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# first-unique-number

## Problem Description

You have a queue of integers, you need to retrieve the first unique integer in the queue.
Implement the FirstUnique class:
FirstUnique(int[] nums) Initializes the object with the numbers in the queue.
int showFirstUnique() returns the value of the first unique integer of the queue, and returns -1 if there is no such integer.
void add(int value) insert value to the queue.
Example 1:
Input:
[[[2,3,5]],[],[5],[],[2],[],[3],[]]
Output:
[null,2,null,2,null,3,null,-1]
Explanation:
FirstUnique firstUnique = new FirstUnique([2,3,5]);
firstUnique.showFirstUnique(); // return 2
firstUnique.add(5); // the queue is now [2,3,5,5]
firstUnique.showFirstUnique(); // return 2
firstUnique.add(2); // the queue is now [2,3,5,5,2]
firstUnique.showFirstUnique(); // return 3
firstUnique.add(3); // the queue is now [2,3,5,5,2,3]
firstUnique.showFirstUnique(); // return -1
Example 2:
Input:
[[[7,7,7,7,7,7]],[],[7],[3],[3],[7],[17],[]]
Output:
[null,-1,null,null,null,null,null,17]
Explanation:
FirstUnique firstUnique = new FirstUnique([7,7,7,7,7,7]);
firstUnique.showFirstUnique(); // return -1
firstUnique.add(7); // the queue is now [7,7,7,7,7,7,7]
firstUnique.add(3); // the queue is now [7,7,7,7,7,7,7,3]
firstUnique.add(3); // the queue is now [7,7,7,7,7,7,7,3,3]
firstUnique.add(7); // the queue is now [7,7,7,7,7,7,7,3,3,7]
firstUnique.add(17); // the queue is now [7,7,7,7,7,7,7,3,3,7,17]
firstUnique.showFirstUnique(); // return 17
Example 3:
Input:
[[[809]],[],[809],[]]
Output:
[null,809,null,-1]
Explanation:
FirstUnique firstUnique = new FirstUnique([809]);
firstUnique.showFirstUnique(); // return 809
firstUnique.add(809); // the queue is now [809,809]
firstUnique.showFirstUnique(); // return -1
Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^8
1 <= value <= 10^8

## Solution

Using HashMap to keep track of number and the frequency, use Queue to store number in order. (FIFO -- first in first out).
when add(value), add value into Map, and check current value's frequency, if map.get(value) > 1, meaning not unique, not need to add into Queue.
when showFirstUnique(), check whether queue is empty:
• if queue.isEmpty(), then return -1.
• if queue is not empty, check queue's first value's frequency:
• if first value's frequency > 1, then discard, until first value's frequency == 1, then return first value in queue.
There are other solutions like using LRU/LFU problem, using doubly linkedlist + hashmap to make all operation in `O(1)` time complexity.
For example:
First Unique Number

### Complexity Analysis

Time Complexity:
• add(value) - `O(1)`
• showFirstUnique() - `O(N)`
Space Complexity: `O(N)`
• N - the length of array nums

### Code

class FirstUnique {
private Map<Integer, Integer> map;
private Queue<Integer> queue;
public FirstUnique(int[] nums) {
map = new HashMap<>();
for (int num : nums) {
}
}
public int showFirstUnique() {
if (queue.isEmpty()) return -1;
while (!queue.isEmpty() && map.get(queue.peek()) > 1) {
queue.poll();
}
return queue.isEmpty() ? -1 : queue.peek();
}