2367. Number of Arithmetic Triplets

Description

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.

Solutions

• 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;
}
}

• class Solution {
public:
int arithmeticTriplets(vector<int>& nums, int diff) {
int ans = 0;
int n = nums.size();
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;
}
};

• 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))


• 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
}

• 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;
}