How hard is LeetCode 2552. Count Increasing Quadruplets?

LeetCode 2552. Count Increasing Quadruplets is rated Hard on LeetCode.

What is the time complexity of LeetCode 2552. Count Increasing Quadruplets?

The Python solution on this page runs in O(n^2).

What is the space complexity of LeetCode 2552. Count Increasing Quadruplets?

The Python solution on this page uses O(n^2) auxiliary space.

What topics does LeetCode 2552. Count Increasing Quadruplets cover?

LeetCode 2552. Count Increasing Quadruplets is tagged Binary Indexed Tree, Array, Dynamic Programming, Enumeration and Prefix Sum on LeetCode.

Count Increasing Quadruplets — LeetCode 2552 Python Solution

HardBinary Indexed TreeArrayDynamic ProgrammingEnumerationPrefix Sum

Problem: #2552
Pattern: Prefix Sum
Reading time: 4 min
Source: leetcode.com

The problem

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].

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].

Python solution

Python

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

Measure	Complexity
Time	O(n^2)
Space	O(n^2) auxiliary

Pattern: Prefix Sum

Precompute running totals once so any range query becomes a single subtraction. LeetCode 2552. Count Increasing Quadruplets is filed here because LeetCode tags it Prefix Sum, which is the vocabulary this hub collects.

The prefix sum guide has the Python template for the pattern and the 157 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2552. Count Increasing Quadruplets?: LeetCode 2552. Count Increasing Quadruplets is rated Hard on LeetCode.
What is the time complexity of LeetCode 2552. Count Increasing Quadruplets?: The Python solution on this page runs in O(n^2).
What is the space complexity of LeetCode 2552. Count Increasing Quadruplets?: The Python solution on this page uses O(n^2) auxiliary space.
What topics does LeetCode 2552. Count Increasing Quadruplets cover?: LeetCode 2552. Count Increasing Quadruplets is tagged Binary Indexed Tree, Array, Dynamic Programming, Enumeration and Prefix Sum on LeetCode.