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# Number of Arithmetic Triplets LeetCode Solution

## Problem – Number of Arithmetic Triplets LeetCode Solution

You are given a 0-indexedstrictly 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.

### Number of Arithmetic Triplets LeetCode Solution in C++

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

### Number of Arithmetic Triplets LeetCode Solution in Java

``````    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;
}
}
return cnt;
}
``````

### Number of Arithmetic Triplets LeetCode Solution in Python

``````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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