How does the three-sum problem use the two pointers pattern?

Sort the array first. For each element, fix it as the first number and use two pointers on the remaining portion to find pairs that sum to the negative of the fixed element, making the total zero. Skip duplicate values for the fixed element to avoid duplicate triplets. Within the two-pointer search, also skip duplicates when a valid triplet is found. The sorting takes O of n log n and the nested two-pointer search takes O of n squared total, making the overall complexity O of n squared. Without sorting and two pointers, the brute force approach would be O of n cubed.