#2552

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Increasing Quadruplets

Given a permutation of numbers, count the number of increasing quadruplets using dynamic programming.

array dynamic programming binary indexed tree enumeration prefix sum

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 nums[i] < nums[j] < nums[k] < nums[l] and i < j < k < l.

To solve this problem, you need to count how many quadruplets (i, j, k, l) satisfy the condition using an efficient approach, considering the constraints where n can be as large as 4000.

Examples

Example 1

Input: nums = [1,3,2,4,5]

Output: 2

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.

Example 2

Input: nums = [1,2,3,4]

Output: 0

There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0.

Constraints

4 <= nums.length <= 4000
1 <= nums[i] <= nums.length
All the integers of nums are unique. nums is a permutation.

Solution Approach

State Transition Dynamic Programming

This problem can be solved using dynamic programming by tracking the number of valid quadruplets up to each index. We use dynamic programming states to count the number of valid i, j, k, l combinations. The solution efficiently counts the quadruplets by iterating over potential (j, k) pairs and using previously computed results to find valid i and l combinations.

Binary Indexed Tree and Prefix Sum Optimization

To handle the large input size, combine the state transition dynamic programming with binary indexed trees (BIT) or prefix sums. Use a BIT to count valid smaller and larger elements dynamically as you process the array, thus improving the time complexity. This allows you to keep track of necessary counts without recomputing them each time.

Efficient Looping Over (j, k) Pairs

Focus on iterating through all possible (j, k) pairs and counting valid (i, l) pairs efficiently. By looping over (j, k), you reduce the complexity and avoid directly checking all possible quadruplets. The key insight is that by considering pairs and calculating valid quadruplets for each pair, we can solve the problem in a more optimal manner.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the specific approach used. The naive brute-force approach would have a time complexity of O(n^4), but by using dynamic programming and binary indexed trees, we can reduce the time complexity to O(n^2 log n). The space complexity also varies, typically being O(n^2) or O(n log n) depending on the chosen optimization technique.

What Interviewers Usually Probe

Understanding dynamic programming and state transitions is crucial for solving this problem.
Ability to optimize with binary indexed trees and prefix sums shows proficiency in handling large input sizes.
The ability to loop efficiently over (j, k) pairs without redundant computations is key to solving the problem within constraints.

Common Pitfalls or Variants

Common pitfalls

Forgetting to check that the indices i < j < k < l, which is a crucial part of the problem's constraints.
Not using optimization techniques like binary indexed trees or prefix sums for larger arrays, which can lead to time limits being exceeded.
Incorrectly counting quadruplets by overlooking the order of indices or invalid combinations (i.e., nums[i] > nums[j]).

Follow-up variants

Variation 1: Consider the problem with duplicates and modify the algorithm to account for them.
Variation 2: Count the quadruplets under additional constraints such as maximum or minimum values for certain elements.
Variation 3: Modify the problem to return the quadruplet values themselves instead of just the count.

FAQ

What is the primary algorithmic pattern in the Count Increasing Quadruplets problem?

The primary pattern is state transition dynamic programming, which efficiently counts valid quadruplets by iterating over pairs and optimizing with advanced techniques like binary indexed trees.

How can I optimize the brute-force approach for this problem?

By using dynamic programming combined with binary indexed trees or prefix sums, you can reduce the time complexity and efficiently handle larger inputs.

What are the constraints of the Count Increasing Quadruplets problem?

The input array has a size of 4 <= nums.length <= 4000, with all integers being unique and forming a permutation from 1 to n.

What are some common pitfalls when solving this problem?

Common pitfalls include forgetting to maintain the order of indices, not optimizing with binary indexed trees, and incorrectly counting quadruplets by overlooking valid combinations.

How can GhostInterview assist with preparing for this problem?

GhostInterview helps by providing optimized solutions, clarifying the problem's approach, and pointing out common mistakes so you can confidently tackle the problem.

terminal

Solution

Solution 1: Enumeration + Preprocessing

We can enumerate $j$ and $k$ in the quadruplet, then the problem is transformed into, for the current $j$ and $k$:

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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)
        )

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