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;

// Using map --> create map // key--> item, value --> occurences of item in input array (a) // iterate through map --> return key whose value is 1

let numMap = new Map()

a.forEach(item=> {
    if(numMap.has(item)){
        numMap.set(item,numMap.get(item)+1)
    } else {
        numMap.set(item,1)
    }
})

for (let [key,value] of numMap) {
    if(value===1) { return key}
}