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 <= 1050 <= nums[i] <= 106
Similar Questions:
// OJ: https://leetcode.com/problems/find-all-lonely-numbers-in-the-array/
// Author: github.com/lzl124631x
// 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;
}
};