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[201]{};
    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;
            }
            seen.add(num);
        }
        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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