3Sum

Try every combination of three indices and check if their values sum to zero. Use a set to deduplicate results.

1. 1Three nested loops over i, j, k with i < j < k.
2. 2If nums[i] + nums[j] + nums[k] == 0, add the sorted triplet to a result set.
3. 3Convert the set to a list and return.

Why it's not optimal: O(n³) — too slow for n up to 3000. Sorting + two-pointer reduces this to O(n²).

def threeSum(nums):
    res = set()
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            for k in range(j + 1, len(nums)):
                if nums[i] + nums[j] + nums[k] == 0:
                    res.add(tuple(sorted([nums[i], nums[j], nums[k]]))) 
    return [list(t) for t in res]

Sort the array first. Fix one element at index i, then use two pointers L and R on the remaining subarray to find pairs that sum to -nums[i]. Skip duplicate values of i, L, and R to avoid duplicate triplets.

1. 1Sort the array. If nums[i] > 0, no triplet can sum to 0 — break early.
2. 2Skip duplicate values of i (same as previous).
3. 3Two-pointer L=i+1, R=n-1. If sum is 0, record triplet and advance both. If sum < 0, increment L. If sum > 0, decrement R.
4. 4After finding a triplet, skip duplicate values of L and R.