#### QUESTION:

You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:

i < j < k, nums[j] - nums[i] == diff, and nums[k] - nums[j] == diff. Return the number of unique arithmetic triplets.

**Example 1:**

```
Input: nums = [0,1,4,6,7,10], diff = 3
Output: 2
Explanation:
(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.
(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.
```

**Example 2:**

```
Input: nums = [4,5,6,7,8,9], diff = 2
Output: 2
Explanation:
(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.
(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.
```

**Constraints:**

```
3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums is strictly increasing.
```

#### EXPLANATION:

easy的题目， 三个for循环嵌套就可以。

#### SOLUTION:

```
class Solution {
func arithmeticTriplets(_ nums: [Int], _ diff: Int) -> Int {
var result:Int = 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[j] - nums[i] == diff && nums[k] - nums[j] == diff {
result += 1
}
}
}
}
return result
}
}
```