Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2367.html

2367. Number of Arithmetic Triplets

• Difficulty: Easy.
• Related Topics: Array, Hash Table, Two Pointers, Enumeration.
• Similar Questions: Two Sum, 3Sum.

Problem

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 (Java, C++, Python)

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

• class Solution {
      public int arithmeticTriplets(int[] nums, int diff) {
          Set<Integer> set = new HashSet<>();
          for (int x : nums) {
              set.add(x);
          }
          int ans = 0;
          for (int x : nums) {
              if (set.contains(x - diff) && set.contains(x + diff)) {
                  ans++;
              }
          }
          return ans;
      }
  }
  
  ############
  
  class Solution {
      public int arithmeticTriplets(int[] nums, int diff) {
          boolean[] vis = new boolean[310];
          for (int v : nums) {
              vis[v] = true;
          }
          int ans = 0;
          for (int v : nums) {
              if (vis[v + diff] && vis[v + diff + diff]) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution:
      def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
          s = set(nums)
          return sum(v + diff in s and v + diff + diff in s for v in nums)
  
  ############
  
  # 2367. Number of Arithmetic Triplets
  # https://leetcode.com/problems/number-of-arithmetic-triplets/
  
  class Solution:
      def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
          n = len(nums)
          res = 0
          
          for i in range(n):
              for j in range(i + 1, n):
                  for k in range(j + 1, n):
                      if nums[j] - nums[i] == diff and nums[k] - nums[j] == diff:
                          res += 1
          
          return res
  
  

• class Solution {
  public:
      int arithmeticTriplets(vector<int>& nums, int diff) {
          vector<bool> vis(310);
          for (int v : nums) vis[v] = true;
          int ans = 0;
          for (int v : nums) ans += vis[v + diff] && vis[v + diff + diff];
          return ans;
      }
  };
  

• func arithmeticTriplets(nums []int, diff int) int {
  	vis := make([]bool, 310)
  	for _, v := range nums {
  		vis[v] = true
  	}
  	ans := 0
  	for _, v := range nums {
  		if vis[v+diff] && vis[v+diff+diff] {
  			ans++
  		}
  	}
  	return ans
  }
  

• function arithmeticTriplets(nums: number[], diff: number): number {
      let vis = new Array(310).fill(false);
      for (const v of nums) {
          vis[v] = true;
      }
      let ans = 0;
      for (const v of nums) {
          if (vis[v + diff] && vis[v + diff + diff]) {
              ++ans;
          }
      }
      return ans;
  }
  
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

All Problems

All Solutions