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Leetcode 2475. Number of Unequal Triplets in Array

Problem Statement

You are given a 0-indexed array of positive integers nums. A triplet of indices (i, j, k) is called unequal if:

0 <= i < j < k < nums.length, and
nums[i], nums[j], and nums[k] are pairwise distinct.

Return the number of unequal triplets in nums.

Clarifying Questions

  1. Can the array contain duplicate values? Yes, the array can contain positive integers, and duplicates are allowed.

  2. What is the expected range of the array length and the values in it? The length of nums will be between 1 and 1000, and each integer in nums will be between 1 and 1000.

  3. How should the function behave if the array length is less than 3? If the length of nums is less than 3, it should return 0, as it’s not possible to form a triplet.

Strategy

  1. Initial Check: If the length of nums is less than 3, return 0 immediately.
  2. Brute Force Approach: Iterate through all possible triplets using three nested loops and check if the elements are distinct.
  3. Count Valid Combos: For each valid triplet, where the elements at indices i, j, and k are distinct, increment a counter.
  4. Return the Counter after iterating through all possible triplets.

This brute-force approach is feasible given the constraints (0 <= len(nums) <= 1000), making the maximum number of combinations ( \binom{1000}{3} \approx 166,167,000 ), which should be manageable.

Code

#include <vector>
#include <unordered_set>
using namespace std;

int unequalTriplets(vector<int>& nums) {
    int n = nums.size();
    if (n < 3) return 0;
    
    int count = 0;
    
    // Iterate through all possible triplets
    for (int i = 0; i < n - 2; ++i) {
        for (int j = i + 1; j < n - 1; ++j) {
            for (int k = j + 1; k < n; ++k) {
                if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) {
                    ++count;
                }
            }
        }
    }
    
    return count;
}

Time Complexity

The brute force approach should be acceptable given the constraints and allows us to directly count and verify the distinct property required for the triplets.

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