How do you solve Count Special Quadruplets on LeetCode?

Count Special Quadruplets (LeetCode #1995) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2) time complexity and O(n) space complexity. Key insight: Using a hash map can significantly reduce the number of checks needed.

What is the brute force solution for Count Special Quadruplets?

The brute force approach for Count Special Quadruplets is Brute Force, with O(n^4) time complexity and O(1) space complexity. The brute-force approach checks every possible combination of quadruplets in the array. Since the problem requires us to find distinct indices where the sum of three numbers equals the fourth, we can simply iterate through all possible combinations. The optimal Optimal Solution approach improves this to O(n^2).

What algorithm or data structure is used to solve Count Special Quadruplets?

Count Special Quadruplets uses the following concepts: Array, Hash Table, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Special Quadruplets?

Always clarify the problem constraints. Think about edge cases and how they might affect your solution. Practice explaining your thought process clearly.

What are common mistakes when solving Count Special Quadruplets?

Not ensuring indices are distinct. Overlooking the order of indices.

#1995

Count Special Quadruplets

Easy

ArrayHash TableEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^4)	O(n^2)
Space	O(1)	O(n)

💡

Intuition

Time O(n^2)Space O(n)

Instead of checking every combination of quadruplets, we can use a hash map to store the sums of pairs of numbers. This allows us to quickly check if the sum of any three numbers equals a fourth number, significantly reducing the number of checks we need to perform.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store the frequency of sums of pairs of indices.
2Step 2: Iterate through the array, and for each index d, check if there exists a sum of two previous indices (a, b) such that nums[a] + nums[b] == nums[d] - nums[c].
3Step 3: Count valid quadruplets based on the frequency of these sums.

solution.py17 lines

1# Full working Python code
2from collections import defaultdict
3
4def countQuadruplets(nums):
5    count = 0
6    sum_map = defaultdict(int)
7    n = len(nums)
8    for d in range(n):
9        for c in range(d):
10            sum_map[nums[d] - nums[c]] += 1
11        for a in range(d):
12            for b in range(a + 1, d):
13                if nums[a] + nums[b] in sum_map:
14                    count += sum_map[nums[a] + nums[b]]
15    return count
16
17print(countQuadruplets([1, 2, 3, 6]))

ℹ

Complexity note: The time complexity is O(n^2) because we iterate through pairs of indices to build our hash map and then check sums. The space complexity is O(n) due to the hash map storing sums.

1Using a hash map can significantly reduce the number of checks needed.
2Understanding the relationship between indices and sums is crucial.

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