I was wondering how to code the game, Mastermind, in Java, but with things up a notch (I want to inform the user not only how many pegs they got right or wrong, but also how many they guessed correctly in the wrong slots).
For instance, say the RNG answer of 5 digits of numbers 1-6 is:
22354
... and the user's guess is:
32624
Resulting in:
two guessed correctly (2 and 4)
two guessed partially correct (2 and 3)
one guessed incorrectly (6).
Here's my code for informing the user what they got correct:
String answer = "22354";
String guess = "32624";
int correctPegs = 0;
for (int i = 0; i < 5; i++) {
char a = answer.charAt(i);
char g = guess.charAt(i);
if (a == g) {
correctPegs++;
}
}
System.out.println(correctPegs);
How would I find the partially correct ones?
... and for calculating how many the user guessed incorrectly, I was thinking of using basic algebra to find the remaining characters after finding the correct and partially correct ones.
I would put '0' into the strings where they match. Then for every char that is not a zero, I would look through the answer string again. If any character in the answer string matches, I would make them both zero, and increment outOfPlacePegs.
This guards against two corner cases I am assuming you do not want:
One is when you have '323' as the answer and '223' as the input. You don't want the first '2' to be recorded as out of place.
Second is when you have '223' as the answer and '442' as the input. You don't want this recorded as two numbers out of place.
Related
I have program in which 4 random numbers are set to 4 buttons (one no. for each button). There are two random variables (rbvalue and loadG4). The rbvalue number is different for each button, but the loadG4 value overrides a button's value and replaces it. The idea is, none of the rbvalue numbers are meant to be equal to loadG4, and a part of the program ensures they are never equal. Here's the code:
Random GenerateG4 = new Random();
int loadG4 = GenerateG4.nextInt(10);
Random randoms1 = new Random();
final TextView number = (TextView) findViewById(R.id.number);
number.setText(""+loadG4);
for(int allrbA=0; allrbA<4; allrbA++) {
int rbvalue = randoms1.nextInt(10 - 1);
if (rbvalue==loadG4) rbvalue=9;
selectrb[allrbA].setText(""+rbvalue);
}
selectrb[rbselector].setText(""+loadG4);
The part that makes this work is:
if (rbvalue==loadG4) rbvalue=9;
Simply adding that line of code in did the job. Now out of all numbers generated from 0 to 9, there is only one of which is the value of loadG4. How did that one line do this? I always thought that after if (...) there has to be curly brackets for the actual statements e.g if (...) {System.out.println("...")} Why set rbvalue to 9? Why is there (10 -1) for the random rbvalue? What does that -1 even mean in this situation?
I'd appreciate anyone who could clear this up for me, thank you.
You can take 10 - 1 as primary school math, it equals 9, so randoms1.nextInt(10 - 1) returns a number from 0 through 8 inclusive. I don’t know why it’s not written as 9. It might be an attempt to stress that the value is 1 smaller than the 10 used in GenerateG4.nextInt(10).
When only one statement is executed when the if condition is true, the curly braces are a matter of opinion. Very many prefer to have them there; in my workplace no one gets away with leaving them out.
The if statement sets rbvalue to 9 if it was equal to loadG4. That will make sure the values differ. Since rbvalue was at most 8, we know that when they are equal, setting one of them to 9 will ensure they become different. It’s a bit subtle for my taste, I understand you needed to ask.
The code doesn’t appear to do anything to make the rbvalue numbers different for each button.
Forgive me if this is not formatted properly, this is my first post. I looked to see if this issue has been found before and I cannot find anyone who has had the same problem I am having.
I am trying to learn Java and cannot for the life of me figure out why my for loops are not outputting the last iteration. I am going through codeabbey's exercises and completed the first two relatively easily. However on the third and fourth problems, I cant get my for loop to output during the last iteration.
I began looking on google and thought I would compare my answer to someone else's. I couldn't see why mine wouldn't work when my code was almost identical to the person I found. So I copied their code and to my surprise I had the same problem when this code also would not output on the last iteration.
So, here is the context.
The website gives you a single number first which is the number of sets of the following numbers. For the third problem, you are to add the sets of two, output the sum followed by a space and loop through the entire batch. For the fourth problem, it is similar where the first number is the number of sets in the batch but you are to compare the two numbers and output the lower number. I will copy my code here for the third problem because the code is simpler.
Here is the code:
import java.util.Scanner;
public class Summation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 0; i < n;i++){
int a = in.nextInt();
int b = in.nextInt();
System.out.println(a + b + " ");
}
}
}
Here is the input you are to copy and paste:
3
100 8
15 245
1945 54
and this is my output:
108 260
So, as you can see we are missing the last output here. I tried changing the for loop to (i < (n+1) ) which still didn't change anything. Any help would be GREATLY appreciated!
Ok so I tested it, and with your numbers, typing them in one by one it works. Copy pasting them, press enter one more time at the end of the copy. If you don't press enter, the scanner thinks you're still adding to the second number so it won't continue until enter is pressed.
I would try using println() as someone else suggested, or calling flush() at the end of the program to make sure something isn't being held in a buffer and not being written.
I'm attempting to write a Java program that searches for a specific substring (xyz) within a user-entered string, and keeping a running count, unless that substring is preceded by a period. At this point in the class, we've only used charAt and length, so if possible I need to stick to that. Additionally, we haven't used regular expressions at all, so that's out the window too.
I've managed to get the program working as desired, with one notable exception: if the String entered begins with a period, it fails to count any successive matches. This is what I've got so far:
System.out.println("Give me a String:");
String s1 = kb.nextLine();
int index = 0;
int count = 0;
while(index <= s1.length() - 1 && s1.charAt(index) != '.')
{
if(s1.charAt(index) == 'x' && s1.charAt(index + 2) == 'z')
{
count++;
}
index++;
}
System.out.println(count);
You can simply check the input string whether it starts with period. If so then you can use the following piece of code to handle the validation.
if(s1.charAt(0)!='.')
{
while(index <= s1.length() - 1 && s1.charAt(index) != '.')
{
if(s1.charAt(index) == 'x' && s1.charAt(index + 2) == 'z')
{
count++;
}
index++;
}
}
else
{
index=1;
while(index <= s1.length() - 1 && s1.charAt(index) != '.')
{
if(s1.charAt(index) == 'x' && s1.charAt(index + 2) == 'z')
{
count++;
}
index++;
}
}
System.out.println(count);
}
As this seems like a homework type of question I will attempt to guide you in the right direction first and provide a solution at a later time. I strongly encourage you to work through the problem on your own first to the best of your ability before you look at my solution (once I post it) and to read this page before going ANY further
First, consider the kinds of inputs you could receive. Since you didn't specify any limitations you could get things like:
"" (empty string)
"\n" (whitespace)
"x" (a single character)
"xx" (two characters string)
"abc" (string of correct length, but not containing your substring)
".xyz" (the substring to be ignored)
I could go on, but I'm sure you can come up with all the various combinations of weird things you might receive. These are just a few examples to get you started (along with those I posted in the comments already)
Next, think about what you need your algorithm to do. As I said in the comments it sounds like you want to count the occurrences of the substring "xyz" while ignoring the occurrences of the substring ".xyz". Now consider how you're going to look for these substrings - you're going to advance one character at a time from left to right across the String looking for a substring that matches one of these two possibilities. When you find one of them you'll either ignore it or count it.
Hopefully this helps and as I said, I will post a solution later after you've had some time to wrestle with the code. If you do solve it go ahead and post your solution (maybe edit your question to add the new code or add an answer) Finally I once again strongly urge you to read this page if you have not already.
EDIT #1:
I wanted to add a little more information and that is: you already have a pretty good idea of what you need to do in order to count your "xyz" substring at this point - despite the small flaw in the logic for inputs like "xaz", which is easily fixable. What you need to focus on is how to ignore the substring ".xyz" so think about how you could implement the ignore logic, what does it mean to ignore it? Once you answer that it should start coming together for you.
EDIT #2:
Below you will find my solution to the problem. Once again it's important to understand how the solution works not just copy and paste it. If you simply copy my code without understanding it you're cheating yourself out of the education that you're trying to gain. I don't have time at the moment to describe in detail why and how this code works, but I do plan to edit again later to add those details.
import java.util.Scanner;
public class Main {
private static Scanner scan = new Scanner(System.in);
public static void main(String[] args) {
System.out.println("Give me a String:");
String s1 = scan.nextLine();
System.out.println(countSubstrings(s1));
}
public static int countSubstrings(String s1){
int index = 0;
int count = 0;
while (index < s1.length()-2) {
if(s1.charAt(index) == '.' && s1.charAt(index+1) != '.'){
index++;
}
else if (index+2 < s1.length() && s1.charAt(index) == 'x' && s1.charAt(index + 1) == 'y'
&& s1.charAt(index + 2) == 'z') {
count++;
index+=2;
}
index++;
}
return count;
}
}
EDIT #3:
Here is the nuts and bolts of why the above code does what it does. First, we think about the fact that we're looking for 3 items (a triple) in a specific order within an array and if we see a fourth item (a period) immediately preceding the first item of the triple then we need to ignore the triple.
Per my previous edit we need to define what it means to ignore. In this case what we mean is to simply not count it and move on with our search for valid substrings to count. The simplest way to do that is to advance the index without incrementing the count.
So, ask yourself the following:
When should my loop stop? Since we're looking for triples we know we can stop if the length of the input String is less than 3 or when there are less than 3 characters left in the String that we have not examined yet. For example if the input is "xyzab" by the time we get to index 3 we know there's no possible way to form a triple where "a" is the first character in the triple and that therefore our counting is done.
Is there ever a time when I would not want to skip the next 3 characters after a period? After all the goal is to look for triples so wouldn't I want to skip 3 characters not just 1? Yes there is a time when you do NOT want to skip 3 characters and that's when you have something like ".axyz" because a valid triple could start as soon as the 2nd character past the period. So in fact you want to skip only 1 character.
This, and the fact that index is always incremented by 1 at the end of the loop (more on this later), is why the first condition inside the while only advances the index by 1:
if(s1.charAt(index) == '.' && s1.charAt(index+1) != '.'){
index++;
}
Is there ever a time when I would see a period and not wish to ignore (skip) the next character? Yes, when the next character is another period because it could indicate that another triple needs to be skipped. Consider the input "..xyz" which would result in a wrong answer if you encounter the first period and skip the second period since your algorithm could see the next three characters as a valid triple but in fact it is invalid because of the second period.
This is why the second half of the above condition exists:
`&& s1.charAt(index+1) != '.'`
Now ask yourself how to identify a valid triple. I'm sure by now you can see how to do this - check the current character, the next character, and the character after that for the values you want. This logic is the latter portion of the second if condition within the while:
s1.charAt(index) == 'x' && s1.charAt(index + 1) == 'y'
&& s1.charAt(index + 2) == 'z'
Whenever you're using calculations like index +1 or index +2 inside of a loop that is incrementing the index until it reaches a boundary you have to consider the possibility that your calculation will exceed the boundary because you can't rely on the loop to check this for you as the loop will not perform that check until either the end or beginning of the loop (depending on which kind of loop it is)
Considering the above you must ask yourself: How do I prevent out of boundary scenarios when I use these index+1, index+2, etc types of calculations? The answer is to add another piece to your condition:
index+2 < s1.length()
You may be wondering - why not add two checks since we're using index+1 and index+2? We only need one check to see if the greatest index we use will exceed the boundary in this case. If index +2 is beyond the bounds we don't care if index+1 is or is not because it won't matter we can't possibly have a matching substring.
Next, inside of the second if inside the while you see there is code to increment the index by 2: index+=2; This is done for efficiency since once we have identified a triple we know there is no way to form another triple with characters that are already part of another triple. Therefore we want to skip over them and just like the first bullet point we take advantage of the loop incrementing the index so we only need to increment by 2 and let the loop add the extra 1 later.
Finally we reach the end of the logic within the loop. This part you're already familiar with and that's the index++; which simply increments the position within the String that we're currently examining. Note that this works in tandem with the first bullet point. Take the example from the first bullet point of ".axyz". There is a period in index 0 and the character in index 1 is not another period so the logic from the first bullet point will increment index by 1, making it 1. At the end of the loop index is incremented again making it 2 thereby skipping over the period - at the start of the next loop index is 2, it was never 1 at the start of the loop.
Well, I hope this helps to explain how it all works and illustrate how to think about these sorts of problems. The basic principle is to visualize where your current element is and how you can use that to achieve your goal. At the same time think about what kinds of properties the different elements of your program have and how you can take advantage of them - such as the fact that once you identify a triple it is safe to skip over those characters because they have the property of only being usable once. As with any program you always want to try to create as many test inputs as you can to test all the weird boundary cases that might occur to ensure the correctness of your code. I realize you probably are not familiar with JUnit but it is a very useful tool and you might try researching the basics of using it when you have a little spare time, and the bonus is that if you use the Eclipse IDE it has integrated JUnit functionality you can use.
This question already has answers here:
Replace a character at a specific index in a string?
(9 answers)
Closed 8 years ago.
Can someone help me understand why I'm getting an "Unexpected Type Error"?
if(s.charAt(i-1) == ' '){
String sub = s.substring(i, s.indexOf(' ')+1);
for(int j = 0; j < sub.length()/2; j++){
char temp;
temp = sub.charAt(j);
sub.charAt(j) = sub.charAt(sub.length()-1-j);
sub.charAt(sub.length()-1-j) = temp;
sub = sub+" ";
complete = complete+sub;
}
}
I'm getting the error on lines 6 and 7. I can't figure out why and I'd appreciate the help!
charAt() returns the character. It is not a left side operand aka you cannot assign a value to it.
Strings are immutable, which means you cannot change them (this seems to be your intention).
Instead: create a new String and add to that.
If this confused you, I try to elaborate a little: the assignment operator takes whatever is on the right and tries to shove it into whatever is on the left of it.
The problem here is that some things do not like it when you try to shove other things into them. You cannot put everything on the left that you want. Well, you can try:
"everything" = 5;
but this does not work, neither does this:
"everything" = 42;
Set aside what that last snippet failing implies to the universe and everything, your problem at hand is that charAt() is also one of those things that do not like it on the left side of the assignment operator.
I'm afraid there's no way to turn charAt() into one of those things that like it on the left side. A week after stranding on a deserted island without any plants but only solar powered refrigerators filled with steaks, this probably works:
vegetarian = meat;
Even though the vegetarian doesn't like it there, he'd accept his situation being on the left side of the = operator. He eats some steaks.
This does not hold true for what you are trying, though. There's no such deserted island for charAt().
In these lines you're trying to set the value of functions' return. This is illegal
sub.charAt(j) = sub.charAt(sub.length()-1-j);
sub.charAt(sub.length()-1-j) = temp;
as far as I see you're trying to change the characters of a String, but Strings are imutable objects. So you'll need to use a StringBuffer to set the values.
I don't want the answer, I just don't understand how to scan the first number to tell the program how many pairs there are. If you could nudge me in the right direction I would greatly appreciate it.
"Most programs should be able to make some choices and decisions. And we are going to practice conditional programming now.
This is usually done by a kind of if ... else statements which may look like:
IF some_condition THEN
do_something
ELSE
do_other_thing
ENDIF
Depending on your programming language syntax could be different and else part is almost always optional. You can read more in wikipedia article on Conditional statements.
Of two numbers, please, select one with minimum value. Here are several pairs of numbers for thorough testing.
Input data will contain number of test-cases in the first line.
Following lines will contain a pair of numbers to compare each.
For Answer please enter the same amount of minimums separated by space, for example:
data:
3
5 3
2 8
100 15
answer:
3 2 15 "
Firstly, you might want to format your example data a bit. I understood it, but mostly only because I've seen that question format before.
Well, to answer your question but not the question's question (heh), note this:
Following lines will contain a pair of numbers to compare each.
Note the "lines" (plural) and the "each." We're going to need a loop.
We also know each line is a test case.
So modify the instructions:
Loop over the following test cases, comparing each pair
But how many times do we loop?
Input data will contain number of test-cases in the first line
That's the first number.
So here's our code skeleton:
//We can use a Scanner for convenience, it has a readInt() method
Scanner input = new Scanner(/*your input*/);
int numCases = input.readInt();
for(int i = 0; i < numCases; i++) {
int first = input.readInt(); //readInt() will also skip newlines, just a tip.
int second = input.readInt();
/* Compare two inputs, do stuff*/
}
For the record, you could also simply ignore the first input and just loop until there is no more input, but that's sloppy.
First you need to create a Scanner. If you're reading from the console, then this will work:
Scanner scan = new Scanner(System.in);
If you need to read from a file, then you can add this line as well.
System.setIn(new FileInputStream("inputFileName"));
For your specific case, you can do something like:
int numPairs = scan.nextInt();
You can find out more about Scanner and its methods from the Oracle documentation here.