**Earliest Uniques in a Stream - Amazon Top Interview Questions**

### Problem Statement :

Implement a data structure with the following methods: EarliestUnique(int[] nums) constructs a new instance with the given list of numbers. add(int num) adds num to the data structure. firstUnique() returns the first unique number. If there's no unique number, return -1. Constraints n ≤ 100,000 where n is the number of calls to add and firstUnique. Example 1 Input methods = ["constructor", "add", "earliestUnique", "add", "earliestUnique"] arguments = [[[1, 2, 3]], [1], [], [2], []]` Output [None, None, 2, None, 3] Explanation e = EarliestUnique([1, 2, 3]) e.add(1) e.earliestUnique() == 2 e.add(2) e.earliestUnique() == 3

### Solution :

` ````
Solution in C++ :
class EarliestUnique {
deque<int> a;
unordered_map<int, int> c;
public:
EarliestUnique(vector<int>& nums) {
for (int x : nums) add(x);
}
void add(int x) {
a.push_back(x);
++c[x];
}
int earliestUnique() {
while (!a.empty() && c[a.front()] > 1) a.pop_front();
return a.empty() ? -1 : a.front();
}
};
```

` ````
Solution in Java :
import java.util.*;
class EarliestUnique {
private Map<Integer, Integer> map;
public EarliestUnique(int[] nums) {
map = new LinkedHashMap();
for (int num : nums) map.put(num, map.getOrDefault(num, 0) + 1);
}
public void add(int num) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
public int earliestUnique() {
for (int key : map.keySet()) {
if (map.get(key) == 1)
return key;
}
return -1;
}
}
```

` ````
Solution in Python :
class EarliestUnique:
def __init__(self, nums):
self.counter = Counter(nums)
self.queue = deque(nums)
self._balance()
def _balance(self):
while self.queue and self.counter[self.queue[0]] > 1:
self.queue.popleft()
def add(self, num):
self.queue.append(num)
self.counter[num] += 1
self._balance()
def earliestUnique(self):
return self.queue[0] if self.queue else -1
```

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