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Count Increasing Quadruplets

Hard

SAP

3 views

Topics:

Arrays

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 1:

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].
- 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
Explanation: 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.

Try coding it in LeetCode

Count Increasing Quadruplets

Hard

SAP

3 views

Topics:

Arrays

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 1:

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].
- 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
Explanation: 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array nums?
Can the array nums contain negative numbers, zeros, or duplicate values?
If no such quadruplets exist, what value should I return?
Are the integers in the array guaranteed to fit within the standard integer range, or should I consider using a larger data type?
Is the ordering of the quadruplets important? (e.g., should I return a list of tuples in a specific order?)

Brute Force Solution

Approach

To find increasing quadruplets, we'll look at absolutely every possible combination of four numbers from the given list. We'll then check if those four numbers form an increasing sequence.

Here's how the algorithm would work step-by-step:

First, pick any four numbers from the list, making sure you pick them in the order they appear in the list.
Next, check if the first number is smaller than the second number, and if the second number is smaller than the third number, and if the third number is smaller than the fourth number.
If all three of those conditions are true, then you've found an increasing quadruplet, so count it.
Repeat this process by choosing a different set of four numbers from the list. Make sure you eventually try every single possible combination of four numbers.
Once you've examined all the combinations, the total count of increasing quadruplets will be your final answer.

Code Implementation

def count_increasing_quadruplets_brute_force(numbers):
    quadruplet_count = 0
    list_length = len(numbers)

    # Iterate through all possible combinations of indices
    for first_index in range(list_length):
        for second_index in range(first_index + 1, list_length):
            for third_index in range(second_index + 1, list_length):
                for fourth_index in range(third_index + 1, list_length):
                    # Check if the quadruplet is increasing
                    if numbers[first_index] < numbers[second_index] and \
                       numbers[second_index] < numbers[third_index] and \
                       numbers[third_index] < numbers[fourth_index]:

                        # Increment the count if it's an increasing quadruplet
                        quadruplet_count += 1

    return quadruplet_count

Big(O) Analysis

Time Complexity

O(n^4) – The algorithm iterates through all possible combinations of four numbers in the list. With an input list of size n, selecting four numbers requires nested loops, each potentially iterating up to n times. Therefore, we have four nested loops, resulting in approximately n * n * n * n operations. The total number of operations can be approximated as n^4, simplifying to a time complexity of O(n^4).

Space Complexity

O(1) – The algorithm iterates through all possible combinations of four numbers within the input list of size N. It doesn't explicitly create any auxiliary data structures to store intermediate results or track visited combinations. The operations involve comparisons and potentially incrementing a counter variable. Therefore, the algorithm uses constant auxiliary space, independent of the input size N.

Optimal Solution

Approach

The efficient solution avoids checking every possible group of four numbers. It cleverly counts the potential middle pairs and uses that information to quickly determine the number of valid quadruplets.

Here's how the algorithm would work step-by-step:

For each number in the sequence, consider it as the second number in a potential quadruplet.
For each number that comes *before* it, count how many are *smaller* than our chosen second number. These are candidates for the first number in the quadruplet.
For each number that comes *after* it, consider it as the third number in a potential quadruplet.
For each number that comes *after* our chosen third number, count how many are *larger* than our chosen third number. These are candidates for the fourth number in the quadruplet.
Now, for each number acting as the second number, we can iterate through the potential third numbers after it. For each of those potential third numbers, multiply the count of potential first numbers (that came before the second) by the count of potential fourth numbers (that come after the third).
The result of this multiplication tells you how many valid quadruplets you can form with that specific pairing of the second and third numbers.
Sum up all these counts (for every possible second and third number pairing) to get the total number of valid quadruplets in the sequence.

Code Implementation

def count_increasing_quadruplets(numbers):
    list_length = len(numbers)
    count = 0

    for second_index in range(1, list_length - 2):
        # Count numbers before second_index that are smaller.
        count_less_than_second = 0
        for first_index in range(second_index):
            if numbers[first_index] < numbers[second_index]:
                count_less_than_second += 1

        for third_index in range(second_index + 1, list_length - 1):
            # Only proceed if the third number is larger than the second.
            if numbers[third_index] <= numbers[second_index]:
                continue

            # Count numbers after third_index that are larger.
            count_greater_than_third = 0
            for fourth_index in range(third_index + 1, list_length):
                if numbers[fourth_index] > numbers[third_index]:
                    count_greater_than_third += 1

            # Accumulate count of quadruplets based on counts.
            count += count_less_than_second * count_greater_than_third

    return count

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through each element as the second number of the quadruplet. Then, for each subsequent element, it considers it as the third number. For each second and third number pair, it calculates the count of potential first numbers preceding the second number and the count of potential fourth numbers succeeding the third number. The crucial part is that these counts are computed in linear time relative to the current second and third numbers. Therefore, the overall time complexity is dominated by the nested loops for second and third numbers, resulting in O(n*n) which simplifies to O(n^2).

Space Complexity

O(1) – The algorithm primarily uses a few integer variables to store counts and indices. It doesn't create any auxiliary data structures like arrays, lists, or hash maps that scale with the input size N (the length of the sequence). Therefore, the extra space used remains constant regardless of the input size.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no quadruplets can be formed.
Input array with size less than 4	Return 0 since we need at least four elements to form a quadruplet.
Input array with all identical values	The count should be 0 as nums[i] < nums[k] < nums[j] < nums[l] cannot hold true.
Input array with a very large size (n approaching integer limit)	Ensure the algorithm's time complexity is optimized (ideally better than O(n^4)) to avoid timeouts, and consider potential integer overflows in counting.
Input array contains negative numbers, zeros, and positive numbers.	The comparison logic should handle all number types correctly without special casing.
Input array contains duplicate values interspersed	The algorithm must correctly count quadruplets even with duplicates, ensuring indices i, j, k, l are distinct.
Integer overflow in intermediate calculations (e.g., counting prefix sums or using large array values).	Use appropriate data types (e.g., long or BigInteger) for intermediate calculations to prevent overflow.
No valid quadruplets exist in the input array	The algorithm should return 0 in cases where no quadruplets satisfying the condition can be found.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array nums?
Can the array nums contain negative numbers, zeros, or duplicate values?
If no such quadruplets exist, what value should I return?
Are the integers in the array guaranteed to fit within the standard integer range, or should I consider using a larger data type?
Is the ordering of the quadruplets important? (e.g., should I return a list of tuples in a specific order?)

Brute Force Solution

Approach

To find increasing quadruplets, we'll look at absolutely every possible combination of four numbers from the given list. We'll then check if those four numbers form an increasing sequence.

Here's how the algorithm would work step-by-step:

First, pick any four numbers from the list, making sure you pick them in the order they appear in the list.
Next, check if the first number is smaller than the second number, and if the second number is smaller than the third number, and if the third number is smaller than the fourth number.
If all three of those conditions are true, then you've found an increasing quadruplet, so count it.
Repeat this process by choosing a different set of four numbers from the list. Make sure you eventually try every single possible combination of four numbers.
Once you've examined all the combinations, the total count of increasing quadruplets will be your final answer.

Code Implementation

def count_increasing_quadruplets_brute_force(numbers):
    quadruplet_count = 0
    list_length = len(numbers)

    # Iterate through all possible combinations of indices
    for first_index in range(list_length):
        for second_index in range(first_index + 1, list_length):
            for third_index in range(second_index + 1, list_length):
                for fourth_index in range(third_index + 1, list_length):
                    # Check if the quadruplet is increasing
                    if numbers[first_index] < numbers[second_index] and \
                       numbers[second_index] < numbers[third_index] and \
                       numbers[third_index] < numbers[fourth_index]:

                        # Increment the count if it's an increasing quadruplet
                        quadruplet_count += 1

    return quadruplet_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

For each number in the sequence, consider it as the second number in a potential quadruplet.
For each number that comes *before* it, count how many are *smaller* than our chosen second number. These are candidates for the first number in the quadruplet.
For each number that comes *after* it, consider it as the third number in a potential quadruplet.
For each number that comes *after* our chosen third number, count how many are *larger* than our chosen third number. These are candidates for the fourth number in the quadruplet.
Now, for each number acting as the second number, we can iterate through the potential third numbers after it. For each of those potential third numbers, multiply the count of potential first numbers (that came before the second) by the count of potential fourth numbers (that come after the third).
The result of this multiplication tells you how many valid quadruplets you can form with that specific pairing of the second and third numbers.
Sum up all these counts (for every possible second and third number pairing) to get the total number of valid quadruplets in the sequence.

Code Implementation

def count_increasing_quadruplets(numbers):
    list_length = len(numbers)
    count = 0

    for second_index in range(1, list_length - 2):
        # Count numbers before second_index that are smaller.
        count_less_than_second = 0
        for first_index in range(second_index):
            if numbers[first_index] < numbers[second_index]:
                count_less_than_second += 1

        for third_index in range(second_index + 1, list_length - 1):
            # Only proceed if the third number is larger than the second.
            if numbers[third_index] <= numbers[second_index]:
                continue

            # Count numbers after third_index that are larger.
            count_greater_than_third = 0
            for fourth_index in range(third_index + 1, list_length):
                if numbers[fourth_index] > numbers[third_index]:
                    count_greater_than_third += 1

            # Accumulate count of quadruplets based on counts.
            count += count_less_than_second * count_greater_than_third

    return count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no quadruplets can be formed.
Input array with size less than 4	Return 0 since we need at least four elements to form a quadruplet.
Input array with all identical values	The count should be 0 as nums[i] < nums[k] < nums[j] < nums[l] cannot hold true.
Input array with a very large size (n approaching integer limit)	Ensure the algorithm's time complexity is optimized (ideally better than O(n^4)) to avoid timeouts, and consider potential integer overflows in counting.
Input array contains negative numbers, zeros, and positive numbers.	The comparison logic should handle all number types correctly without special casing.
Input array contains duplicate values interspersed	The algorithm must correctly count quadruplets even with duplicates, ensuring indices i, j, k, l are distinct.
Integer overflow in intermediate calculations (e.g., counting prefix sums or using large array values).	Use appropriate data types (e.g., long or BigInteger) for intermediate calculations to prevent overflow.
No valid quadruplets exist in the input array	The algorithm should return 0 in cases where no quadruplets satisfying the condition can be found.