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The 2 Missing Duplicate Numbers - Medium


This is a slightly trickier version of The Missing Duplicate Number. Instead of one single missing number, we have 2.

Problem : Given an array of Integers. Each number is repeated even number of times. Except 2 numbers which occur odd number of times. Find the two numbers Ex. [1,2,3,1,2,3,4,5] The Output should be 4,5.


The naive solution is similar to the previous solution. Use a hash table to keep track of the frequencies and find the elements that occur an odd number of times. The algorithm has a Time and Space Complexity of O(n).

Say those two required numbers are a and b
If you remember the previous post, we used the XOR over all the elements to find the required element. If we do the same here, we get a value xor which is actually a ^ b.

So how can we use this? We know that a and b are different, so their xor is not zero. Infact their XOR value has its bit set for all its dissimilar bits in both a and b. (eg. 1101 ^ 0110 = 0011).

It's important to notice here that each bit set in xor corresponds to either a or b, but not both. So we isolate a single bit from xor  , say mask, and run the original XOR algorithm only for those number whose bit  is set the same as mask. This list of elements will either have a or b. depending on whether the bit was set due to a or b. So the xor algorithm will return one of the numbers,say xor_1. To obtain the other we simply xor ^ xor_1.

 Example:
[1,2,3,4,5,1,2,3]

xor = 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 1 ^ 2 ^3 = 1 ( = 4 ^ 5)

 Notice, 4 = 0100  and 5 = 0101 , xor = 0001.

In this case, since there's just one bit set, xor itself is the mask.
Now, we calculate xor_1. Ie XOR all number whose (v & mask) != 0

xor_1 = 1 ^ 3 ^ 5 ^ 1 ^ 3 = 5
xor_2 = xor ^ xor_1 = 1 ^ 5 = 4

We obtain (4,5) which occurred odd number of times.


import random
r = {random.randint(10,100) for i in xrange(10)}
lst = random.sample(r,len(r))
l = lst[:2] + lst[2:] * 2
random.shuffle(l)
def missing_2_numbers(lst):
xor = reduce(lambda a,b : a ^ b, lst)
mask = (xor & -xor) #get the right most set bit.
xor_1 = 0
for v in lst:
if v & mask:
xor_1 = xor_1 ^ v
return (xor_1,xor ^ xor_1)
print l
print missing_2_numbers(l)
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