Is Count Special Quadruplets easy or hard?

Count Special Quadruplets is classified as an Easy problem on LeetCode. The brute force solution is straightforward, but recognizing the equation transformation and applying a hash map to reach O(n^2) efficiency demonstrates stronger problem‑solving skills.

Count Special Quadruplets Python/Java solution

Python and Java implementations typically use a dictionary or HashMap to store differences between elements while iterating from the right side of the array. The algorithm checks pair sums on the left and adds the frequency of matching differences, resulting in an O(n^2) solution.

How to solve Count Special Quadruplets in O(n^2)?

Rearrange the equation to nums[a] + nums[b] = nums[d] - nums[c]. While iterating from the right side of the array, store the differences (nums[d] - nums[c]) in a hash map. Then iterate pairs (a, b) on the left and check whether their sum exists in the map, accumulating the frequency of matches.

What is the best approach for Count Special Quadruplets?

The most efficient approach uses a hash map to store differences between elements on the right side of the array. By transforming the equation nums[a] + nums[b] + nums[c] = nums[d] into nums[a] + nums[b] = nums[d] - nums[c], the algorithm reduces the search to pair sums and hash lookups. This achieves about O(n^2) time with O(n) extra space.

Is Count Special Quadruplets asked at Google/Amazon/Meta?

Count Special Quadruplets is a common array and hash‑table interview pattern seen in coding interviews. While the exact problem may not appear frequently at top companies, similar variations involving pair sums, hash maps, and index constraints appear in interviews at companies like Amazon, Google, and Meta.

What data structure is used in Count Special Quadruplets?

The optimized solution relies on a hash map (dictionary) to store frequencies of value differences such as nums[d] - nums[c]. This allows constant‑time lookups when checking whether a pair sum nums[a] + nums[b] completes the equation.

What is the time complexity of Count Special Quadruplets?

The naive enumeration solution runs in O(n^4) time because it checks every quadruplet combination. With a hash map optimization that stores differences for (c, d) pairs and matches them against sums of (a, b), the complexity improves to O(n^2) time and O(n) space.

Description Hints 2 Solution Video 3:35

1995. Count Special Quadruplets

Easy65.1% acceptance3 companies

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

Asked at 3 companies

Google Microsoft Meta

Topics

Hints open one at a time

Hint 1

N is very small, how can we use that?

Hint 2

Can we check every possible quadruplet?

3:35

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