3Sum

Problem

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.

Examples

The brute-force approach

Try all combinations of three indices (i, j, k). For each, check whether the three values sum to zero. Collect unique triplets using a set.

result = set()

for i in range(len(nums)):
    for j in range(i + 1, len(nums)):
        for k in range(j + 1, len(nums)):                  # ← triple nested loop
            if nums[i] + nums[j] + nums[k] == 0:
                triplet = tuple(sorted([nums[i], nums[j], nums[k]]))
                result.add(triplet)

return [list(t) for t in result]

Three nested loops make this O(n³). For an array of 1000 elements, that's roughly a billion iterations. Worse, the set-based deduplication adds overhead. The fix is to sort the array first, which gives you a structure you can exploit to find pairs in O(n) per fixed element rather than O(n²).

Spotting the pattern

This is a Two Pointers problem. The key question to ask yourself:

Once I fix one number and need two others that complete the sum, how do I find all valid pairs efficiently?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.