QUESTION:
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
EXPLANATION:
直接一个for循环即可.
SOLUTION:
class Solution {
func unequalTriplets(_ nums: [Int]) -> Int {
var result = 0
for i in 0...nums.count - 3 {
for j in i+1...nums.count - 2 {
for k in j+1...nums.count - 1 {
if nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k] {
result += 1
}
}
}
}
return result
}
}