How do you solve Count Good Triplets in an Array on LeetCode?

Count Good Triplets in an Array (LeetCode #2179) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding the relationship between positions in both arrays is crucial for identifying good triplets.

What is the time complexity of Count Good Triplets in an Array?

The optimal solution for Count Good Triplets in an Array runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the use of Fenwick Trees for counting, which allows us to efficiently update and query counts.

What is the brute force solution for Count Good Triplets in an Array?

The brute force approach for Count Good Triplets in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible triplet (x, y, z) to see if they satisfy the conditions of being in increasing order in both arrays. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Count Good Triplets in an Array?

Always start with a brute-force solution to understand the problem better. Be prepared to explain your thought process and how you arrived at the optimal solution. Practice implementing data structures like Fenwick Trees or Segment Trees, as they are commonly used in optimization problems.

What are common mistakes when solving Count Good Triplets in an Array?

Not properly managing the indices when using data structures for counting. Failing to account for the distinctness of triplet values.

#2179

Count Good Triplets in an Array

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetFenwick TreeCountingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

We can optimize the solution by using a combination of counting and data structures to efficiently track the number of valid triplets without checking every combination explicitly.

⚙️

Algorithm

4 steps

1Step 1: Create position mappings pos1 and pos2 for nums1 and nums2 respectively.
2Step 2: For each element y in nums2, count how many elements x come before y in both arrays using a Fenwick Tree (or Binary Indexed Tree).
3Step 3: For each element y, count how many elements z come after y in both arrays using a Fenwick Tree.
4Step 4: Multiply the counts from Step 2 and Step 3 for each y to get the total number of good triplets.

solution.py28 lines

1def countGoodTriplets(nums1, nums2):
2    n = len(nums1)
3    pos1 = {v: i for i, v in enumerate(nums1)}
4    pos2 = {v: i for i, v in enumerate(nums2)}
5    fenwick1 = [0] * (n + 1)
6    fenwick2 = [0] * (n + 1)
7
8    def update(fenwick, index, value):
9        while index <= n:
10            fenwick[index] += value
11            index += index & -index
12
13    def query(fenwick, index):
14        total = 0
15        while index > 0:
16            total += fenwick[index]
17            index -= index & -index
18        return total
19
20    count = 0
21    for y in range(n):
22        count_x = query(fenwick1, pos2[y])
23        count_z = query(fenwick2, n) - query(fenwick2, pos2[y])
24        count += count_x * count_z
25        update(fenwick1, pos2[y] + 1, 1)
26        update(fenwick2, pos2[y] + 1, 1)
27
28    return count

ℹ

Complexity note: The complexity is O(n log n) due to the use of Fenwick Trees for counting, which allows us to efficiently update and query counts.

1Understanding the relationship between positions in both arrays is crucial for identifying good triplets.
2Using efficient data structures like Fenwick Trees can significantly reduce the time complexity.

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