This problem is most optimally solved using bit manipulation. The XOR logical function returns 1 when both inputs are diferent and 0 when they're the same, so 3 ^ 3 = 0 because in 11 xor 11 all digits are the same and they cancel out, this way we'll end up with the following: 0 ^ {lonelyInteger} which is the lonelyInteger itself: Java code:

    public static int lonelyinteger(List<Integer> a) {
    // Write your code here
        int res = 0;
        for(Integer i: a) {
            res ^= i;
        }
        return res;
    }