content: package codingquestions.leetcode;
  
  import java.util.ArrayList;
  import java.util.Collections;
  import java.util.List;
  
  /*
   * 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.
   * 
   * 
   */
  public class SumOfThree {
  
      public static void main(String[] args) {
  
      }
  
      public List<List<Integer>> threeSum(int[] nums) {
          List<List<Integer>> list = new ArrayList<>();
  
          for (int i = 0; i < nums.length - 2; i++) {
              for (int j = i + 1; j < nums.length - 1; j++) {
                  for (int k = j + 1; k < nums.length; k++) {
                      if ((nums[i] + nums[j] + nums[k]) == 0) {
                          List<Integer> list1 = new ArrayList<>();
                          list1.add(nums[i]);
                          list1.add(nums[j]);
                          list1.add(nums[k]);
                          Collections.sort(list1);
                          if (!list.contains(list1)) {
                              list.add(list1);
                          }
                      }
                  }
              }
          }
  
          return list;
      }
  }
  