Leetcode #2179: Count Good Triplets in an Array
In this guide, we solve Leetcode #2179 Count Good Triplets in an Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed arrays nums1 and nums2 of length n, both of which are permutations of [0, 1, ..., n - 1]. A good triplet is a set of 3 distinct values which are present in increasing order by position both in nums1 and nums2.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Binary Indexed Tree, Segment Tree, Array, Binary Search, Divide and Conquer, Ordered Set, Merge Sort
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums1 = [2,0,1,3], nums2 = [0,1,2,3]
Output: 1
Explanation:
There are 4 triplets (x,y,z) such that pos1x < pos1y < pos1z. They are (2,0,1), (2,0,3), (2,1,3), and (0,1,3).
Out of those triplets, only the triplet (0,1,3) satisfies pos2x < pos2y < pos2z. Hence, there is only 1 good triplet.
Python Solution
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
def lowbit(x):
return x & -x
def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += BinaryIndexedTree.lowbit(x)
def query(self, x):
s = 0
while x > 0:
s += self.c[x]
x -= BinaryIndexedTree.lowbit(x)
return s
class Solution:
def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
pos = {v: i for i, v in enumerate(nums2, 1)}
ans = 0
n = len(nums1)
tree = BinaryIndexedTree(n)
for num in nums1:
p = pos[num]
left = tree.query(p)
right = n - p - (tree.query(n) - tree.query(p))
ans += left * right
tree.update(p, 1)
return ans
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.