/**
 Sort the array , iterate over the array by having left and right pointers and find the    sum.

*/
public Set<List<Integer>> searchTriplets2(int[] array) {
    Arrays.sort(array);
    Set<List<Integer>> triplets = new HashSet<>();
    // Iterate only length -2 since you cant make triplets with only 2 elements
    for (int i = 0; i < array.length - 2; i++) {
        // Left is 1 ahead of i when it starts
        int right = array.length - 1;
        // Right always starts from the end
        int left = i+1;
        while (left < right) {
            int currentSum = array[i] + array[left] + array[right];
            if (currentSum == 0) {
                triplets.add(Arrays.asList(array[i], array[left], array[right]));
                left++;
                right--;
         

class Solution:
  def searchTriplets(self, arr):
    triplets = []
    # TODO: Write your code here
    arr.sort()
    n=len(arr)
    for i in range(0,n-2):#go upto the third last element only 
      el=arr[i]
      if i>0 and arr[i]==arr[i-1]:
        continue
      j=i+1
      k=n-1
      while(j<k):
        #TODO: logic for skipping second element already chosen
        el2=arr[j]
        el3=arr[k]

        if arr[j]+arr[k]==(0-el):
          triplets.append([el,el2,el3])
          j+=1
          k-=1  
          while(j<k and arr[j]==el2):
            j+=1
          while(j<k and arr[k]==el3):
            k-=1
        elif arr[j]+arr[k]<(0-el):
          j+=1
          while(j<k and arr[j]==el2):
            j+=1
        else:
          k-=1
          while(j<k and arr[k