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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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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