to 
        # check that, because we compare  
        # elements from these two groups. 
        # searching on left 
        elif (index_in_A > 0 and index_in_B < len_B and B[index_in_B] < A[index_in_A - 1]) : 
            max_index = index_in_A - 1
           
        # we have found the 
        # desired halves. 
        else : 
          
            # this condition happens when  
            # we don't have any elements  
            # in the first half from A[]  
            # so we returning the last 
            # element in B[] from the  
            # first half. 
            if (index_in_A == 0) : 
                median = B[index_in_B - 1] 
            # and this condition happens  
            # when we don't have any  
            # elements in the first half  
            # from B[] so we returning the  
            # last element in A[] from the  
            # first half. 
            elif (index_in_B == 0) : 
                median = A[index_in_A - 1]
            else : 
                median = max(A[index_in_A - 1], B[index_in_B - 1])
            break

    # calculating the median. 
    # If number of elements  
    # is odd there is  
    # one middle element. 
   
    if ((len_A + len_B) % 2 == 1) : 
        return median 
   
    # Elements from A[] in the  
    # second half is an empty set.  
    if (index_in_A == len_A) : 
        return ((median + B[index_in_B]) / 2.0) 
   
    # Elements from B[] in the  
    # second half is an empty set. 
    if (index_in_B == len_B) : 
        return ((median + A[index_in_A]) / 2.0) 
       
    return ((median + min(A[index_in_A], B[index_in_B])) / 2.0) 

if __name__ == "__main__":
    # Driver code 
    A = [1, 5, 11, 17] 
    B = [10, 13, 14, 60] 
    len_A = len(A) 
    len_B = len(B) 
       
    # we need to define the  
    # smaller array as the  
    # first parameter to make  
    # sure that the time complexity 
    # will be O(log(min(len_A,len_B))) 
    if (len_A <= len_B) : 
        print ("The median is : {}".format(findMedianSortedArrays(A, len_A, B, len_B))) 
    else : 
        print ("The median is : {}".format(findMedianSortedArrays(B, len_B, A, len_A)))