Maximum Possible Number by Binary Concatenation

class Solution { int max = 0; public int maxGoodNumber(int[] nums) { int visited[] = new int[3]; List<Integer> list = new ArrayList<>(); find(nums,visited,list); return max; } public void find(int nums[], int visited[],List<Integer> list){ //base case if(list.size()==3){ //System.out.println(list); max = Math.max(max,Integer.parseInt(Integer.toBinaryString(list.get(0)) + Integer.toBinaryString(list.get(1)) + Integer.toBinaryString(list.get(2)),2)); return; } for(int i = 0;i<3;i++){ if(visited[i]!=1){ visited[i] =1; list.add(nums[i]); find(nums,visited,list); visited[i] = 0; list.remove(list.size()-1); } } } }