JS: Finding unpaired elements in an array

Question

I have the following question (this is not school -- just code site practice questions) and I can't see what my solution is missing.

A non-empty array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

Assume that:

• *N is an odd integer within the range [1..1,000,000];
• *each element of array A is an integer within the range [1..1,000,000,000];
• *all but one of the values in A occur an even number of times.

EX: A = [9,3,9,3,9,7,9] Result: 7

The official solution is using the bitwise XOR operator :

function solution(A) {
    
    var agg = 0;
    
    for(var i=0; i<A.length; i++) {
        agg ^= A[i];
    }
    
    return agg;
}

My first instinct was to keep track of the occurrences of each value in a Map lookup table and returning the key whose only value appeared once.

function solution(A) {

    if (A.length < 1) {return 0}
    let map = new Map();
    let res = A[0]
    for (var x = 0; x < A.length; x++) {
        if (map.has(A[x])) {
            map.set(A[x], map.get(A[x]) + 1)
        } else {
            map.set(A[x], 1)
        }
    }
    for ([key,value] of map.entries()) {
        if (value===1) {
            res = key
        } 
    }

    return res;
}

I feel like I handled any edge cases but I'm still failing some tests and it's coming back with a 66% correct by the automated scorer.

Answer 1

You could use a Set and check if deletion deletes an item or not. If not, then add the value to the set.

function check(array) {
    var s = new Set;
    
    array.forEach(v => s.delete(v) || s.add(v));
    
    return s.values().next().value;
}

console.log(check([9, 3, 9, 7, 3, 9, 9])); // 7

Answer 2

You're not accounting for cases like this:

[ 1, 1, 2, 2, 2 ] => the last 2 is left unpaired

So your condition should be if ( value % 2 ) instead of if ( value === 1 ).

I think also there is not much benefit to using a Map rather than just a plain object.