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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 <= 105
• 0 <= nums[i] <= 106

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;
}
};

• class Solution:
def findLonely(self, nums: List[int]) -> List[int]:
counter = Counter(nums)
ans = []
for num, cnt in counter.items():
if cnt == 1 and counter[num - 1] == 0 and counter[num + 1] == 0:
ans.append(num)
return ans

############

# 2150. Find All Lonely Numbers in the Array
# https://leetcode.com/problems/find-all-lonely-numbers-in-the-array/

class Solution:
def findLonely(self, nums: List[int]) -> List[int]:
counter = Counter(nums)
res = []

for x in counter.keys():
if counter[x] == 1 and x - 1 not in counter and x + 1 not in counter:
res.append(x)

return res


• class Solution {
public List<Integer> findLonely(int[] nums) {
Map<Integer, Integer> counter = new HashMap<>();
for (int num : nums) {
counter.put(num, counter.getOrDefault(num, 0) + 1);
}
List<Integer> ans = new ArrayList<>();
counter.forEach((k, v) -> {
if (v == 1 && !counter.containsKey(k - 1) && !counter.containsKey(k + 1)) {
}
});
return ans;
}
}

• func findLonely(nums []int) []int {
counter := make(map[int]int)
for _, num := range nums {
counter[num]++
}
var ans []int
for k, v := range counter {
if v == 1 && counter[k-1] == 0 && counter[k+1] == 0 {
ans = append(ans, k)
}
}
return ans
}

• function findLonely(nums: number[]): number[] {
let hashMap: Map<number, number> = new Map();
for (let num of nums) {
hashMap.set(num, (hashMap.get(num) || 0) + 1);
}
let ans: Array<number> = [];
for (let [num, count] of hashMap.entries()) {
if (count == 1 && !hashMap.get(num - 1) && !hashMap.get(num + 1)) {
ans.push(num);
}
}
return ans;
}



## Discuss

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