3Sum
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LeetCode #15: 3Sum (C/C++).

medium

source: https://leetcode.com/problems/3sum/
C/C++ Solution to LeetCode problem 15. 3Sum.

Problem

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Examples

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints

• 3 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Solution

We will use a Two Pointers technique.

• First we sort the numbers.
• Set a pointer next to this element (left pointer).
• Set a pointer to the end of the array (right pointer).
• We add together the 3 elements (the number where we are in the iteration, plus the numbers at the two pointers indexes).
• If the sum is 0 we add the numbers to the result.
• If result is < 0 we move the left pointer.
• If result is > 0 we move the right pointer.

To optimize, after procesing a set of numbers, we can move the pointers until the new number is different to the last processed.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> result; int l; int r; sort(nums.begin(), nums.end()); for(int i=0; i<nums.size()-2; i++) { if(i>0 && nums[i] == nums[i-1]) continue; l=i+1; r=nums.size()-1; while(l<r) { int res = nums[i] + nums[l] + nums[r]; if (res == 0) { result.push_back({nums[i], nums[l], nums[r]}); l += 1; r -= 1; while (l<r && nums[l-1] == nums[l]) l += 1; while (r>l && nums[r+1] == nums[r]) r -= 1; } else if (res > 0) r -= 1; else l += 1; } } return result; } };