public static int lonelyinteger(List<Integer> a) {
        // Sort list
        Collections.sort(a);
        //   Return first elem if size is one or second elem doesn't match
        int ans=a.get(0);
        if(a.size()==1 || a.get(0) != a.get(1))
                return ans;
        // Iterate through list
        for(int i=1;i<a.size()-1;i++) {
        // Compare num before and after from index 1 - size-1
                if(a.get(i)!= a.get(i-1) && a.get(i)!= a.get(i+1)){
        // If unequal return elem
                        return a.get(i);                
                }
        }
        // If each elem has match but last return last
        return a.get(a.size()-1);            
}

List< Integer> passedNumbers = new ArrayList<>(); for (int i = 0; i a.size(); i++) { if(!passedNumbers.contains(a.get(i))) { passedNumbers.add(a.get(i)); } else { passedNumbers.remove(a.get(i)); } } return passedNumbers.get(0)

JAVA

for(Integer n: a){ int frec = Collections.frequency(a, n); if(frec==1){ num=n; } } return num;

Scala Solution

def lonelyinteger(a: Array[Int]): Int = {
    // Write your code here)
        a.reduce(_^_)
    }

My Python solution using Mathematical logic:

def lonelyinteger(a):
    '''
    Mathematical solution
    
    The sum of the elements of a is the sum of the
    doubles of each repeated element, plus the lonely (b)
    sum(a) = sum[2.x] + b
    
    The sum of the unique elements is the sum of all
    the unique elements that are repeated, plus the lonely (b)
    sum(set(a)) = sum[x] + b
    
    Then, 2*sum(set(a)) - sum(a) = b
    
    Time complexity: O(n)
    Space complexity: O(1)
    '''
    
    return 2*sum(set(a)) - sum(a)