Jobs

Interviews

Discussions

Count Special Quadruplets

Easy

Asked by:

54 views

Topics:

Arrays

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

Constraints:

4 <= nums.length <= 50
1 <= nums[i] <= 100

Try coding it in LeetCode

Count Special Quadruplets

Easy

Asked by:

54 views

Topics:

Arrays

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

Constraints:

4 <= nums.length <= 50
1 <= nums[i] <= 100

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 possible size of the input array `nums`?
Can the integers in `nums` be negative, zero, or only positive?
Are duplicate numbers allowed within the input array `nums`?
If multiple quadruplets satisfy the condition, should I return the count of all such quadruplets?
Is the order of the quadruplets in the count significant, or is only the total count important?

Brute Force Solution

Approach

The goal is to find special groups of four numbers from a larger collection of numbers. A brute-force approach means we will simply try every possible combination of four numbers and check if it meets our requirement. It's like trying every possible team of four players from a group and seeing if that team wins.

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

Pick any four numbers from the collection. It doesn't matter how you pick them, just make sure you have four.
Check if these four numbers are a 'special quadruplet' according to the rules. This means seeing if the sum of the first three numbers is equal to the fourth number.
If they are a special quadruplet, make a note of it, like adding one to a counter.
Repeat these steps for every single possible combination of four numbers you can make from the collection.
Once you've checked every possible combination, the counter will tell you the total number of special quadruplets that exist.

Code Implementation

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

    # Iterate through all possible combinations of four 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 sum of the first three equals the fourth element
                    if numbers[first_index] + numbers[second_index] + numbers[third_index] == numbers[fourth_index]:

                        quadruplet_count += 1

    return quadruplet_count

Big(O) Analysis

Time Complexity

O(n^4) – The problem involves iterating through all possible quadruplets (groups of four) from an array of size n. This means we have four nested loops, each potentially iterating up to n times. Therefore, the total number of operations is proportional to n * n * n * n, which is n to the power of 4. Thus, the time complexity is O(n^4).

Space Complexity

O(1) – The provided plain English explanation describes a brute-force approach that iterates through all possible quadruplets within the input array. It only mentions counting the 'special quadruplets' using a counter. This counter is a single integer variable. Therefore, the algorithm only requires a constant amount of extra space, independent of the input size N, to store the count of special quadruplets. Hence the auxiliary space complexity is O(1).

Optimal Solution

Approach

Instead of checking every possible combination of four numbers, we can use a clever trick to speed things up. We'll rearrange the equation we're checking and use a pre-calculated count to directly find the solutions.

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

First, rearrange the question: instead of checking if a + b + c = d, we'll check if a + b = d - c.
Next, we'll go through the list of numbers. For each number, we'll remember how many times we've seen it.
For each possible pair of numbers (a and b), calculate their sum.
Then, we'll look ahead in the list. For each number we see (d), we'll consider all numbers that come after it (c).
Calculate the difference between d and c (d - c).
Check how many times this difference appeared as a sum of a and b. This tells us how many quadruplets satisfy our rearranged equation for that particular d.
Add this count to our total count of special quadruplets.
Continue this process until we have considered all possible values of a, b, c, and d.

Code Implementation

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

    for second_index in range(list_length):
        sum_count = {}
        for first_index in range(second_index):
            sum_of_first_pair = numbers[first_index] + numbers[second_index]
            if sum_of_first_pair not in sum_count:
                sum_count[sum_of_first_pair] = 0
            sum_count[sum_of_first_pair] += 1

        for third_index in range(second_index + 1, list_length):
            # Iterate only on elements right of third_index
            for fourth_index in range(third_index + 1, list_length):

                # Calculating the difference.
                difference = numbers[fourth_index] - numbers[third_index]

                # Find the count where the difference is already seen
                if difference in sum_count:
                    count += sum_count[difference]

    return count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array nums of size n. For each pair of numbers (a, b), their sum a + b is calculated, taking O(1) time. Then, for each element d, it iterates through all subsequent elements c to calculate d - c, which also takes O(1) time. These nested loops dominate the runtime. In the worst case, where we consider every possible quadruplet (a, b, d, c) according to the problem description, the number of operations approximates n * n/2, thus the overall time complexity simplifies to O(n²).

Space Complexity

O(N) – The provided solution uses a data structure (specifically, a counter or hash map) to store the frequencies of sums of pairs (a + b). In the worst-case scenario, where all sums a + b are distinct, the size of this counter will be proportional to the number of possible pairs, which is related to N, the size of the input array nums. Therefore, the auxiliary space used by this counter can grow up to O(N^2) in the worst-case. Thus, the space complexity is O(N^2).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no quadruplets can be formed.
Array with fewer than 4 elements	Return 0 immediately since a quadruplet requires at least 4 elements.
Array with all identical elements	The solution should correctly count the number of quadruplets (i, j, k, l) such that nums[i] + nums[j] + nums[k] == nums[l] where i < j < k < l, even if all nums are the same.
Array with extremely large numbers causing integer overflow during addition	Use a larger data type (e.g., long in Java/C++) or check for potential overflow before addition to prevent incorrect counts.
Array with a large number of elements (performance consideration)	Optimize the solution to avoid brute-force O(n^4) complexity by using a hash map or other data structure to improve efficiency, preferably to O(n^3).
Array contains negative numbers	The solution should correctly handle negative numbers as addition and comparison operations work for both positive and negative values.
No quadruplets satisfy the condition	Return 0 if no quadruplets satisfy the condition nums[i] + nums[j] + nums[k] == nums[l].
Array with a single valid quadruplet	The solution must return 1 when only a single valid quadruplet exists in the array.

Null or empty input array

How to Handle:

Return 0 immediately as no quadruplets can be formed.

Array with fewer than 4 elements

How to Handle:

Return 0 immediately since a quadruplet requires at least 4 elements.

Array with all identical elements

How to Handle:

The solution should correctly count the number of quadruplets (i, j, k, l) such that nums[i] + nums[j] + nums[k] == nums[l] where i < j < k < l, even if all nums are the same.

Array with extremely large numbers causing integer overflow during addition

How to Handle:

Use a larger data type (e.g., long in Java/C++) or check for potential overflow before addition to prevent incorrect counts.

Array with a large number of elements (performance consideration)

How to Handle:

Optimize the solution to avoid brute-force O(n^4) complexity by using a hash map or other data structure to improve efficiency, preferably to O(n^3).

Array contains negative numbers

How to Handle:

The solution should correctly handle negative numbers as addition and comparison operations work for both positive and negative values.

No quadruplets satisfy the condition

How to Handle:

Return 0 if no quadruplets satisfy the condition nums[i] + nums[j] + nums[k] == nums[l].

Array with a single valid quadruplet

How to Handle:

The solution must return 1 when only a single valid quadruplet exists in the array.

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 possible size of the input array `nums`?
Can the integers in `nums` be negative, zero, or only positive?
Are duplicate numbers allowed within the input array `nums`?
If multiple quadruplets satisfy the condition, should I return the count of all such quadruplets?
Is the order of the quadruplets in the count significant, or is only the total count important?

Brute Force Solution

Approach

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

Pick any four numbers from the collection. It doesn't matter how you pick them, just make sure you have four.
Check if these four numbers are a 'special quadruplet' according to the rules. This means seeing if the sum of the first three numbers is equal to the fourth number.
If they are a special quadruplet, make a note of it, like adding one to a counter.
Repeat these steps for every single possible combination of four numbers you can make from the collection.
Once you've checked every possible combination, the counter will tell you the total number of special quadruplets that exist.

Code Implementation

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

    # Iterate through all possible combinations of four 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 sum of the first three equals the fourth element
                    if numbers[first_index] + numbers[second_index] + numbers[third_index] == numbers[fourth_index]:

                        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:

First, rearrange the question: instead of checking if a + b + c = d, we'll check if a + b = d - c.
Next, we'll go through the list of numbers. For each number, we'll remember how many times we've seen it.
For each possible pair of numbers (a and b), calculate their sum.
Then, we'll look ahead in the list. For each number we see (d), we'll consider all numbers that come after it (c).
Calculate the difference between d and c (d - c).
Check how many times this difference appeared as a sum of a and b. This tells us how many quadruplets satisfy our rearranged equation for that particular d.
Add this count to our total count of special quadruplets.
Continue this process until we have considered all possible values of a, b, c, and d.

Code Implementation

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

    for second_index in range(list_length):
        sum_count = {}
        for first_index in range(second_index):
            sum_of_first_pair = numbers[first_index] + numbers[second_index]
            if sum_of_first_pair not in sum_count:
                sum_count[sum_of_first_pair] = 0
            sum_count[sum_of_first_pair] += 1

        for third_index in range(second_index + 1, list_length):
            # Iterate only on elements right of third_index
            for fourth_index in range(third_index + 1, list_length):

                # Calculating the difference.
                difference = numbers[fourth_index] - numbers[third_index]

                # Find the count where the difference is already seen
                if difference in sum_count:
                    count += sum_count[difference]

    return count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no quadruplets can be formed.
Array with fewer than 4 elements	Return 0 immediately since a quadruplet requires at least 4 elements.
Array with all identical elements	The solution should correctly count the number of quadruplets (i, j, k, l) such that nums[i] + nums[j] + nums[k] == nums[l] where i < j < k < l, even if all nums are the same.
Array with extremely large numbers causing integer overflow during addition	Use a larger data type (e.g., long in Java/C++) or check for potential overflow before addition to prevent incorrect counts.
Array with a large number of elements (performance consideration)	Optimize the solution to avoid brute-force O(n^4) complexity by using a hash map or other data structure to improve efficiency, preferably to O(n^3).
Array contains negative numbers	The solution should correctly handle negative numbers as addition and comparison operations work for both positive and negative values.
No quadruplets satisfy the condition	Return 0 if no quadruplets satisfy the condition nums[i] + nums[j] + nums[k] == nums[l].
Array with a single valid quadruplet	The solution must return 1 when only a single valid quadruplet exists in the array.

Null or empty input array

How to Handle:

Return 0 immediately as no quadruplets can be formed.

Array with fewer than 4 elements

How to Handle:

Return 0 immediately since a quadruplet requires at least 4 elements.

Array with all identical elements

How to Handle:

The solution should correctly count the number of quadruplets (i, j, k, l) such that nums[i] + nums[j] + nums[k] == nums[l] where i < j < k < l, even if all nums are the same.

Array with extremely large numbers causing integer overflow during addition

How to Handle:

Use a larger data type (e.g., long in Java/C++) or check for potential overflow before addition to prevent incorrect counts.

Array with a large number of elements (performance consideration)

How to Handle:

Optimize the solution to avoid brute-force O(n^4) complexity by using a hash map or other data structure to improve efficiency, preferably to O(n^3).

Array contains negative numbers

How to Handle:

The solution should correctly handle negative numbers as addition and comparison operations work for both positive and negative values.

No quadruplets satisfy the condition

How to Handle:

Return 0 if no quadruplets satisfy the condition nums[i] + nums[j] + nums[k] == nums[l].

Array with a single valid quadruplet

How to Handle:

The solution must return 1 when only a single valid quadruplet exists in the array.