So I'm trying to solve the Longest Substring Without Repeating Character problem in a webpage and when I'm trying to upload it it will show me this bug:
class Solution {
public int lengthOfLongestSubstring(String s) {
HashSet<Character> hash = new HashSet<>();
int count = 0, finalCount = 1;
char prevChar = s.charAt(0);
hash.add(prevChar);
for (int i = 1; i < s.length(); i++)
{
char character = s.charAt(i);
if (!hash.contains(character)){
hash.add(character);
count++;
if (count > finalCount) finalCount = count;
}
else{
hash.clear();
hash.add(character);
count = 1;
}
prevChar = character;
}
return finalCount;
} }
Is there anything wrong with it?
If not, do you think my algorithm was efficient? I can't compare its performance since the webpage won't let me upload it.
You call s.charAt(0) in line 5. I imagine they pass in the empty string as a test case and you are getting an out of bounds exception. Prior to line 5 add a check to see if the string length is 0 and if it is return 0.
According to the error description it's doing a dummy-spit at line 5 of the Solution class.
Based on the picture that's:
char prevChar = s.charAt(0);
The error is ArrayIndexOutOfBounds which generally indicates you tried to get more out of something than was actually there (e.g. running over the end of an array).
Here I'd suggest maybe putting in some System.out.println lines at line 3 to sanity check the method parameter, e.g.:
(a) if the input String s is null
or
(b) if the input String s is empty (e.g. "")
charAt(0) will get the first character, but if there are zero characters then trying to get the 1th character is an error, no?
NB: something like this:
System.out.println("Input was :" + s + ":");
Will show both of those conditions, as either:
Input was ::
for an empty String
Input was :null:
for a null String
The problem is:
Client accounts are filed under a classification system using codes eg MA400. I need a method that will reset the original MA400 to an updated code such as MA400.4. If the new code has 5 characters to which the original is reset then the method returns true. Not the best wording but that is all I have right now.
It hasn't been specified if the characters need to be in the same order, eg.
String str = "abc123";
String newStr = "xyz123abc";
I am assuming they need to be in the same order. So the above strings would only have 3 like characters.
char[]array = str.toCharArray();
char[]array2 = newStr.toCharArray();
I am thinking now to use a compareTo method on the two arrays, but I am not sure how this would work exactly. Perhaps I could use a for loop to stop comparing after the final element in the shortest string but not entirely sure if I can do much with that.
I feel like I am going about this in the wrong way and there is a less complicated way to check for like characters in a string?
From what I understand something like this will work. Remember this will only count unique characters. Order does not matter
public static boolean matchingChar(final String st1, final String st2) {
if(st1 == null || st2 == null || st1.length() < 5 || st2.length() < 5) {
return false;
}
//This is if you wish unique characters to be counted only
//Otherwise you can use simple int count = 0
HashSet<Character> found = new HashSet<Character>();
//found.size() < 5 so the loop break as soon as the condition is met
for(int i = 0; i < st1.length() && found.size() < 5; i++) {
if(st2.indexOf(st1.charAt(i)) != -1) {
found.add(st1.charAt(i));
}
}
return found.size() >= 5;
}
OK so I'm trying to design a simple program that checks to see if a substring of length 4 characters is within all initial strings. Here is my code as follows:
public class StringSearch{
private String[] s1Array = {"A","C","T","G","A","C","G","C","A","G"};
private String[] s2Array = {"T","C","A","C","A","A","C","G","G","G"};
private String[] s3Array = {"G","A","G","T","C","C","A","G","T","T"};
//{for (int i = 0; i < s1Array.length; i++){
// System.out.print(s1Array[i]);
//}}//check if Array loaded correctly
/**
* This is the search method.
*
* #param length length of sub string to search
* #param count counter for search engine
* #param i for-loop counter
* #return subStr returns strings of length = 4 that are found in all 3 input strings with at most
* one mismatched position.
*/
public String Search()
{
int length = 4;
int count = 0;
int i = 0;
ArrayList<StringSearch> subStr = new ArrayList<StringSearch>();
//String[] subStr = new String[4];
do
{
for (i = count; i < length; i++){
subStr.add(s1Array[i]); // cant find .add method???
count = count + 1;
}
if (s2Array.contains(subStr) && s3Array.contains(subStr)){ //can't find .contains method???
System.out.println(subStr + "is in all 3 lists.");
}
if (count = s1Array.length){
System.out.println("Task complete.");
}
else{
count = count - length;
count = count + 1;
}
}while (count <= s1Array.length);
}
}
For some reason, Java cannot seem to find the .add or .contains methods and I have no idea why. So my approach was to turn the initial Strings each into an array (since the assignment specified each string would be exactly N elements long, in this case N = 10) where 1 letter would be 1 element. The next thing I did was set up a for loop that would scan s1Array and add the first 4 elements to an ArrayList subStr which is used to search s2Array and s3Array. Here is where .add isn't a valid method, for whatever reason. Commenting that out and compiling again, I also ran into an issue with the .contains method not being a valid method. Why won't this work? What am I missing? Logically, it seems to make sense but I guess maybe I'm missing something in the syntax? Help would be appreciated, as I'm a Java novice.
There are lots of errors and misunderstandings here.
Let's start with #1
private String[] s1Array = {"A","C","T","G","A","C","G","C","A","G"};
Making an array of strings is just silly, you should either use a single string or an array of characters.
private String s1 = "ACTGACGCAG";
Or
private char[] s1Array = {'A','C','T','G','A','C','G','C','A','G'};
Now #2
ArrayList<StringSearch> subStr = new ArrayList<StringSearch>();
This means you are trying to make an ArrayList that contains objects of type StringSearch. StringSearch is a class that contains your three arrays and your Search function so I don't think this is what you want.
If you wanted to make a list of 3 strings you might do something like this:
ArrayList<String> stringList = new ArrayList<String>();
stringList.add(s1);
stringList.add(s2);
stringList.add(s3);
Now say you defined s1, s2 and s3 as strings you can do something like this.
for(int i = 0; i <= s1.length() - 4; i++)
{
String subStr = s1.substring(i, i + 4);
if(s2.contains(subStr) && s3.contains(subStr))
{
System.out.println(subStr + " is in all 3 lists.");
}
}
System.out.println("Task Complete.");
The above code should achieve what it looks like you are trying to do. However, it should be noted that this isn't the most efficient way, just a way, of doing it. You should start with some more basic concepts judging by the code you have so far.
After declaring subStr as ArrayList you can call add or contains only with StringSearch objects as parameters.
Instead of:
ArrayList<StringSearch> subStr = new ArrayList<StringSearch>();
Replace it with:
String subStr = "";
And within the for loop to get the first 4 letters in s1 to be in its own string (subStr) add the line:
subStr += s1Array[i];
Also, s1Array is a String array, and not a String. The .contains method is a method that belongs to String variables, so for eg. the way you have it implemented, you can say s1Array[i].contains. But you cannot say s1Array.contains. If you change your String arrays to Strings and edit your code to suit, everything should work the way you expect it to work.
First of all you need to educate yourself on the concept of Java generics.
The most basic thing about generics is that once you declare a collection, here it is the arraylist, as you can only add objects of StringSearch.
Second of all, logically what you can do is to implement an algorithm called
Longest Common Subsequence. Check in pairs whether the longest subsequeces are 4 or not on the arrays.
I have a string: LOAN,NEFT,TRAN. I want to substring the string based on getting a , during traversing the string. So I tried to first get a count for how many , are there. but not sure what function to user to get what I want. Also this should be dynamic, meaning I should be able to create as many substrings as required based on number of ,s. I tried the following code:
package try1;
public class StringTest {
public static void main(String[] args) {
String str="LOAN,NEFT,TRAN";
int strlen=str.length();
int count=0;
for(int i=0;i<strlen;i++)
{
if(str.contains("'"))
count++;
}
System.out.println(""+count);
for (int j=0;j<count;j++)
{
//code to create multiple substrings out of str
}
}
}
But I do not think contains() is the function I am looking for because value of count here is coming 0. What should I use?
Your code doesn't actually count the , characters because 1) contains doesn't take into account your loop variable 2) it's searching for ', not ,
Assuming you want to work at a low level rather than using high level functions like .split(), then I'd recommend the following.
for(char c : str.toCharArray()) {
if (c == ',') {
count++;
}
}
You can use split to get substrings directly:
String[] substrings = str.split(",");
Is this what you want as an output: (shown below)?
["LOAN", "NEFT", "TRAN"] // Array as an output
Or to just get the count of the splitting char, you can use the same line as above with this:
int count = substrings.length - 1;
I know there is an easier way of making things work to check whether a string is a palindrome or not, but I wanted to try it using library functions and I came up with the code below.
public boolean isPalindrome1(String input)
{
int length = input.length()/2;
if(input.length()%2!=0)
{
length = length + 1;
}
return(input.substring(0,length).equals(new StringBuilder(input.substring(length, input.length())).reverse().toString()));
}
I'm trying to check whether half the string is equal to the other half's reverse. But it is getting messed up for odd and even lengths. Can someone point corrections in this where it will work for odd, even lengths as well as empty string and string of length = 1.
You're already using reverse(). Why can you not compare the input String with the reverse? Isn't that exactly what you're wanting in the first place? No need to be splitting things in half in a complicated way.
you can:
return new StringBuilder(input).reverse().toString().equals(input);
here you go as you want to know :
public boolean isPalindrome(String input) {
for (int i = 0; i < input.length() / 2; i++) {
if (input.charAt(i) != input.charAt(input.length() - 1 - i)) {
return false;
}
}
return true;
}