Count Inversions with MergeSort

import sys

def mergecount(b,c):
    N = len(b) + len(c)
    ret = [0]*N

    bi = 0
    cj = 0
    num_inv = 0
    for k in range(0,N):
        if bi == len(b):
            assert cj < len(c)
            ret[k] = c[cj]
            cj+=1

        elif cj == len(c):
            assert bi < len(b)
            ret[k] = b[bi]
            bi += 1

        elif b[bi] < c[cj]:
            ret[k] = b[bi]
            bi +=1

        else:
            ret[k] = c[cj]
            cj+=1
            num_inv += len(b)- bi

    b = c = None
    #print 'num_inv:', num_inv, ret
    return num_inv, ret

def sortcount(arr):

    N = len(arr)
    #print N, arr

    if N==1:
        return 0, arr
    else:
        na, a = sortcount(arr[:N/2])
        nb, b = sortcount(arr[N/2:])
        nc, c = mergecount(a,b)
        return na+nb+nc, c


def main(argv):
    fp = open(argv[1],'rt')
    arr = []
    for L in fp:
        arr.append(int(L))

    begin = 0
    end = 10 # len(arr)
    num_inv, arr = sortcount(arr)
    print num_inv, len(arr)

if __name__ == '__main__':
    main(sys.argv)