LeetCode in Kotlin

2367. Number of Arithmetic Triplets

Easy

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.

Solution

class Solution {
    fun arithmeticTriplets(nums: IntArray, diff: Int): Int {
        val set: MutableSet<Int> = HashSet()
        for (x in nums) {
            set.add(x)
        }
        var ans = 0
        for (x in nums) {
            if (set.contains(x - diff) && set.contains(x + diff)) {
                ans++
            }
        }
        return ans
    }
}

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