LeetCode-in-Java
2475. Number of Unequal Triplets in Array
Easy
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.lengthnums[i],nums[j], andnums[k]are pairwise distinct.- In other words,
nums[i] != nums[j],nums[i] != nums[k], andnums[j] != nums[k].
- In other words,
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 <= 1001 <= nums[i] <= 1000
Solution
public class Solution {
public int unequalTriplets(int[] nums) {
int trips = 0;
int pairs = 0;
int[] count = new int[1001];
for (int i = 0; i < nums.length; ++i) {
trips += pairs - count[nums[i]] * (i - count[nums[i]]);
pairs += i - count[nums[i]];
count[nums[i]] += 1;
}
return trips;
}
}