You have L, a list containing some digits (0 to 9). Write a function answer(L) which finds the largest number that can be made from some or all of these digits and is divisible by 3. If it is not possible to make such a number, return 0 as the answer. L will contain anywhere from 1 to 9 digits. The same digit may appear multiple times in the list, but each element in the list may only be used once.

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Test cases

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Inputs:

(int list) l = [3, 1, 4, 1]

Output:

(int) 4311

Inputs:

(int list) l = [3, 1, 4, 1, 5, 9]

Output:

(int) 94311

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My Solution:

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package com.google.challenges;

import java.util.ArrayList;

import java.util.Arrays;

import java.util.Collections;

public class Answer {

public static int answer(int[] l) {

// Your code goes here.

ArrayList<Integer> list0 = new ArrayList<>();

ArrayList<Integer> list1 = new ArrayList<>();

ArrayList<Integer> list2 = new ArrayList<>();

int sum =0;

Arrays.sort(l);

for(int i = 0; i<l.length; i++){

if(l[i] % 3 == 0){

list0.add(l[i]);

}else if(l[i] % 3 == 1){

list1.add(l[i]);

}else{

list2.add(l[i]);

}

sum += l[i];

}

if(sum%3==0){

StringBuilder strNum = new StringBuilder();

for(int i = l.length-1; i >= 0; i--)

{

strNum.append(l[i]);

}

return Integer.parseInt(strNum.toString());

}else if(sum%3 == 1){

if(list1.size()>0){

Collections.sort(list1);

list1.remove(0);

}else if(list2.size() >= 2){

Collections.sort(list2);

list2.remove(1);

list2.remove(0);

}else{

return -1;

}

}else if(sum%3 == 2){

if(list2.size()>0){

Collections.sort(list2);

list2.remove(0);

}else if(list1.size() >= 2){

Collections.sort(list1);

list1.remove(1);

list1.remove(0);

}else{

return -1;

}

}

list0.addAll(list1);

list0.addAll(list2);

StringBuilder strNum = new StringBuilder();

Collections.sort(list0);

for(int i = list0.size()-1; i >= 0; i--)

{

strNum.append(list0.get(i));

}

return strNum.length() > 0 ? Integer.parseInt(strNum.toString()) : -1;

}

}

}}

But here I am able to pass 4 test cases out of 5. Therefore I am looking for scenario which is left to check.

Can someone help me?