What algorithm or data structure is used to solve Count Good Triplets?

Count Good Triplets uses the following concepts: Array, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Good Triplets?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process while coding to show your understanding. Test your solution with sample inputs before finalizing it.

What are common mistakes when solving Count Good Triplets?

Not considering the order of indices i, j, k properly. Overlooking the absolute value conditions.

Count Good Triplets — LeetCode #1534 (Easy)

Tags: Array, Enumeration

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking every possible triplet in the array to see if it meets the conditions of being a good triplet. Given the constraints, this method is straightforward and will work within the limits.

The time complexity is O(n²) because we are using three nested loops to iterate through the array, where n is the length of the array. The space complexity is O(1) since we are using a constant amount of extra space.

Step 1: Initialize a counter to zero for counting good triplets.
Step 2: Use three nested loops to iterate through all possible triplets (i, j, k) such that 0 <= i < j < k < arr.length.
Step 3: For each triplet, check if the conditions |arr[i] - arr[j]| <= a, |arr[j] - arr[k]| <= b, and |arr[i] - arr[k]| <= c are satisfied. If they are, increment the counter.
Step 4: Return the counter after checking all triplets.

For arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3: 1. i=0, j=1, k=2: (3,0,1) -> |3-0|=3 <= 7, |0-1|=1 <= 2, |3-1|=2 <= 3 -> count=1 2. i=0, j=1, k=3: (3,0,1) -> same as above -> count=2 3. i=0, j=2, k=3: (3,1,1) -> |3-1|=2 <= 7, |1-1|=0 <= 2, |3-1|=2 <= 3 -> count=3 4. i=1, j=2, k=3: (0,1,1) -> |0-1|=1 <= 7, |1-1|=0 <= 2, |0-1|=1 <= 3 -> count=4 Final count = 4.

Common Mistakes

Not considering the order of indices i, j, k properly.
Overlooking the absolute value conditions.

Interview Tips

Always clarify the constraints and edge cases with the interviewer.
Explain your thought process while coding to show your understanding.
Test your solution with sample inputs before finalizing it.

#1534

Count Good Triplets

Easy

Array↗Enumeration↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter to zero for counting good triplets.
2Step 2: Use three nested loops to iterate through all possible triplets (i, j, k) such that 0 <= i < j < k < arr.length.
3Step 3: For each triplet, check if the conditions |arr[i] - arr[j]| <= a, |arr[j] - arr[k]| <= b, and |arr[i] - arr[k]| <= c are satisfied. If they are, increment the counter.
4Step 4: Return the counter after checking all triplets.

solution.py9 lines

1def countGoodTriplets(arr, a, b, c):
2    count = 0
3    n = len(arr)
4    for i in range(n):
5        for j in range(i + 1, n):
6            for k in range(j + 1, n):
7                if abs(arr[i] - arr[j]) <= a and abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
8                    count += 1
9    return count

ℹ

Complexity note: The time complexity is O(n²) because we are using three nested loops to iterate through the array, where n is the length of the array. The space complexity is O(1) since we are using a constant amount of extra space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.