3 Sum Leetcode solution

15. 3Sum

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.

 

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 : 

Leetcode 15  explanation:  

• we iterate the array nums using iterator `i` from 0 to sizeof(nums) -2 .
• for every element , nums[i] we apply the apply the twoSums problem concept using two pointer approach. 
• we take two pointers l = i+1 and m = size -1 .
• `while(l < m && nums[l] == nums[l+1]   ) l++` and `while(l < m && nums[m] == nums[m-1]  ) m--` are used to avoid repeatitions.

class Solution {

public:

    vector<vector<int>> threeSum(vector<int>& nums) {

           

           int size = nums.size();

           vector<vector<int>>result;

           sort(nums.begin(), nums.end());

           for(int i = 0 ; i < size-2 ; i++)

           {   if(i ==0 || nums[i] != nums[i-1] )

              {

               

               int l = i+1;

               int m = size -1;

               while(l<  m)

               {

                   if(nums[l] + nums[m] + nums[i] == 0)

                   {    

                       vector<int>sub_result;

                       sub_result.push_back(nums[i]);

                       sub_result.push_back(nums[l]);

                       sub_result.push_back(nums[m]);

                       result.push_back(sub_result);

                      while(l < m && nums[l] == nums[l+1]   ) l++ ;

                     while(l < m && nums[m] == nums[m-1]  ) m--;

                      l++; m--;

                   }

                   else

                   if(nums[l] + nums[m] < -nums[i] )

                   {

                       l++;

                   }

                   else

                    m--;

               }

              }

           }

           return result;

    }

};

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