How do you solve Number of Unequal Triplets in Array on LeetCode?

Number of Unequal Triplets in Array (LeetCode #2475) 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: Count distinct numbers efficiently using a hashmap.

What is the brute force solution for Number of Unequal Triplets in Array?

The brute force approach for Number of Unequal Triplets in Array is Brute Force, with O(n³) time complexity and O(1) space complexity. We can simply check all possible triplets in the array using three nested loops. This approach is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Unequal Triplets in Array?

Number of Unequal Triplets in Array uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Unequal Triplets in Array?

Explain your thought process clearly. Discuss the brute-force approach first before optimizing. Be ready to justify your choice of data structures.

What are common mistakes when solving Number of Unequal Triplets in Array?

Not checking for distinctness properly. Overlooking the need for combinatorial calculations.

#2475

Number of Unequal Triplets in Array

Easy

ArrayHash TableSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of checking all triplets, we can count the occurrences of each number and use combinatorial math to find the number of distinct triplets. This is more efficient and leverages the properties of combinations.

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Algorithm

4 steps

1Step 1: Count the frequency of each distinct number in the array using a hashmap.
2Step 2: Calculate the total number of triplets that can be formed from the array using the formula nC3 = n! / (3!(n-3)!).
3Step 3: Subtract the number of triplets that can be formed with the same numbers (using the frequencies from the hashmap).
4Step 4: Return the final count of distinct triplets.

solution.py14 lines

1# Full working Python code
2
3def countTriplets(nums):
4    from collections import Counter
5    freq = Counter(nums)
6    totalTriplets = len(nums) * (len(nums) - 1) * (len(nums) - 2) // 6
7    sameTriplets = 0
8    for count in freq.values():
9        if count >= 3:
10            sameTriplets += count * (count - 1) * (count - 2) // 6
11    return totalTriplets - sameTriplets
12
13# Example usage
14print(countTriplets([4,4,2,4,3]))  # Output: 3

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Complexity note: The time complexity is O(n) because we traverse the array once to count frequencies and then again to calculate the triplets. The space complexity is O(n) due to the hashmap storing the frequencies of distinct numbers.

1Count distinct numbers efficiently using a hashmap.
2Use combinatorial math to calculate triplet combinations.

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