Simple Python 3 solution

Here We are using XOR operator (^) so as to identify unique element in the array. Give a like as support and learn

def lonelyinteger(a):

unique=0
for i in a:
    unique ^=i
return(unique)

yo guys this is simpler and more effecient.....

from collections import Counter as c

at #write your code replace it with res=c(a) for i in a: if res[i]==1: return i

The solution to the problem is by summing xor values;

return std::accumulate(a.begin(), a.end(), 0, std::bit_xor<int>{});

** int lonelyinteger(vector a) { map freq; for(int i=0;i for(auto i:freq) { if(i.second == 1) { return (i.first); }

}
return -1;

} **

Save time and space I believe

int result = 0; Map count = new HashMap<>();

for(int num : a) { if(count.containsKey(num)) { count.remove(num); } else { count.put(num, 0); } } for(Map.Entry entry : count.entrySet()) { result = entry.getKey(); }

    return result;