24x7 Google chrome technical support number 1-888-201-2039 Interview Questions
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Answer123+123+123
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abc 24x7 Google chrome technical support number 1-888-201-2039 - 0of 4 votes
AnswersYou 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.
- Parth Patel February 21, 2017 in United States
{{
Test cases
==========
Inputs:
(int list) l = [3, 1, 4, 1]
Output:
(int) 4311
Inputs:
(int list) l = [3, 1, 4, 1, 5, 9]
Output:
(int) 94311
}}
My Solution:
{{
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?| Report Duplicate | Flag | PURGE
Google Software Engineer Google FooBar 24x7 Google chrome technical support number 1-888-201-2039 Arrays Computer Science Java Problem Solving - 0of 0 votes
AnswerYou have a cycled doubly linked list (meaning there is a cycle and each node has prev() and next() method).
- Patrick February 21, 2017 in United States
You can set/check the value of each node in the list to be 0/1 (method setValue(0/1) getValue())
Find how many elements there are in the list.
You start from the some node and you don’t know the status of the nodes value, could be any combination of 1’s and 0’s)..| Report Duplicate | Flag | PURGE
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