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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 $O(1)$.

Solution 2: Sorting + Enumeration of Middle Elements + Binary Search

We can also sort the array $nums$ first.

Then traverse $nums$, enumerate the middle element $nums[j]$, and use binary search to find the nearest index $i$ on the left side of $nums[j]$ such that $nums[i] < nums[j]$; find the nearest index $k$ on the right side of $nums[j]$ such that $nums[k] > nums[j]$. Then the number of triples with $nums[j]$ as the middle element and meeting the conditions is $(i + 1) \times (n - k)$, which is added to the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

Solution 3: Hash Table

We can also use a hash table $cnt$ to count the number of each element in the array $nums$.

Then traverse the hash table $cnt$, enumerate the number of middle elements $b$, and denote the number of elements on the left as $a$. Then the number of elements on the right is $c = n - a - b$. At this time, the number of triples that meet the conditions is $a \times b \times c$, which is added to the answer. Then update $a = a + b$ and continue to enumerate the number of middle elements $b$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

  • 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;
    }
};

  • 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


  • 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(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n - 2; i++) {
        for (let j = i + 1; j < n - 1; j++) {
            for (let k = j + 1; k < n; k++) {
                if (nums[i] !== nums[j] && nums[j] !== nums[k] && nums[i] !== nums[k]) {
                    ans++;
                }
            }
        }
    }
    return ans;
}


  • impl Solution {
    pub fn unequal_triplets(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = 0;
        for i in 0..n - 2 {
            for j in i + 1..n - 1 {
                for k in j + 1..n {
                    if nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k] {
                        ans += 1;
                    }
                }
            }
        }
        ans
    }
}

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