#2150

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find All Lonely Numbers in the Array

Find lonely numbers in an array where each lonely number appears once and has no adjacent values.

Problem Statement

You are given an integer array nums. A number x is considered lonely if it appears exactly once in the array and neither x - 1 nor x + 1 exists in the array. Your task is to return all lonely numbers found in the given array. The answer can be returned in any order.

For example, given nums = [10, 6, 5, 8], the correct output is [10, 8] since both 10 and 8 appear exactly once, with no adjacent values in the array. On the other hand, 5 is not lonely because 6 is present in the array, which is adjacent to 5.

Examples

Example 1

Input: nums = [10,6,5,8]

Output: [10,8]

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]

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

Solution Approach

Using Hash Table for Fast Lookup

A hash table can be used to store the frequency of each element in the array. By scanning the array once, we can then check if the adjacent numbers (x-1 and x+1) for any number are present in the hash table. This method ensures we efficiently check each element without unnecessary reiterations.

Single Pass with Adjacent Check

Another approach is to first count the occurrences of each number in a hash table. Then, as we traverse the array, we check if a number appears once and if neither its previous (x-1) nor next (x+1) number is in the hash table. This two-step process eliminates the need for double iteration, achieving optimal time complexity.

Handling Edge Cases

Edge cases such as arrays with duplicate elements, arrays with only one element, or arrays where all numbers are lonely need to be carefully handled. The solution should ensure the correct identification of lonely numbers in various edge scenarios, including when adjacent values are sparse.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n), where n is the length of the input array, due to the single pass to fill the hash table and another pass to check the lonely numbers. The space complexity is O(n) due to the hash table used for counting the occurrences of each element in the array.

What Interviewers Usually Probe

Look for an efficient way to handle large arrays by minimizing unnecessary iterations.
Test for understanding of hash table usage in combination with array scanning techniques.
Assess the candidate's ability to optimize for time complexity without overcomplicating the solution.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the problem by using nested loops or inefficient searching techniques.
Forgetting to check both the presence of a number and its adjacent values in the array.
Not handling edge cases such as arrays with a single element or duplicate values properly.

Follow-up variants

What if you need to find lonely numbers in a sorted array?
How would the solution change if the numbers in the array can be negative?
Can you optimize the solution further if the array is known to have a limited range of numbers?

FAQ

What is the primary approach to solving the 'Find All Lonely Numbers in the Array' problem?

The primary approach involves using a hash table to count occurrences and check adjacent numbers, ensuring efficient lookup and single-pass scanning.

How can you improve the time complexity for the 'Find All Lonely Numbers' problem?

By using a hash table for counting and checking adjacent values, the solution can be optimized to O(n) time complexity, where n is the array length.

What should you do when dealing with edge cases in this problem?

Edge cases like arrays with a single element or duplicates should be handled by ensuring the lonely number condition is still met and no adjacent numbers exist.

Can the solution for finding lonely numbers be adapted for sorted arrays?

Yes, a sorted array can simplify the checking of adjacent values, allowing for potential optimizations that reduce the space complexity.

How does GhostInterview help in solving the 'Find All Lonely Numbers in the Array' problem?

GhostInterview offers hints for optimizing your approach and checks for edge cases, ensuring a correct and efficient solution using array scanning and hash table lookup.

terminal

Solution

Solution 1: Hash Table

We use a hash table $\textit{cnt}$ to record the occurrence count of each number. Then, we iterate through the hash table. For each number and its occurrence count $(x, v)$, if $v = 1$ and $\textit{cnt}[x - 1] = 0$ and $\textit{cnt}[x + 1] = 0$, then $x$ is a lonely number, and we add it to the answer array.

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

Continue Practicing

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