Three sum

On a sorted array, fix one value and use two pointers to find pairs that cancel it out to zero.

Time O(n²)Space O(1) extra, O(n log n) for the sortCode C#

Watch it animatedStep through it, with the running line highlighted. Free with an account.

The idea

Sort first, then fix the smallest number of a triplet and run the two-pointer pair-sum search for "add up to -a[i]" on the rest of the array. Sorting also makes duplicate triplets sit next to each other so they are easy to skip.

Code

public static List<(int, int, int)> ThreeSum(int[] a) {
    Array.Sort(a);
    var res = new List<(int, int, int)>();
    for (int i = 0; i < a.Length - 2; i++) {
        if (i > 0 && a[i] == a[i - 1]) continue;
        int l = i + 1, r = a.Length - 1;
        while (l < r) {
            int sum = a[i] + a[l] + a[r];
            if (sum == 0) { res.Add((a[i], a[l], a[r])); l++; r--; }
            else if (sum < 0) l++;
            else r--;
        }
    }
    return res;
}

Tips

Skip a repeated value for i: using the same first number again would only rediscover a triplet already found.
This is the natural next step after two-pointer pair sum: the same l/r closing-in idea, run once per fixed index.

In the .NET library

The O(n²) here is as good as it gets with plain comparisons; a hash-based variant can find pairs faster but still needs care to avoid duplicate triplets.