15. 3Sum

Problem statement:

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Solution:

Which is different with 1. Two Sum, this question just need all three numbers whose sum is 0.

The basic idea:

• Sort this array by ascending order.
• Loop from the first element in array. it becomes a two sum problem. For i element, we should find two elements from rest of the array whose sum is -nums[i].

NOTE: In order to remove the duplicate element, we do duplication check when update the new value for the index, left and right pointers.

Time complexity O(n * n). Space complexity is O(1).

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> triplet;
        // get the size of the array and convert it to integer type
        int size = nums.size();
        for(int i = 0; i < size - 1; i++){
            int left = i + 1;
            int right = size - 1;
            while(left < right){
                // no need do duplicate check
                // For the same number, it will do same operation until it changes
                if(nums[left] + nums[right] == -nums[i]){
                    triplet.push_back({nums[i], nums[left], nums[right]});
                    // check for duplicated
                    while(left < right && left + 1 < size && nums[left] == nums[left + 1]){
                        left++;
                    }
                    // check for duplicated
                    while(left < right && right - 1 >= 0 && nums[right - 1] == nums[right]){
                        right--;
                    }
                    // normal pointers move
                    left++;
                    right--;
                } else if (nums[left] + nums[right] > -nums[i]) {
                    right--;
                } else {
                    left++;
                }
            }
            // remove duplicate candidates
            while(i + 1 < size && nums[i] == nums[i + 1]){
                i++;
            }
        }
        return triplet;
    }
};