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 &lt; j &lt; k,
	nums[j] - nums[i] == diff, and
	nums[k] - nums[j] == diff.


Return the number of unique arithmetic triplets.

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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.


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Constraints:


	3 &lt;= nums.length &lt;= 200
	0 &lt;= nums[i] &lt;= 200
	1 &lt;= diff &lt;= 50
	nums is strictly increasing.


class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        
        ans = 0
        n = len(nums)
        for i in range(n):
            if nums[i] + diff in nums and nums[i] + 2 * diff in nums:
                ans += 1
        
        return ans
class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int result = 0;
        int[] map = new int[201];

        for(int num: nums) {
            map[num] = 1;

            if(num - diff >= 0) {
                map[num] += map[num - diff];
            }

            if(map[num] >= 3) result += 1;
        }

        return result;
    }
}
class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        int ans=0;
        for(int i=0;i<nums.size();i++){
            for(int j=i+1;j<nums.size();j++){
                for(int k=j+1;k<nums.size();k++){
                    if((nums[j]-nums[i])==diff && (nums[k]-nums[j])==diff){
                        ans++;
                    }
                }
            }
        }
        return ans;
    }
};
/**
 * @param {number[]} nums
 * @param {number} diff
 * @return {number}
 */
var arithmeticTriplets = function(nums, diff) {
    count = 0
    for(let i = 0; i < nums.length - 2; i++){
       for(let j = i + 1; j < nums.length - 1; j++){
           for(let k = j + 1; k < nums.length; k++){
            if(i < j && j < k && nums[j] - nums[i] === diff && nums[k] - nums[j] === diff){
                count++
            }
         }
      }
    }
    return count
};
Number of Arithmetic Triplets