find_unfaired_num.js

function solution(A){
    if(A.length === 1)
        return A[0];
    A = A.sort();
    var xor = 0;
    for(var i=0; i<A.length; i++){
        console.log('xor' , xor^A[i], 'xor:',xor, A[i]);
        xor = xor ^ A[i];
        
    }
    return xor
}
console.log(solution([9,3,9,3,7]));

https://codility.com/programmers/lessons/14

First, I thought how I can do this in the O(1) space complexity until I notice that there is only one element left unpaired in this problem.

If all the elements except one is paired, we can use xor for this problem and just one scanning is okay. Since A xor A = 0, and B xor 0 = B, if we keep on performing xor for all the values, the one left after iteration is the value unpaired.