Leetcode #2552: Count Increasing Quadruplets
In this guide, we solve Leetcode #2552 Count Increasing Quadruplets 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
Given a 0-indexed integer array nums of size n containing all numbers from 1 to n, return the number of increasing quadruplets. A quadruplet (i, j, k, l) is increasing if: 0 <= i < j < k < l < n, and nums[i] < nums[k] < nums[j] < nums[l].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Binary Indexed Tree, Array, Dynamic Programming, Enumeration, Prefix Sum
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [1,3,2,4,5]
Output: 2
Explanation:
- When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l].
- When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l].
There are no other quadruplets, so we return 2.
Python Solution
class Solution:
def countQuadruplets(self, nums: List[int]) -> int:
n = len(nums)
f = [[0] * n for _ in range(n)]
g = [[0] * n for _ in range(n)]
for j in range(1, n - 2):
cnt = sum(nums[l] > nums[j] for l in range(j + 1, n))
for k in range(j + 1, n - 1):
if nums[j] > nums[k]:
f[j][k] = cnt
else:
cnt -= 1
for k in range(2, n - 1):
cnt = sum(nums[i] < nums[k] for i in range(k))
for j in range(k - 1, 0, -1):
if nums[j] > nums[k]:
g[j][k] = cnt
else:
cnt -= 1
return sum(
f[j][k] * g[j][k] for j in range(1, n - 2) for k in range(j + 1, n - 1)
)
Complexity
The time complexity is , and the space complexity is . 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.