My Java solution in O(n) time complexity and O(1) space complexity using XOR

  • 47
    public class Solution {
        public int singleNumber(int[] nums) {
            int res = 0;
            for(int num : nums) {
                res ^= num;
            return res;

  • 5

    I'm sure for a lot of people this may be confusing as to why this works. The idea hinges on 3 properties of xor. (1) that its a commutative operation (i.e. a xor b = b xor a). (2) that something xor itself is 0. So a xor a = 0. And (3) 0 xor a = a. These three properties mean that

    a xor b xor a = a xor a xor b = 0 xor b = b.

    Thus it doesn't matter the order of the numbers. If something only occurs once it won't get negated.

  • 0

    I wonder if the complexity of this algo is really O(n). It seems like O(n * length of binary number)

  • 0

    We only need to calculate the times that res ^= num; is called. We do not need to know the times of the bit operation for each bit so the time complexity is O(n).

  • 0

    nice explain, thanks

  • 0

    The XOR operation is one single machine operation, it doesn't matter the length of binary number.

    Also, in Big O(), O(n) is equivalent as O(x*n), the complexity remains linear which is what we care most of the time.

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