Find all unique triplets in the array which gives the sum of zero

I am having a difficult time with this question. So I am trying to display the correct solutions in forms of strings but having a difficult time with the last for loop. Also is there a way to sort these arrays or just simply add the Arrays.sort function?

The questions follows:

//Given an array nums of n integers, are there elements a, b, c in nums 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.

//Example:

//Given array nums = [-1, 0, 1, 2, -1, -4],

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

and this is what I have so far

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        //Arrays.sort(nums);
        int isZero = 0;
        
        for(int i = 0; i < nums.length; i++)
        {
            for(int j = i+1; j< nums.length; j++)
            {
                for(int x = i + 2; x < nums.length;x++ )
                {
                    if(nums[i] + nums[j]+ nums[x] == isZero)
                    {
                        
                       
                    }
                }
            }
        }
        return Collections.emptyList();
        
    }
}

You need an outer List to store the arrays, and at each match save the 3 values

public List<List<Integer>> threeSum(int[] nums) {
    int isZero = 0;
    List<List<Integer>> result = new ArrayList<>();

    for(int i = 0; i < nums.length; i++){
        for(int j = i+1; j< nums.length; j++){
            for(int x = i + 2; x < nums.length;x++ ){
                if(nums[i] + nums[j]+ nums[x] == isZero){
                    result.add(Arrays.asList(nums[i], nums[j], nums[x]));
                }
            }
        }
    }
    return result;        
}

If you meant so sort the triplets, and so no have duplicate,

• use a Set
• sort the inner list before
• you can remove useless iteration with -2 and -1 on the end bound of the 2 first loops

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public static Set<List<Integer>> threeSum(int[] nums) {
    Set<List<Integer>> result = new HashSet<>();
    int isZero = 0;
    for (int i = 0; i < nums.length - 2; i++) {
        for (int j = i + 1; j < nums.length - 1; j++) {
            for (int x = i + 2; x < nums.length; x++) {
                if (nums[i] + nums[j] + nums[x] == isZero) {
                    List<Integer> tmp = Arrays.asList(nums[i], nums[j], nums[x]);
                    tmp.sort(Comparator.naturalOrder());
                    result.add(tmp);
                }
            }
        }
    }
    return result;
}

The below code passes all tests. The only issue is the time complexity of O(n*2), where time exceeds for very LARGE inputs. Welcome, if someone improves the algorithm.

class Solution {
public List<List<Integer>> threeSum(int[] A) {
    
 if ( A.length <= 2 ) return List.of();
	
	Set<List<Integer>> set = new HashSet<>();
	
	for (int i = 0; i < A.length; ++i) {
		
		for (int j = i + 1; j < A.length ; ++j) {
			int thirdNumber = - ( A[i] + A[j] ) ;
			List<Integer> tempp = Arrays.stream(A).boxed().collect(Collectors.toList());
			tempp.remove(Integer.valueOf(A[i]));
			tempp.remove(Integer.valueOf(A[j]));
			if (tempp.contains(Integer.valueOf(thirdNumber))) {
				List<Integer> temp = Arrays.asList(A[i], A[j], thirdNumber);
				Collections.sort(temp);
				set.add(temp);
			}				
		}
	}
	
	return new ArrayList<>(set);
}

}

You can sort the array before collecting those triplets, then those triplets are also sorted.

  public Set<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);  // Sort the array
    for (int i = 0; i < nums.length; i++) {
      for (int j = i + 1; j < nums.length; j++) {
        for (int x = j + 1; x < nums.length; x++) {
          if (nums[i] + nums[j] + nums[x] == 0) {
            res.add(Arrays.asList(nums[i], nums[j], nums[x]));
          }
        }
      }
    }
    return res;
  }