How do you solve Count Increasing Quadruplets on LeetCode?

Count Increasing Quadruplets (LeetCode #2552) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Fixing two indices and counting valid combinations for the remaining two can significantly reduce complexity.

What is the brute force solution for Count Increasing Quadruplets?

The brute force approach for Count Increasing Quadruplets is Brute Force, with O(n^4) time complexity and O(1) space complexity. The brute-force approach involves checking all possible combinations of indices (i, j, k, l) to see if they form an increasing quadruplet. This is straightforward but inefficient as it examines every possible combination. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Count Increasing Quadruplets?

Always start with a brute-force approach to understand the problem deeply. Think about how you can reduce the number of checks by fixing some indices and counting valid combinations. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Increasing Quadruplets?

Overlooking the order of indices when checking conditions for quadruplets. Failing to optimize the nested loops, leading to excessive time complexity.

#2552

Count Increasing Quadruplets

Hard

ArrayDynamic ProgrammingBinary Indexed TreeEnumerationPrefix SumEnumerationPrefix SumDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^4)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution leverages pre-computation and efficient counting to avoid the nested loops. By fixing j and k, we can count valid i and l indices using prefix sums and suffix counts.

⚙️

Algorithm

3 steps

1Step 1: Create an array to count how many valid i's exist for each j and k.
2Step 2: For each pair (j, k), count valid i's before j and valid l's after k.
3Step 3: Sum the products of valid i's and l's for each (j, k) pair to get the total count of quadruplets.

solution.py21 lines

1def countQuadruplets(nums):
2    n = len(nums)
3    count = 0
4    left_count = [0] * n
5    right_count = [0] * n
6
7    for j in range(1, n - 2):
8        for i in range(j):
9            if nums[i] < nums[j]:
10                left_count[j] += 1
11
12    for k in range(n - 2, 1, -1):
13        for l in range(k + 1, n):
14            if nums[k] < nums[l]:
15                right_count[k] += 1
16
17    for j in range(1, n - 2):
18        for k in range(j + 1, n - 1):
19            count += left_count[j] * right_count[k]
20
21    return count

ℹ

Complexity note: The complexity is O(n²) due to the nested loops for counting valid indices, but the overall number of iterations is significantly reduced compared to the brute-force approach.

1Fixing two indices and counting valid combinations for the remaining two can significantly reduce complexity.
2Using prefix and suffix counts allows us to efficiently compute valid pairs without redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.