2475. Number of Unequal Triplets in Array

Problem

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

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

• In other words, nums[i] != nums[j], nums[i] != nums[k], and nums[j] != nums[k].

Return the number of triplets that meet the conditions.

  Example 1:

Input: nums = [4,4,2,4,3]
Output: 3
Explanation: The following triplets meet the conditions:
- (0, 2, 4) because 4 != 2 != 3
- (1, 2, 4) because 4 != 2 != 3
- (2, 3, 4) because 2 != 4 != 3
Since there are 3 triplets, we return 3.
Note that (2, 0, 4) is not a valid triplet because 2 > 0.

Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: No triplets meet the conditions so we return 0.

  Constraints:

• 3 <= nums.length <= 100

• 1 <= nums[i] <= 1000

Solution (Java)

class Solution {
    public int unequalTriplets(int[] nums) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] == nums[j]) {
                    continue;
                }
                for (int k = j + 1; k < nums.length; k++) {
                    if (nums[j] == nums[k]) {
                        continue;
                    }
                    if (nums[i] != nums[j] && nums[i] != nums[k] && nums[j] != nums[k]) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).