Example 1:
Input: ["FirstUnique","showFirstUnique","add","showFirstUnique","add","showFirstUnique","add","showFirstUnique"]
[[[2,3,5]],[],[5],[],[2],[],[3],[]]
Output: [null,2,null,2,null,3,null,-1]
Explanation:
Example 2:
Input: ["FirstUnique","showFirstUnique","add","showFirstUnique"]
[[[809]],[],[809],[]]
Output: [null,809,null,-1]
Explanation:
You have a queue of integers, and 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 in the queue, or -1 if there is no unique integer.void add(int value) Inserts value into the queue.A number is unique if it appears exactly once among all numbers currently in the queue.
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
A number stops being unique as soon as its frequency reaches two.
1 <= nums.length <= 10^51 <= nums[i] <= 10^81 <= value <= 10^85 * 10^4 calls will be made to showFirstUnique and add.Reported as a LeetCode-style design question for software engineer phone screen and onsite rounds, with discussion centered on maintaining insertion order while updating duplicate counts efficiently.