How do you solve Find All Lonely Numbers in the Array on LeetCode?

Find All Lonely Numbers in the Array (LeetCode #2150) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Lonely numbers must appear exactly once.

What is the brute force solution for Find All Lonely Numbers in the Array?

The brute force approach for Find All Lonely Numbers in the Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each number in the array to see if it is lonely by iterating through the entire array multiple times. This is straightforward but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find All Lonely Numbers in the Array?

Find All Lonely Numbers in the Array uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find All Lonely Numbers in the Array?

Explain your thought process clearly. Consider edge cases, such as arrays with all duplicates. Discuss time and space complexities as you go.

What are common mistakes when solving Find All Lonely Numbers in the Array?

Not checking for both adjacent numbers. Confusing frequency count with presence check.

#2150

Find All Lonely Numbers in the Array

Medium

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a HashMap allows us to efficiently count occurrences of each number and then check for adjacent numbers in constant time. This significantly reduces the number of operations needed compared to the brute force approach.

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Algorithm

4 steps

1Step 1: Create a frequency map to count occurrences of each number in the array.
2Step 2: Create a set from the array to allow O(1) lookups for adjacent numbers.
3Step 3: Iterate through the frequency map and for each number, check if it appears once and if its adjacent numbers are absent in the set.
4Step 4: Collect all such numbers into the result list.

solution.py9 lines

1def findLonelyNumbers(nums):
2    from collections import Counter
3    freq = Counter(nums)
4    num_set = set(nums)
5    lonely_numbers = []
6    for num in freq:
7        if freq[num] == 1 and (num - 1) not in num_set and (num + 1) not in num_set:
8            lonely_numbers.append(num)
9    return lonely_numbers

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a couple of times (once for counting and once for checking conditions). The space complexity is O(n) due to the storage of the frequency map and the set.

1Lonely numbers must appear exactly once.
2Adjacent numbers must not be present in the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.