Find triplets in an array that add up to zero

Sort the numbers. For each index i, skip it if it equals the previous anchor, then look for two values summing to -nums[i] using left=i+1 and right=n-1. If the sum is too small, increment left; if too large, decrement right; on equality, record the triplet and advance both pointers past duplicate values. Sorting makes pointer moves monotonic, and duplicate skipping ensures unique value triplets rather than repeated index combinations. You can stop early when nums[i] is positive because later values are no smaller. Test fewer than three values, all zeros, many duplicates, no solution, and overflow when adding fixed-width integers; use a wider sum if needed. Sorting costs O(n log n), and the nested anchor plus pointer scans cost O(n squared) total time. Extra space is O(1) apart from the output if sorting in place is allowed.