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Question about the task

Streets and houses

Java Syntax, Level 7, Lesson 4

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heip in finding error

1. Create an array of 15 integers.
2. Populate it with values from the keyboard.
3. Let the array index represent the house number. The array value at a particular index represents the number of people living in the corresponding house.
Houses with odd numbers are located on one side of the street. Those with even numbers are on the other side. Find out which side of the street has more people living on it.
4. Display the following message: "Odd-numbered houses have more residents." or "Even-numbered houses have more residents."

Note:
The house at index 0 is considered even.

5. If the number of residents on each side of the street is the same, then output nothing.

Requirements:

The program must create an array of 15 integers.
The program should read numbers for the array from the keyboard.
The program should display "Odd-numbered houses have more residents." if the sum of odd array elements is greater than the sum of even ones.
The program should display "Even-numbered houses have more residents." if the sum of even array elements is greater than the sum of odd ones.
The program should not output text to the console if the number of residents on each side is the same.

package com.codegym.task.task07.task0706; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; /* Streets and houses */ public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int[] n=new int[15]; for(int i=0;i<n.length;i++) { n[i]=Integer.parseInt(br.readLine()); } int e=0, o=0; for(int i=0; i<n.length; i++) { if(n[i]%2 == 0) e++; else o++; } if(e > o) { System.out.println("Even-numbered houses have more residents."); } else { System.out.println("Odd-numbered houses have more residents."); } } }

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ThomasLC

Level 16 , Singapore, Singapore

24 May 2019, 11:24

Your algorithm / understanding of the conditions is wrong. Right now you're just adding even and odd counter by one each time you see an even or odd number of people in the list. It's asking you to total up the numbers in even and odd positions instead. I.e. 5 3 7 8 Even number houses are position 0 and 2, which has 5 and 7 ppl. Even has 12 ppl in total Odd number houses are position 1 and 3 which is 3 and 8.odd has 11 ppl in total The answer should then be even numbered houses have more people. Your current algorithm will answer odd because there's 3 odd numbers and 1 even number in the array.