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.

#2552

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

统计上升四元组

给你一个长度为 n 下标从 0 开始的整数数组 nums ，它包含 1 到 n 的所有数字，请你返回上升四元组的数目。如果一个四元组 (i, j, k, l) 满足以下条件，我们称它是上升的： 0 且 nums[i] 。示例 1：输入： nums = [1,3,2,4,5] 输出： 2 解释：…

数组动态规划树状数组枚举前缀和

题目描述

给你一个长度为 n 下标从 0 开始的整数数组 nums ，它包含 1 到 n 的所有数字，请你返回上升四元组的数目。

如果一个四元组 (i, j, k, l) 满足以下条件，我们称它是上升的：

0 <= i < j < k < l < n 且
nums[i] < nums[k] < nums[j] < nums[l] 。

示例 1：

输入：nums = [1,3,2,4,5]
输出：2
解释：
- 当 i = 0 ，j = 1 ，k = 2 且 l = 3 时，有 nums[i] < nums[k] < nums[j] < nums[l] 。
- 当 i = 0 ，j = 1 ，k = 2 且 l = 4 时，有 nums[i] < nums[k] < nums[j] < nums[l] 。
没有其他的四元组，所以我们返回 2 。

示例 2：

输入：nums = [1,2,3,4]
输出：0
解释：只存在一个四元组 i = 0 ，j = 1 ，k = 2 ，l = 3 ，但是 nums[j] < nums[k] ，所以我们返回 0 。

提示：

4 <= nums.length <= 4000
1 <= nums[i] <= nums.length
nums 中所有数字 互不相同 ，nums 是一个排列。

lightbulb

解题思路

方法一：枚举 + 预处理

我们可以枚举四元组中的 $j$ 和 $k$ ，那么问题转化为，对于当前的 $j$ 和 $k$ ：

统计有多少个 $l$ 满足 $l \gt k$ 且 $nums[l] \gt nums[j]$ ；
统计有多少个 $i$ 满足 $i \lt j$ 且 $nums[i] \lt nums[k]$ 。

我们可以使用两个二维数组 $f$ 和 $g$ 分别记录这两个信息。其中 $f[j][k]$ 表示有多少个 $l$ 满足 $l \gt k$ 且 $nums[l] \gt nums[j]$ ，而 $g[j][k]$ 表示有多少个 $i$ 满足 $i \lt j$ 且 $nums[i] \lt nums[k]$ 。

那么答案就是所有的 $f[j][k] \times g[j][k]$ 的和。

时间复杂度为 $O(n^2)$ ，空间复杂度为 $O(n^2)$ 。其中 $n$ 是数组的长度。

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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)
        )

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

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