Number of Arithmetic Triplets

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Problem Highlights

• 🔗 Leetcode Link: Number of Arithmetic Triplets
• 💡 Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Array, Hashtable
• 🗒️ Similar Questions: Two Sum

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• How do you calculate subsequent terms?
  • Observe that if you know the smallest term in an arithmetic triplet you can calculate the the subsequent terms.
• Are the elements in the array already sorted?
  • Yes. Because the numbers in original array are sorted, if for any index i we find nums[i] + diff and nums[i] + (2 * diff) in the set, we can be sure that they exist at indexes that satisfy the constraint of i < j < k since any value greater than current index number's value would be at a higher index than current index (nums[i] + diff and nums[i] + (2 * diff) will always be greater indexes than nums[i] since according to problem constraints 1 <= diff <= 50).

HAPPY CASE
Example 1:

Input: nums = [0,1,4,6,7,10], diff = 3
Output: 2

Example 2:

Input: nums = [4,5,6,7,8,9], diff = 2
Output: 2

EDGE CASE
Example 3:

Input: nums = [0,100,101,102,103,200], diff = 3
Output: 0

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Array problems, we want to consider the following approaches:

• Sort. Sorting does not help solve this specific problem, because the ordering of the letters is important.
• Two pointer solutions (left and right pointer variables).Two pointer approach does not help solve this specific problem, because ordering of the letters is unidirectional. Take a variable count to keep a count of the triplets.
• Storing the elements of the array in a Hashtable. This could potentially work if know exactly what to look for after we store elements. Create dictionary where we are going to store the value and the index of each list element as a key-pair respectively. Then we iterate through the indices and values of the list containing our numbers. If the difference between the target and the current value in the list is already included as a key in the dictionary, then it means that the current value and the value stored in the dictionary is the solution to our problem. To speed up and simplify the looking up work, we can convert the nums array into a set (hashtable) data structure which has O(1) look up time complexity.
• Traversing the array with a sliding window. A solution may not be two adjacent numbers, so a fixed sliding window won’t solve the problem in all cases.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea:

1. Store all the elements of array in a map or a set
2. Traverse the array and see for every element if there are two more elements present in map 
3. If present, then increment the count by 1
4. Return the count variable

⚠️ Common Mistakes

• Avoid misunderstanding the problem by reframing the problem: to find 3 numbers (i, j , k) such that i < j < k, nums[j] - nums[i] == diff, and nums[k] - nums[j] == diff.

We can rearrange the last 2 constraints to define nums[j] and nums[k] in terms of nums[i] as follows:

nums[j] - nums[i] == diff ---> nums[j] = nums[i] + diff

nums[k] - nums[j] == diff ---> nums[k] = nums[j] + diff ---> nums[k] = (nums[i] + diff) + diff) ---> nums[k] = nums[i] + (2 * diff)

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:

        # create variable to count the number of triplets
        ans = 0

        # loop through nums backwards
        for i in range(len(nums)-1, -1, -1):
            
            # the current number minus diff and diff*2
            a = nums[i] - (diff*1)
            b = nums[i] - (diff*2)

            # if both of these are in nums, we have a triplet 
            if (a in nums) and (b in nums):
                ans +=1

        return ans

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        # create variable to count the number of triplets
        final var arr = new boolean[201];

        # loop through nums backwards
        for (int i : nums)
            arr[i] = true;
        int cnt = 0, t = 0;

        # if both of these are in nums, we have a triplet
        for (int i : nums)
            if ( (t = i - diff) >= 0 && arr[t] && (t -= diff) >= 0 && arr[t])
                cnt++;
        return cnt;
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume n is the length of nums,

• Time Complexity: O(n) - we use two separate loops that iterate over the same number of elements

• Space Complexity: O(n) - we need to store the mapping from numbers to their positions in the array