How do you solve Smallest Pair With Different Frequencies on LeetCode?

Smallest Pair With Different Frequencies (LeetCode #3852) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Use frequency counting to simplify comparisons.

What is the time complexity of Smallest Pair With Different Frequencies?

The optimal solution for Smallest Pair With Different Frequencies runs in O(n log n) time and uses O(n) space. The complexity is linearithmic due to sorting the unique elements.

What is the brute force solution for Smallest Pair With Different Frequencies?

The brute force approach for Smallest Pair With Different Frequencies is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all pairs of distinct values and compare their frequencies. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Smallest Pair With Different Frequencies?

Smallest Pair With Different Frequencies uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Pair With Different Frequencies?

Clarify constraints and edge cases. Explain your thought process while coding. Discuss time and space complexities.

What are common mistakes when solving Smallest Pair With Different Frequencies?

Not considering distinct values properly. Forgetting to handle cases with no valid pairs.

#3852

Smallest Pair With Different Frequencies

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Use a frequency map to count occurrences and then find the smallest pair with different frequencies efficiently.

⚙️

Algorithm

3 steps

1Step 1: Count frequencies of each number using a hash map.
2Step 2: Sort the unique numbers based on their values.
3Step 3: Iterate through sorted unique numbers to find the first pair with different frequencies.

solution.py8 lines

1def smallest_pair(nums):
2    from collections import Counter
3    freq = Counter(nums)
4    unique = sorted(freq.keys())
5    for i in range(len(unique) - 1):
6        if freq[unique[i]] != freq[unique[i + 1]]:
7            return [unique[i], unique[i + 1]]
8    return [-1, -1]

ℹ

Complexity note: The complexity is linearithmic due to sorting the unique elements.

1Use frequency counting to simplify comparisons.
2Sorting helps in efficiently finding pairs.

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