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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.

```
int arithmeticTriplets(vector<int>& nums, int diff) {
int dp[201]{};
int res=0;
for(auto i:nums){
dp[i]= i>=diff? dp[i-diff]+1 : 1;
if(dp[i]>=3) res++;
}
return res;
}
```

```
public int arithmeticTriplets(int[] nums, int diff) {
int cnt = 0;
Set<Integer> seen = new HashSet<>();
for (int num : nums) {
if (seen.contains(num - diff) && seen.contains(num - diff * 2)) {
++cnt;
}
seen.add(num);
}
return cnt;
}
```

```
class Solution:
def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
res = 0
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[j] - nums[i] == diff:
for k in range(j+1, len(nums)):
if nums[k] - nums[j] == diff:
res += 1
return res
```

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