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

medium

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

Problem

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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

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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

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

Solution

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We will use a Two Pointers technique.

• First we sort the numbers.
• We start with the first element.
  • 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.

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;
  }
};