question about missing element in array

Question

i have following problem from book introduction algorithm second edition by MIT university

problem is following

An array A[1 . . n] contai

Answer 1

Call your missing number M.

You can split your array into two parts depending on whether the least significant bit of A[i] is a 1 or a 0. The smaller of the two parts (call it P_1) is at most (n-1)/2 elements in size, and it tells you whether M's least significant bit is a 1 or a 0.

Now consider the 2nd bit for the elements of P_1. Again, this part can be split in two, and the smaller of the two parts (P_2) tells you whether this bit should be a 1 or a 0.

Carry on going (P_3, P_4, ...) until you've worked out what all the bits are.

You can prove that this is O(n) because you are essentially looking at n + n/2 + n/4 + ... different individual bits in your array, and this sum is less than 2n.

Answer 2

Here is a Python implementation:

def bit_at(n, bit):
    return (n>>bit) & 1

def find_missing(a, bits):

    indexes = range(len(a))
    missing = 0

    for bit in range(bits):

        ones = [i for i in indexes if bit_at(a[i], bit)==1]
        zeroes = [i for i in indexes if bit_at(a[i], bit)==0]

        if len(ones) <= len(zeroes):
            indexes = ones
            missing |= (1<<bit) 
        else:
            indexes = zeroes

    return missing

print find_missing([7,2,6,4,1,5,0], 3)