Please help me, the code runs according to the task conditions but it is not verifying.

Question about the task

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I spent a lot of time testing the code with different inputs, i got it to run according to the task conditions but it still fails to verify. I am out of options, i need your help caring Java community. Thanks.

1. Create a list of numbers.
2. Use the keyboard to add 10 numbers to the list.
3. Display the length of the longest sequence of repeating numbers in the list.

Example for the list 2, 4, 4, 4, 8, 8, 4, 12, 12, 14:
3

The value is 3, because the longest sequence of repeating numbers is three fours.

Requirements:

The program must display a number on the screen.
The program should read values from the keyboard.
In the main method, declare an ArrayList variable with Integer elements and immediately initialize it.
The program should add 10 numbers to the collection in accordance with the conditions.
The program should display the length of the longest sequence of repeating numbers in the list.

package com.codegym.task.task08.task0812; import java.io.*; import java.util.ArrayList; /* Longest sequence */ public class Solution { static int count = 1; public static int sequence() throws IOException{ BufferedReader read = new BufferedReader(new InputStreamReader(System.in)); ArrayList<Integer> list = new ArrayList<Integer>(); for(int i= 0; i < 10; i++){ list.add(i, Integer.parseInt(read.readLine())); } for(int i = 0; i < list.size()-1; i++){ if (list.get(i) == list.get(i + 1)) { count++; //System.out.println(i); // System.out.println(count); } } return count; } public static void main(String[] args) throws IOException { //write your code here // System.out.println(); System.out.println(sequence()); //System.out.println(); } }

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Godwin

Level 14 , Sango, Nigeria

23 May 2019, 18:12

i tested the code, it won't work if the sequence is at the end of the list, the sequence is always 2, how do i correct this?

Godwin

Level 14 , Sango, Nigeria

23 May 2019, 17:07

package com.codegym.task.task08.task0812; import java.io.*; import java.util.ArrayList; /* Longest sequence */ public class Solution { public static int sequence() throws IOException { BufferedReader read = new BufferedReader(new InputStreamReader(System.in)); ArrayList<Integer> list = new ArrayList<Integer>(); for (int i = 0; i < 10; i++) { list.add(i, Integer.parseInt(read.readLine())); } int count = 1; int finalCount = 0; for (int i = 0; i < list.size() - 1; i++) { if (list.get(i) == list.get(i + 1)) { count++; } if (count > finalCount) { finalCount = count; } count = 1; } return finalCount; } public static void main(String[] args) throws IOException { System.out.println(sequence()); } } It still wont verify, but it works as you have suggested. What else can i do?

ThomasLC

Level 16 , Singapore, Singapore

23 May 2019, 00:52

You're almost there! Your algorithm is wrong as it's now just one "count" variable that increases everytime there's a match. i.e. input: 2445566 Count goes up to "2" at the third digit, cause 4 repeats, then goes up to "3" at the fifth digit cause 5 repeats, then "4" at the last digit cause 6 repeats. The correct output is supposed to be "2" cause the longest running sequence is 2 numbers only. What you can do is create another variable called "finalCount" which checks if it's more than count, if it is, set finalCount as count and use finalCount as your final result. You then need to reset your count counter i.e. count = 1, finalCount = 0 input: 2445566 at the third digit, count increases to 2. Count 2 is more than finalCount 0, so finalCount is now set to same as Count 2. Reset the initial count counter back to 1. at fifth digit, count is now 2 again, but 2 is not more than the current finalCount 2, so finalCount is still 2. Hope this helps! :)