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 :



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                        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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