CodeWithYash

 class Solution {     ArrayList<Integer> commonElements(int arr1[], int arr2[], int arr3[], int n1, int n2, int n3)      {                   HashSet<Integer> set1=new HashSet<Integer>(); HashSet<Integer> set2=new HashSet<Integer>(); HashSet<Integer> set3=new HashSet<Integer>(); for(int i=0;i<n1;i++){ set1.add(arr1[i]); } for(int i=0;i<n2;i++){ if(set1.contains(arr2[i])){ set2.add(arr2[i]); } } for(int i=0;i<n3;i++){ if(set2.contains(arr3[i])){ set3.add(arr3[i]); } } /*System.out.println("set1: "); for (Integer i : set1){             System.out.println(i); } System.out.println("set2: "); for (Integer i : set2){             System.out.println(i); }*/ Set<Integer> set4 = new TreeSet<Integer>(set3); /*System.out...

  public   static  List<Integer> reverseArray(List<Integer> a) { Approach 1 Time O(N) Space O(N)        List<Integer> aa= new  ArrayList<Integer>();         for ( int  i= 0 ;i<a.size();i++){            aa.add(a.get(i));        }         Collections.reverse(aa);          return  aa;     } /* Given is the list so we have two choices 1.either to use arrays for storing data 2.either to use ArrayList for storing data The thing is we need to return the List in the end so we can use directly the ArrayList. */ Approach 2 Time O(N) Space O(1)     Collections.reverse(a);     return a;