Question

2367. Number of Arithmetic Triplets

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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

Solution 1: Brute Force

We notice that the length of the array $nums$ is no more than $200$. Therefore, we can directly enumerate $i$, $j$, $k$, and check whether they meet the conditions. If they do, we increment the count of the triplet.

The time complexity is $O(n^3)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

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))

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;
  }
};

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;
}

Solution 2: Array or Hash Table

We can first store the elements of $nums$ in a hash table or array $vis$. Then, for each element $x$ in $nums$, we check if $x+diff$ and $x+diff+diff$ are also in $vis$. If they are, we increment the count of the triplet.

After the enumeration, we return the answer.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.

class Solution:
  def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
    vis = set(nums)
    return sum(x + diff in vis and x + diff * 2 in vis for x in nums)

class Solution {
  public int arithmeticTriplets(int[] nums, int diff) {
    boolean[] vis = new boolean[301];
    for (int x : nums) {
      vis[x] = true;
    }
    int ans = 0;
    for (int x : nums) {
      if (vis[x + diff] && vis[x + diff + diff]) {
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int arithmeticTriplets(vector<int>& nums, int diff) {
    bitset<301> vis;
    for (int x : nums) {
      vis[x] = 1;
    }
    int ans = 0;
    for (int x : nums) {
      ans += vis[x + diff] && vis[x + diff + diff];
    }
    return ans;
  }
};

func arithmeticTriplets(nums []int, diff int) (ans int) {
  vis := [301]bool{}
  for _, x := range nums {
    vis[x] = true
  }
  for _, x := range nums {
    if vis[x+diff] && vis[x+diff+diff] {
      ans++
    }
  }
  return
}

function arithmeticTriplets(nums: number[], diff: number): number {
  const vis: boolean[] = new Array(301).fill(false);
  for (const x of nums) {
    vis[x] = true;
  }
  let ans = 0;
  for (const x of nums) {
    if (vis[x + diff] && vis[x + diff + diff]) {
      ++ans;
    }
  }
  return ans;
}