LeetCode 1429: First Unique Number

Explanation:

To solve this problem, we can use a combination of a queue and a hashmap. The queue will store the order of appearance of each unique number, while the hashmap will store the frequency of each number.

1. Initialize a queue and a hashmap.
2. Iterate over the given array:
   • If the number is not in the hashmap, add it to the queue and set its frequency to 1 in the hashmap.
   • If the number is already in the hashmap, increment its frequency.
   • While the queue is not empty and the front element's frequency is more than 1, remove it from the queue.
3. If the queue is not empty, return the front element as the first unique number. Otherwise, return -1.

Time complexity: O(n) where n is the number of elements in the array. Space complexity: O(n) for the hashmap and the queue.

:

import java.util.*;

class FirstUnique {
    Queue<Integer> queue;
    Map<Integer, Integer> frequency;

    public FirstUnique(int[] nums) {
        queue = new LinkedList<>();
        frequency = new HashMap<>();

        for (int num : nums) {
            add(num);
        }
    }

    public int showFirstUnique() {
        while (!queue.isEmpty() && frequency.get(queue.peek()) > 1) {
            queue.poll();
        }
        return queue.isEmpty() ? -1 : queue.peek();
    }

    public void add(int value) {
        frequency.put(value, frequency.getOrDefault(value, 0) + 1);
        if (frequency.get(value) == 1) {
            queue.offer(value);
        }
    }
}