We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Count Increasing Quadruplets

Difficulty: Hard

Problem Description

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].

Key Insights

  • We need to find quadruplets in the form of indices that satisfy both the index and value conditions.
  • A brute-force solution would be too slow (O(n^4)), hence we need a more efficient approach.
  • Using a combination of prefix sums and binary indexed trees can help count valid quadruplets efficiently.
  • The uniqueness of the numbers in the array ensures that we don't have to deal with duplicates.

Space and Time Complexity

Time Complexity: O(n^2 log n)
Space Complexity: O(n)

Solution

To solve the problem, we can utilize a combination of prefix sums and data structures like Binary Indexed Trees (BIT) to efficiently count the number of valid quadruplets. The approach involves:

  1. Iterating through all possible pairs (j, k) where j < k.
  2. For each pair, we count how many valid indices i exist such that nums[i] < nums[k] and i < j using a BIT or similar structure.
  3. Similarly, we count how many valid indices l exist such that nums[l] > nums[j] and l > k.
  4. The number of valid quadruplets for each (j, k) pair can then be determined by multiplying the counts from steps 2 and 3.

Code Solutions

def countQuadruplets(nums):
    n = len(nums)
    count = 0
    # Use a BIT or a similar structure to count valid i and l
    # Initialize BITs for counting
    bit_left = [0] * (n + 1)
    bit_right = [0] * (n + 1)

    def update(bit, idx, val):
        while idx <= n:
            bit[idx] += val
            idx += idx & -idx

    def query(bit, idx):
        total = 0
        while idx > 0:
            total += bit[idx]
            idx -= idx & -idx
        return total

    # Count valid i for each j and k
    for j in range(n):
        for k in range(j + 1, n):
            if nums[j] > nums[k]:
                continue
            count_left = query(bit_left, nums[k] - 1)
            count_right = query(bit_right, n) - query(bit_right, nums[j])
            count += count_left * count_right
        update(bit_left, nums[j], 1)

    # Count valid l for each j
    for k in range(n - 1, -1, -1):
        update(bit_right, nums[k], 1)

    return count
← Back to All Questions