Formatted question description: https://leetcode.ca/all/2150.html

# 2150. Find All Lonely Numbers in the Array (Medium)

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}

**Similar Questions**:

## Solution 1. Hash Map

```
// OJ: https://leetcode.com/problems/find-all-lonely-numbers-in-the-array/
// Time: O(N)
// Space: O(U) where U is the number of unique elements in `A`
class Solution {
public:
vector<int> findLonely(vector<int>& A) {
unordered_map<int, int> m;
for (int n : A) m[n]++;
vector<int> ans;
for (int n : A) {
if (m[n] == 1 && m.count(n + 1) == 0 && m.count(n - 1) == 0) {
ans.push_back(n);
}
}
return ans;
}
};
```

## Discuss

https://leetcode.com/problems/find-all-lonely-numbers-in-the-array/discuss/1711230/C%2B%2B-Hash-Map-O(N)