2150. Find All Lonely Numbers in the Array

Problem

You are given an integer array nums. A number x is lonely when it appears only once, and no adjacent numbers (i.e. x + 1 and x - 1) appear in the array.

Return **all lonely numbers in **nums. You may return the answer in *any order*.

  Example 1:

Input: nums = [10,6,5,8]
Output: [10,8]
Explanation: 
- 10 is a lonely number since it appears exactly once and 9 and 11 does not appear in nums.
- 8 is a lonely number since it appears exactly once and 7 and 9 does not appear in nums.
- 5 is not a lonely number since 6 appears in nums and vice versa.
Hence, the lonely numbers in nums are [10, 8].
Note that [8, 10] may also be returned.

Example 2:

Input: nums = [1,3,5,3]
Output: [1,5]
Explanation: 
- 1 is a lonely number since it appears exactly once and 0 and 2 does not appear in nums.
- 5 is a lonely number since it appears exactly once and 4 and 6 does not appear in nums.
- 3 is not a lonely number since it appears twice.
Hence, the lonely numbers in nums are [1, 5].
Note that [5, 1] may also be returned.

  Constraints:

• 1 <= nums.length <= 10^5

• 0 <= nums[i] <= 10^6

Solution (Java)

class Solution {
    public List<Integer> findLonely(int[] nums) {
        List<Integer> ans = new ArrayList<>();
        HashMap<Integer, Integer> m = new HashMap<>();
        for (int i : nums) {
            m.put(i, m.getOrDefault(i, 0) + 1);
        }
        for (int i : nums) {
            if (m.get(i) == 1 && !m.containsKey(i - 1) && !m.containsKey(i + 1)) {
                ans.add(i);
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).