I came across a fascinating code snippet that helps find a unique number in a list of duplicate numbers (where each number appears twice, except for one).
function findNonPaired(listOfNumbers) {
let nonPairedNumber = 0
listOfNumbers.forEach((n) => {
nonPairedNumber ^= n
})
return nonPairedNumber
}
const x = [1,5,4,3,9,2,3,1,4,5,9]
console.log(findNonPaired(x))This solution seems quite elegant, but I'm curious about the functionality of the ^= operator in this context.
x ^= y is equivalent to x = x ^ y where the operator ^ represents bitwise XOR.
0 ^ 0 === 0
1 ^ 0 === 1
0 ^ 1 === 1
1 ^ 1 === 0
This algorithm is interesting. Let's walk through an example using the sequence [8, 12, 8]:
0 ^ 8 = 0000 ^ 1000 = 1000
1000 ^ 1100 = 0100
0100 ^ 1000 = 1100 = 12
The term "duplicate" may be misleading. The algorithm actually checks for parity. A simple explanation could be "when everything forms a pair". pair...parity.
In the case of [2,2,2,3,3,5,5,5,5], the result will be 2 because all other elements form pairs.
For [3,4,5], the output will be 2 (011^100^101 -> 010) as that is the xor of the unpaired numbers.
As mentioned by other responses, the operation being performed is a bitwise XOR.
In terms of the algorithm itself, it works well when dealing with duplicates in even numbers. When a number is XOR-ed with x, and then XOR-ed again with x, it will revert back to its original value. However, if a number appears for the third time in this sequence, it can disrupt the algorithm.
Additionally, if there is another unique value within the chain such as:
a, b, c, X, a, c, b, Y
the final result will be (X ^ Y), making it difficult to determine if there is only one unique value or more than one.
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