2475. Number of Unequal Triplets in Array

Description

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

Solutions

Solution 1: Brute Force Enumeration

We can directly enumerate all triples $(i, j, k)$ and count all the ones that meet the conditions.

The time complexity is $O(n^3)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def unequalTriplets(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    ans += (
                        nums[i] != nums[j] and nums[j] != nums[k] and nums[i] != nums[k]
                    )
        return ans

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

class Solution {
public:
    int unequalTriplets(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j + 1; k < n; ++k) {
                    if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) {
                        ++ans;
                    }
                }
            }
        }
        return ans;
    }
};

func unequalTriplets(nums []int) (ans int) {
    n := len(nums)
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            for k := j + 1; k < n; k++ {
                if nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k] {
                    ans++
                }
            }
        }
    }
    return
}

function unequalTriplets(