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CodeAbbey Challenge 14 data entry issues?

int fnum = Integer.parseInt(split[0]);// holds 5
//split[] holds each line of the file.
double sum = fnum;// sum = 5
double i = 0.0;
double last = 0.0;
for(int j = 1; j<(split.length-1);j++)
{
i = Integer.parseInt(split[j].replaceAll("[^0-9]", ""));
if(split[j].charAt(0) == '*')
{
sum = sum * i;
}
else if(split[j].charAt(0) == '/')
{
sum = sum / i;
}
else if(split[j].charAt(0) == '+')
{
sum = sum + i;
}
else if(split[j].charAt(0) == '-')
{
sum = sum - i;
}
else if(split[j].charAt(0) == '%')
{
sum = sum % i;
}
}
System.out.println(sum);// Prints 1.0
}
}
/*
Actual Data File Imported
5
+ 3
* 7
+ 10
* 2
* 3
+ 1
% 11
Answer should be : 1
*/
Alright My code may look messy, but I tried hard on it. Gave up a few times but tried again. My question is for smaller data sets such as the one I imported and commented out on the code on the last few lines, work fine. But for bigger data sets it's all wrong why is that? I've tried making al my data sets double to get bigger values but somehow it's wrong?
I'm a beginner so far, any help would be greatly appreciated.
To be more specific on the problem I imported the file, I made it all a String, line by line, then I added it all in a String array so each line was in a string array for example split[1] would print + 3. Now after that I isolated the number and the symbol in the if loop wrapped in a forloop to go over all the sets. Now the if loop captures the symbols and then does the appropriate arithmetic. SomeHow it didn't though? And I used a double instead of an int for sum. That didn't help.I believe the if statement could be the issue.

Not sure if you still need the answer but, here's a tip:
The whole point of that specific exercise was to learn the modular arithmetic, which is that if you sum up/multiply the remainders of all of the numbers, you get the same answer as you would using the numbers given, that is if you apply the same number that's after % for all of them.
For example:
14
+ 78
* 9
* 3
+ 4
% 3
After all of the applied operations, the numbers above % 3 result in 2488.
And so 2488 % 3 = 1.
So if you apply % 3 to each one of the numbers, including the initial one, you get the same answer, using the same operations on their remainders of course and dividing the sum again by 3.
14 % 3 = 2
78 % 3 = 0
9 % 3 = 0
3 % 3 = 0
4 % 3 = 1
So, you get 2 + 0 * 0 * 0 + 1 which equals to 1.
And 1 % 3 = 1 which is the same as 2488 % 3 = 1.
My point being, you should apply modulo to every one of the numbers, so you get little numbers and don't even have the big ones you're having problems with.
Hope this was clear enough and hope it helps.

Need to sum digits in random numbers from loop

I'm randomly generating 3 digit numbers using java.util.Random(). The number of random 3-digit numbers that the system prints is controlled by user input. Here's the code:
//If/then statement for determining game type (determines 3 digit numbers)
if (gameType == 3)
//for loop to generate random numbers (numGames is how many 3 digit numbers to print)
for (int i = 1; i <=numGames; i = i + 1){
int ranNums = randomGenerator.nextInt((999 - 100) + 1) + 100;
//Print random numbers to user
System.out.println(ranNums);
sumNums = ranNums % 10;
Basically, I want the code to add the DIGITS that are printed from the random number generator. For example, if the ranNums output is 111 and 111, I want the variable sumNums to equal 6, not 222. With the code I have now, it only adds the last digit of each number, (in this case 1 + 1 = 2).

You could do this:
sumNums = (n / 100) + (n / 10 % 10) + (n % 10);

Longest Snake Sequence

Question : A set of numbers separated by space is passed as input. The program must print the largest snake sequence present in the numbers. A snake sequence is made up of adjacent numbers such that for each number, the number on the right or left is +1 or -1 of it's value. If multiple snake sequences of maximum length is possible print the snake sequence appearing in the natural input order.
Example Input/Output 1:
Input:
5 6 7 9 8 8
Output:
5 6 7 8 9 8
8 9 8 7 6 5
Example Input/Output 2:
Input:
9 8 7 5 3 0 1 -2 -3 1 2
Output:
3 2 1 0 1
void doPermute(int[] in, StringBuffer out, boolean[] used, int length, int level, StringBuffer max) {
if (level == length) {
int count = 0;
for (int i = 1; i < out.length(); i++) {
if (Math.abs(Character.getNumericValue(out.charAt(i)) - Character.getNumericValue(out.charAt(i - 1))) != 1) {
//System.out.println(Character.getNumericValue(i) - Character.getNumericValue(i - 1) + " " + i + " yes");
count++;
break;
}
}
if (count == 0) {
max.append(out + " ");
}
return;
}
for (int i = 0; i < length; ++i) {
if (used[i]) {
continue;
}
out.append(in[i]);
used[i] = true;
doPermute(in, out, used, length, level + 1, max);
used[i] = false;
out.setLength(out.length() - 1);
}
}
As i am using StringBuffer my code passed the test cases that contains positive value (first test case) but failed in test cases containing negative values(second test case).
Update:-
I replaced stringbuffer with Integer[] and made few changes.it works fine for smaller inputs of length 8 or 9. How to make it fast for larger inputs of length 13 to 15?

Have you tried doing the process using an array of integers?
Scanner sc = new Scanner(System.in);
String s = sc.nextLine(); //The numbers entered in string format separated by spaces
String ss = s.split(" "); //Numbers separated by space will be put as individual numbers in a String array but each number is still in string format
int l = ss.length, i = 0;
int[] n = new int[l]; //The integer array which will store the values
for(i = 0; i < l; i++)
{
n[i] = Integer.parseInt(ss[i]); //Has integers now instead of string numbers
}
There might be creation of a few extra arrays but then calling the Character.getNumericValue() function repeatedly can also reduce efficiency. Also might solve your StringBuffer problem.
But SkillRack is very annoying anyway.

Your comparison isn't finding adjacent numbers for negative values.
For example: Abs(-2) - (-3) = 5 but -2 and -3 should be adjacent.
Ok. I see you're parsing - and digit separately.
Given the requirement of what a snake sequence is, the longest snake sequence for "5 6 7 9 8 8" is "5 6 7". The output listed above does not correspond to the definition: " adjacent numbers such that for each number, the number on the right or left is +1 or -1 of it's value". How does "5 6 7 8 9 8" meet the definition of snake sequence for "5 6 7 9 8 8"? Sorry I couldn't help.
You might want to parse the code into Integers, store the longest sequences in a map).
#Test
public void testSnake(){
String t2 = "9 8 7 5 3 0 1 -2 -3 1 2";
List<String> numsStr = Arrays.asList(t2.split(" "));
List<Integer> nums = new ArrayList();
HashMap<Integer,List<Integer> > numMap = new HashMap();
numsStr.forEach((s) -> {
Integer val = Integer.decode(s);
nums.add(val);
});
nums.forEach((num) -> {
System.out.println("num: " + num);
// track longest sequence, store in numMap
});
// Print numMap
}

using mod 3 in Java

So, I want to find what numbers between 1 and 100 are divisible by 3 and 7. I got it to work, except for one of the numbers. For some reason, 3 % 3 is giving me 3 as a remainder, but 6 % 3 is giving me 0. This is my code:
public class factors
{
public static void main(System args[])
{
//Variables
int integer, remainder;
//Displays header
System.out.print("Integers less than 100 that are \nevenly divisible by 3 or 7");
//Loops through each integer
for (integer = 1; integer <= 100; integer++)
{
remainder = integer % 3; //determines if 3 is a factor
if (remainder == 0) //displays integer
{
System.out.println(integer + " is divisible by 3");
}
remainder = integer % 7; //determines if 7 is a factor
if (remainder == 0) //displays integer
{
System.out.println(integer + " is divisible by 7");
}
}
}
}Does anyone know why this isn't working for the number 3?

You code is actually doing
remainder = 3 % 7; // equals 3.
The best way to determine why your code is not doing what you think is to step through your code using a debugger.
All the multiples of 3 & 7 will be multiples of 21, i.e. 21, 42, 63, 84.

Your 3 is getting tacked onto the end of the line of text above. You'll be seeing
Integers less than 100 that are
evenly divisible by 3 or 73
because you wrote print instead of println for this line of text. The % operator is working just fine, and 3 % 3 is indeed 0, not 3.

You are not outputting a remainder - you are displaying integer. So for 3 it should print 3.
Make you print statements more definite:
System.out.println(integer + " is divisible by 3"); // for the first `if`
and
System.out.println(integer + " is divisible by 7"); // for the second `if`
This should clear your confusion.
Your logic prints number divisible by 3 or 7.

Firstly, your code can be shortened to:
//and
for (int i = 1; i <= 100; i++){
if(i % 3 == 0 && i % 7 == 0) {
System.out.println(i);
}
}
//or
for (int i = 1; i <= 100; i++){
if(i % 3 == 0 || i % 7 == 0) {
System.out.println(i);
}
}
Also I note you're not declaring a type for your integer, remainder variables. I didn't attempt to recreate with those issues; start by solving that.

Why is the solution like this?

From Java Malik textbook- determine if an number is divisible by 11..
Code Solution provided:
import java.util.*;
public class Divby11
{
static Scanner console = new Scanner(System.in);
public static void main (String[] args)
{
int num, temp, sum;
char sign;
System.out.print("Enter a positive integer: ");
num = console.nextInt();
System.out.println();
temp = num;
sum = 0;
sign = '+';
do
{
switch (sign)
{
case '+' :
sum = sum + num % 10;
sign = '-';
break;
case '-' :
sum = sum - num % 10;
sign = '+';
}
num = num / 10; //remove the last digit
}
while (num > 0);
if (sum % 11 == 0)
System.out.println(temp + " is divisible by 11");
else
System.out.println(temp + " is not divisible by 11");
}
Why go through all the effort above and just say...
if (sum % 11 == 0)
System.out.println(temp + " is divisible by 11");
else
System.out.println(temp + " is not divisible by 11");
Can any of you experts see why the author would do it this way (long way)?

for the Divisibility Rule of 11:
form the alternating sum of the digits
if this sum is divisible for 11 then the number is divisible for 11
Examples
68090 = 0 - 9 + 0 - 8 + 6 = -11 => TRUE
493827 = 7 - 2 + 8 - 3 + 9 - 4 = 15 = 4 => FALSE

This code example isn't actually dividing by eleven. If you see, it's alternating between adding and subtracting each digit, then checks at the very end if the result is divisible by 11.
For example, look at the following number and how this algorithm works with it:
Start with sum=0, sign='+', num=517
First iteration: sum=7, sign='-', num=51
Second iteration: sum=6, sign='+', num=5
Final iteration: sum=11, sign='-', num=0
The final result is divisible by eleven.
EDIT: The algorithm does indeed look to be implementing the divisibility rule for 11 as dfa mentions in his answer.

You will have to provide more context from the book as to what the author was trying to demonstrate. This code example does not check to see if the number entered is divisible by 11. What it does is it adds every other digit, subtracts every other digit and then checks THAT number to see if it's divisible by 10.
EX
Entered number is 4938
It takes the 8 adds it to sum
Divides by ten giving 493
Takes the 3 subtracts it from sum: sum = 5
Divides by ten giving 49
Takes 9 and adds it to sum: sum = 14
Divides by ten giving 4
Takes 4 subtracts it from sum: sum = 10
THEN it checks if this is divisible by 11.
Ok, I know why now. He/she's trying to teach you something besides computing about numbers

I suspect it is simulating the manual test that digits in the odd positions and the digits in the even positions differ by a factor of 11. In practice using %11 would be the way to go.
EDIT: If the example were truly trying to avoid doing % 11, it should send the sum through again until it is 0.

It an example to show how to implement that particular check. Using your example would not demonstrate the same code methodologies.