Number of Arithmetic Triplets

// Accepted solution for LeetCode #2367: Number of Arithmetic Triplets
class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j + 1; k < n; ++k) {
                    if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) {
                        ++ans;
                    }
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2367: Number of Arithmetic Triplets
func arithmeticTriplets(nums []int, diff int) (ans int) {
	n := len(nums)
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			for k := j + 1; k < n; k++ {
				if nums[j]-nums[i] == diff && nums[k]-nums[j] == diff {
					ans++
				}
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2367: Number of Arithmetic Triplets
class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        return sum(b - a == diff and c - b == diff for a, b, c in combinations(nums, 3))

// Accepted solution for LeetCode #2367: Number of Arithmetic Triplets
/**
 * [2367] Number of Arithmetic Triplets
 *
 * 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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-arithmetic-triplets/
// discuss: https://leetcode.com/problems/number-of-arithmetic-triplets/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn arithmetic_triplets(nums: Vec<i32>, diff: i32) -> i32 {
        nums.iter().fold(0, |acc, &x| {
            if nums.contains(&(x + diff)) && nums.contains(&(x + diff * 2)) {
                acc + 1
            } else {
                acc
            }
        })
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2367_example_1() {
        let nums = vec![0, 1, 4, 6, 7, 10];
        let diff = 3;

        let result = 2;

        assert_eq!(Solution::arithmetic_triplets(nums, diff), result);
    }

    #[test]
    fn test_2367_example_2() {
        let nums = vec![4, 5, 6, 7, 8, 9];
        let diff = 2;

        let result = 2;

        assert_eq!(Solution::arithmetic_triplets(nums, diff), result);
    }
}

// Accepted solution for LeetCode #2367: Number of Arithmetic Triplets
function arithmeticTriplets(nums: number[], diff: number): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            for (let k = j + 1; k < n; ++k) {
                if (nums[j] - nums[i] === diff && nums[k] - nums[j] === diff) {
                    ++ans;
                }
            }
        }
    }
    return ans;
}