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You are given a 0-indexed array of positive integers nums. Find the number of triplets (i, j, k) that meet the following conditions:

In other words, return the number of triplets where every element in the triplet is distinct.

Clarifying Questions

  1. Can the array contain duplicate numbers?
    • Yes, the array can contain duplicate numbers.
  2. What should be the output if the array has fewer than 3 elements?
    • The output should be 0, as triplets cannot be formed.
  3. Can the array contain negative numbers or zeros?
    • No, the array consists of positive integers only.

Strategy

To solve this problem, we can use a brute force approach to check all possible triplets and count those that satisfy the given conditions. Here’s a step-by-step strategy:

  1. Initialization: Set a counter to zero to keep track of the number of valid triplets.
  2. Triple Loop: Use three nested loops to iterate through all possible triplets in the array and check if they satisfy the conditions.
  3. Condition Check: For each triplet (i, j, k), check whether nums[i], nums[j], and nums[k] are distinct.
  4. Count Valid Triplets: Increment the counter for each valid triplet found.
  5. Return Result: Finally, return the counter.

Code

Here is the implementation of the above strategy:

def unequalTriplets(nums):
    count = 0
    n = len(nums)
    
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if nums[i] != nums[j] and nums[j] != nums[k] and nums[i] != nums[k]:
                    count += 1
    
    return count

# Example usage:
print(unequalTriplets([4, 4, 2, 4, 3]))  # Output: 3
print(unequalTriplets([1, 1, 1, 1]))     # Output: 0

Time Complexity

The time complexity of this approach is (O(n^3)), which is derived from the three nested loops. Each loop runs through the length of the array n.

Therefore, the total number of operations is (O(n^3)). This can be inefficient for large arrays, so optimizations might be required for large input sizes. However, for typical interview problems, this complexity might be acceptable depending on the size constraints of the input.

Conclusion

The provided solution correctly counts the number of triplets in the array that meet the specified conditions by explicitly checking every possible triplet. While this approach is straightforward and guarantees correctness, it may not be efficient for larger arrays. Further optimizations may need to be considered for larger datasets.

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