Find duplicate element in array in time O(n)

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余生分开走
余生分开走 2020-11-27 10:07

I have been asked this question in a job interview and I have been wondering about the right answer.

You have an array of numbers from 0 to n-1, one o

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  •  清酒与你
    2020-11-27 11:07

    Scan the array 3 times:

    1. XOR together all the array elements -> A. XOR together all the numbers from 0 to N-1 -> B. Now A XOR B = X XOR D, where X is the removed element, and D is the duplicate element.
    2. Choose any non-zero bit in A XOR B. XOR together all the array elements where this bit is set -> A1. XOR together all the numbers from 0 to N-1 where this bit is set -> B1. Now either A1 XOR B1 = X or A1 XOR B1 = D.
    3. Scan the array once more and try to find A1 XOR B1. If it is found, this is the duplicate element. If not, the duplicate element is A XOR B XOR A1 XOR B1.

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