The Question

3Sum - Unique Triplets with Zero Sum

Given an integer array nums, return all unique triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and the sum of the elements is exactly zero: nums[i] + nums[j] + nums[k] == 0. Constraints: - 3 <= nums.length <= 3000 - -10^5 <= nums[i] <= 10^5 - The solution set must not contain duplicate triplets.

Java

Two Pointers

Sorting

Questions & Insights

Clarifying Questions

What is the maximum length of the input array nums? (Assumption: Up to 3,000, suggesting an

O(n^2)

solution is acceptable).

Can the array contain duplicate numbers? (Assumption: Yes, and the result must contain only unique triplets).

What is the range of the integers in the array? (Assumption: Standard 32-bit signed integers; the sum of three won't overflow a 64-bit long, and usually fits in a 32-bit int).

Is the order of triplets in the output or the order of elements within the triplets important? (Assumption: No, but typical solutions return sorted triplets or unique sets).

Thinking Process

Sort and Two-Pointer: Sorting the array is the most efficient precursor. It allows us to use a two-pointer approach to find pairs and easily skip duplicate values to ensure triplet uniqueness.

Fix One, Find Two: Iterate through the array, fixing one element nums[i] as the potential first element of the triplet. The problem then reduces to finding two other numbers in the remaining subarray nums[i+1...n-1] that sum to -nums[i].

Handling Duplicates:

Skip the current nums[i] if it's the same as nums[i-1] to avoid redundant triplets starting with the same value.

Inside the two-pointer loop, after finding a valid triplet, increment the left pointer and decrement the right pointer while they point to values identical to the ones just processed.

Optimization: If the fixed element nums[i] is greater than zero, we can stop immediately because the array is sorted and any subsequent numbers will also be positive, making a zero sum impossible.

Implementation Breakdown

Problem Set

Functional Requirement: Find all unique triplets

(a, b, c)

such that

a + b + c = 0

and indices are distinct.

Constraints:

Time complexity should be

O(n^2)

Space complexity should be

O(1)

O(\log n)

(excluding output storage), depending on the sorting algorithm's stack space.

Approach

Algorithm: Sorting + Two Pointers

Data Structure: java.util.ArrayList, java.util.Arrays

Complexity:

Time:

O(n^2)

due to the nested loops (one iteration, one two-pointer scan).

Space:

O(\log n)

O(n)

for the sorting algorithm's internal space.

Implementation

Wrap Up

Advanced Topics

Performance vs. Memory: While the

O(n^2)

sorting approach is standard, a Hash-based approach (using a HashSet for the inner "Two Sum") can work in

O(n^2)

time as well but generally uses more memory and is harder to manage for duplicate removal.

K-Sum Generalization: This problem is a specific case of the K-Sum problem. For

k > 3

, the problem can be solved recursively by reducing

k

down to 2 (Two Sum), maintaining

O(n^{k-1})

complexity.

Parallelization: For very large datasets, the outer loop (fixing nums[i]) can be parallelized, provided the shared result list is handled with proper synchronization or by merging local lists from different threads.