I have a piece of code and I'm a bit confused how to deal with my issue so, please review method below. I was trying to search for a solution but unfortunately none of them fit my needs so I am looking for an advice here. The method is taking a String and removing duplicated characters so for example - input: ABBCDEF should return ABCDEF, but when entering i+1 in the last iteration I got IndexOutOfBound Exception, so I can iterate until string.length-1 but then I loose the last element, what is the SMARTEST solution in your opinion, thanks.
public String removeDuplicates(String source){
if(source.length() < 2){
return source;
}
StringBuilder noDuplicates = new StringBuilder();
char[] string = source.toCharArray();
for(int i = 0; i < string.length-1; i++){
if(string[i] != string[i+1]){
noDuplicates.append(string[i]);
}
}
return noDuplicates.toString();
}
You could do this like so: append the first character in source, and then only append subsequent characters if they are not equal to the previously-appended character.
if (source.isEmpty()) {
return source; // Or "", it doesn't really matter.
}
StringBuilder sb = new StringBuilder();
sb.append(source.charAt(0));
for (int i = 1; i < source.length(); ++i) {
char c = source.charAt(i);
if (c != sb.charAt(sb.length() - 1)) {
sb.append(c);
}
}
return sb.toString();
But if you wanted to do this more concisely, you could do it with regex:
return source.replaceAll("(.)\\1+", "$1");
You could simply append the last character after the loop:
public String removeDuplicates(String source){
...
noDuplicates.append(string[string.length - 1]);
return noDuplicates.toString();
}
You have a simple logic error:
You make your string to a char array.
That is fine, but the length property of any array will show you the
human way of counting if someting is in it.
If there is 1 element the length will be 1
2 -> 2
3 -> 3
etc.
You get the idea.
So when you go string[i + 1] you go one character to far.
You could just change the abort condition to
i < = string.length - 2
Or you could write a string iterator, to be able to access the next element, but
that seems like overkill for this example
This is just what LinkedHashSet was made for! Under the hood it's a HashSet with an iterator to keep track of insertion order, so you can remove duplicates by adding to the set, then reconstruct the string with guaranteed ordering.
public static String removeDuplicates(String source) {
Set<String> dupeSet = new LinkedHashSet<>();
for (Character v : source.toCharArray()) {
dupeSet.add(v.toString());
}
return String.join("", dupeSet);
}
If you wish to remove all repeating characters regardless of their position in the given String you might want to consider using the chars() method which provides a IntStream of the chars and that has the distinct() method to filter out repeating values. You can then put them back together with a StringBuilder like so:
public class RemoveDuplicatesTest {
public static void main(String[] args) {
String value = "ABBCDEFE";
System.out.println("No Duplicates: " + removeDuplicates(value));
}
public static String removeDuplicates(String value) {
StringBuilder result = new StringBuilder();
value.chars().distinct().forEach(c -> result.append((char) c));
return result.toString();
}
}
Related
I am trying to write a program for these instructions:
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note: For the purpose of this problem, we define empty String as valid palindrome.
https://leetcode.com/problems/valid-palindrome/
For some reason, the .reverse() in the last line doesn't reverse it. I've tried debugging by adding print statements and I see that the string DOES reverse earlier on. What is going on here? Please guide me!
public static boolean isPalindrome(String s) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i))) {
sb.append(s.charAt(i));
}
}
String x = sb.reverse().toString();
System.out.println(x);
if (sb.length() == 0)
return true;
else
return sb.toString().equals(x);
}
The problem is that reverse() changes the StringBuilder itself. So you are comparing the reverse, with the reverse again. Instead change your method to:
String x = sb.toString();
return sb.reverse().toString().equals(x);
Which will compare the StringBuilder before reversing, with the reversed StringBuilder. Note that the if(sb.length == 0) is not necessary, as if it is empty, sb.reverse().toString().equals(x) will still return true.
Also your search is case sensitive, when the problem statement says that it should match regardless of case. Change where you append to:
if (Character.isLetter(s.charAt(i))) {
sb.append(Character.toLowerCase(s.charAt(i)));
}
Also you can take advantage of replaceAll and toLowerCase() to shorten your method to:
public static boolean pali(String s) {
String copy = s.toLowerCase().replaceAll("[^a-z]", "");
return new StringBuilder(copy).reverse().toString().equals(copy);
}
Because after first invocation of sb.reverse() sb has changed it state.
i get termination due to timeout error when i compile. Please help me
Given two strings, determine if they share a common substring. A substring may be as small as one character.
For example, the words "a", "and", "art" share the common substring "a" . The words "be" and "cat" do not share a substring.
Input Format
The first line contains a single integer , the number of test cases.
The following pairs of lines are as follows:
The first line contains string s1 .
The second line contains string s2 .
Output Format
For each pair of strings, return YES or NO.
my code in java
public static void main(String args[])
{
String s1,s2;
int n;
Scanner s= new Scanner(System.in);
n=s.nextInt();
while(n>0)
{
int flag = 0;
s1=s.next();
s2=s.next();
for(int i=0;i<s1.length();i++)
{
for(int j=i;j<s2.length();j++)
{
if(s1.charAt(i)==s2.charAt(j))
{
flag=1;
}
}
}
if(flag==1)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
n--;
}
}
}
any tips?
Below is my approach to get through the same HackerRank challenge described above
static String twoStrings(String s1, String s2) {
String result="NO";
Set<Character> set1 = new HashSet<Character>();
for (char s : s1.toCharArray()){
set1.add(s);
}
for(int i=0;i<s2.length();i++){
if(set1.contains(s2.charAt(i))){
result = "YES";
break;
}
}
return result;
}
It passed all the Test cases without a time out issue.
The reason for the timeout is probably: to compare two strings that each are 1.000.000 characters long, your code needs 1.000.000 * 1.000.000 comparisons, always.
There is a faster algorithm that only needs 2 * 1.000.000 comparisons. You should use the faster algorithm instead. Its basic idea is:
for each character in s1: add the character to a set (this is the first million)
for each character in s2: test whether the set from step 1 contains the character, and if so, return "yes" immediately (this is the second million)
Java already provides a BitSet data type that does all you need. It is used like this:
BitSet seenInS1 = new BitSet();
seenInS1.set('x');
seenInS1.get('x');
Since you're worried about execution time, if they give you an expected range of characters (for example 'a' to 'z'), you can solve it very efficiently like this:
import java.util.Arrays;
import java.util.Scanner;
public class Whatever {
final static char HIGHEST_CHAR = 'z'; // Use Character.MAX_VALUE if unsure.
public static void main(final String[] args) {
final Scanner scanner = new Scanner(System.in);
final boolean[] characterSeen = new boolean[HIGHEST_CHAR + 1];
mainloop:
for (int word = Integer.parseInt(scanner.nextLine()); word > 0; word--) {
Arrays.fill(characterSeen, false);
final String word1 = scanner.nextLine();
for (int i = 0; i < word1.length(); i++) {
characterSeen[word1.charAt(i)] = true;
}
final String word2 = scanner.nextLine();
for (int i = 0; i < word2.length(); i++) {
if (characterSeen[word2.charAt(i)]) {
System.out.println("YES");
continue mainloop;
}
}
System.out.println("NO");
}
}
}
The code was tested to work with a few inputs.
This uses a fast array rather than slower sets, and it only creates one non-String object (other than the Scanner) for the entire run of the program. It also runs in O(n) time rather than O(n²) time.
The only thing faster than an array might be the BitSet Roland Illig mentioned.
If you wanted to go completely overboard, you could also potentially speed it up by:
skipping the creation of a Scanner and all those String objects by using System.in.read(buffer) directly with a reusable byte[] buffer
skipping the standard process of having to spend time checking for and properly handling negative numbers and invalid inputs on the first line by making your own very fast int parser that just assumes it's getting the digits of a valid nonnegative int followed by a newline
There are different approaches to solve this problem but solving this problem in linear time is a bit tricky.
Still, this problem can be solved in linear time. Just apply KMP algorithm in a trickier way.
Let's say you have 2 strings. Find the length of both strings first. Say length of string 1 is bigger than string 2. Make string 1 as your text and string 2 as your pattern. If the length of the string is n and length of the pattern is m then time complexity of the above problem would be O(m+n) which is way faster than O(n^2).
In this problem, you need to modify the KMP algorithm to get the desired result.
Just need to modify the KMP
public static void KMPsearch(char[] text,char[] pattern)
{
int[] cache = buildPrefix(pattern);
int i=0,j=0;
while(i<text.length && j<pattern.length)
{
if(text[i]==pattern[j])
{System.out.println("Yes");
return;}
else{
if(j>0)
j = cache[j-1];
else
i++;
}
}
System.out.println("No");
return;
}
Understanding Knuth-Morris-Pratt Algorithm
There are two concepts involved in solving this question.
-Understanding that a single character is a valid substring.
-Deducing that we only need to know that the two strings have a common substring — we don’t need to know what that substring is.
Thus, the key to solving this question is determining whether or not the two strings share a common character.
To do this, we create two sets, a and b, where each set contains the unique characters that appear in the string it’s named after.
Because sets 26 don’t store duplicate values, we know that the size of our sets will never exceed the letters of the English alphabet.
In addition, the small size of these sets makes finding the intersection very quick.
If the intersection of the two sets is empty, we print NO on a new line; if the intersection of the two sets is not empty, then we know that strings and share one or more common characters and we print YES on a new line.
In code, it may look something like this
import java.util.*;
public class Solution {
static Set<Character> a;
static Set<Character> b;
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
for(int i = 0; i < n; i++) {
a = new HashSet<Character>();
b = new HashSet<Character>();
for(char c : scan.next().toCharArray()) {
a.add(c);
}
for(char c : scan.next().toCharArray()) {
b.add(c);
}
// store the set intersection in set 'a'
a.retainAll(b);
System.out.println( (a.isEmpty()) ? "NO" : "YES" );
}
scan.close();
}
}
public String twoStrings(String sOne, String sTwo) {
if (sOne.equals(sTwo)) {
return "YES";
}
Set<Character> charSetOne = new HashSet<Character>();
for (Character c : sOne.toCharArray())
charSetOne.add(c);
Set<Character> charSetTwo = new HashSet<Character>();
for (Character c : sTwo.toCharArray())
charSetTwo.add(c);
charSetOne.retainAll(charSetTwo);
if (charSetOne.size() > 0) {
return "YES";
}
return "NO";
}
This must work. Tested with some large inputs.
Python3
def twoStrings(s1, s2):
flag = False
for x in s1:
if x in s2:
flag = True
if flag == True:
return "YES"
else:
return "NO"
if __name__ == '__main__':
q = 2
text = [("hello","world"), ("hi","world")]
for q_itr in range(q):
s1 = text[q_itr][0]
s2 = text[q_itr][1]
result = twoStrings(s1, s2)
print(result)
static String twoStrings(String s1, String s2) {
for (Character ch : s1.toCharArray()) {
if (s2.indexOf(ch) > -1)
return "YES";
}
return "NO";
}
I am practicing Strings programming examples. i would like to reduce the given strings. it should eliminate a character if its in even numbers
example: Input - aaabbc, Output should be: ac
I have used HashMap to count and store character and count value and computing using value % 2 then continue or else print the output. But some of the test cases are failing in Hackerrank. Could you please help me identify the problem?
static String super_reduced_string(String s){
HashMap<Character, Integer> charCount = new HashMap<Character, Integer>();
StringBuilder output = new StringBuilder();
if (s == null || s.isEmpty()) {
return "Empty String";
}
char[] arr = s.toCharArray();
for (int i = 0; i < s.length(); i++) {
char c = arr[i];
if (!charCount.containsKey(c)) {
charCount.put(c,1);
} else {
charCount.put(c,charCount.get(c)+1);
}
}
for (char c:charCount.keySet()) {
if (charCount.get(c) % 2 != 0) {
output.append(c);
}
}
return output.toString();
}
The problem lies in how you are selecting the output. HashSets and HashMaps do not have any ordering associated with them. In your test case, the output could be either ac OR ca .
To solve this, you can do a variety of things. The quickest way I see is to take your orignal string, lets say s, and call
s.replace(c,"")
for ever char you need to remove.
I doubt this is anywhere near the most, or even mildly, efficient way to solve this, but it should work.
I'm new to java and I wrote this method to input a string word and output the word spelled backwards. The intent is to create a method and not use an already existing method such as the simple reverse. Please help point me in the direction of how to do this to reverse a word. I'm also trying to determine/count if there are palindromes. Please help! I've read other questions and I can't find anything specific enough to my case. I know that my code doesn't run, though I'm unsure how to fix it to get the correct output.
An example would be the word "backwards" to go to "sdrawkcab".
public static int reverseWord(String word) {
int palindromes = 0;
for (int i = word.length(); i >= 0; i--) {
System.out.print(i);
word.equalsIgnoreCase();
if (word.charAt(i)) == index(word.charAt(0 && 1))) {
palindromes++
System.out.println(palindromes)
}
return i;
}
}
There are multiple problems with your code.
1.The prototype of equalsIgnoreCase is
public boolean equalsIgnoreCase(String str);
So this method expect a String to be passed,but your not not passing anything here.To fix this,pass another string with whom you want to match your word like this..
word.equalsIgnoreCase("myAnotherString");
2.word.charAt(i);
Suppose word="qwerty",so indexing of each character will be like this
/* q w e r t y
0 1 2 3 4 5 */
So when you use i = word.length();i will 6 since word is of length 6.So
word.charAt(i) will search for character at index 6,but since there is not index 6,it will return an exception ArrayIndexOutOfBound.To fix this,start i from word.length()-1.
3.if (word.charAt(i));
This extra " ) ".Remove it.
Is Index() your own method?.If Yes,then check that also.
the below code prints the reverse of the input string and checks if it is a palindrome
public static void main(String[] args) {
String input = "dad";
char temp[] = input.toCharArray();//converting it to a array so that each character can be compared to the original string
char output[] = new char[temp.length];//taking another array of the same size as the input string
for (int i = temp.length - 1, j = 0; i >= 0; i--, j++) {//i variable for iterating through the input string and j variable for inserting data into output string.
System.out.print(temp[i]);//printing each variable of the input string in reverse order.
output[j] = temp[i];//inserting data into output string
}
System.out.println(String.valueOf(output));
if (String.valueOf(output).equalsIgnoreCase(input)) {//comparing the output string with the input string for palindrome check
System.out.println("palindrome");
}
}
Because your question about what is wrong with your code was already answered here is another way you could do it by using some concepts which are somewhat less low level than directly working with character arrays
public static boolean printWordAndCheckIfPalindrome(final String word) {
// Create a StringBuilder which helps when building a string
final StringBuilder reversedWordBuilder = new StringBuilder("");
// Get a stream of the character values of the word
word.chars()
// Add each character to the beginning of the reversed word,
// example for "backwards": "b", "ab", "cab", "kcab", ...
.forEach(characterOfString -> reversedWordBuilder.insert(0, (char) characterOfString));
// Generate a String out of the contents of the StringBuilder
final String reversedWord = reversedWordBuilder.toString();
// print the reversed word
System.out.println(reversedWord);
// if the reversed word equals the given word it is a palindrome
return word.equals(reversedWord);
}
I'm using codingbat.com to get some java practice in. One of the String problems, 'withoutString' is as follows:
Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed (not case sensitive).
You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with "xxx" removing "xx" leaves "x".
This problem can be found at: http://codingbat.com/prob/p192570
As you can see from the the dropbox-linked screenshot below, all of the runs pass except for three and a final one called "other tests." The thing is, even though they are marked as incorrect, my output matches exactly the expected output for the correct answer.
Here's a screenshot of my output:
And here's the code I'm using:
public String withoutString(String base, String remove) {
String result = "";
int i = 0;
for(; i < base.length()-remove.length();){
if(!(base.substring(i,i+remove.length()).equalsIgnoreCase(remove))){
result = result + base.substring(i,i+1);
i++;
}
else{
i = i + remove.length();
}
if(result.startsWith(" ")) result = result.substring(1);
if(result.endsWith(" ") && base.substring(i,i+1).equals(" ")) result = result.substring(0,result.length()-1);
}
if(base.length()-i <= remove.length() && !(base.substring(i).equalsIgnoreCase(remove))){
result = result + base.substring(i);
}
return result;
}
Your solution IS failing AND there is a display bug in coding bat.
The correct output should be:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours is:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours fails because it is removing spaces, but also, coding bat does not display the correct expected and run output string due to HTML removing extra spaces.
This recursive solution passes all tests:
public String withoutString(String base, String remove) {
int remIdx = base.toLowerCase().indexOf(remove.toLowerCase());
if (remIdx == -1)
return base;
return base.substring(0, remIdx ) +
withoutString(base.substring(remIdx + remove.length()) , remove);
}
Here is an example of an optimal iterative solution. It has more code than the recursive solution but is faster since far fewer function calls are made.
public String withoutString(String base, String remove) {
int remIdx = 0;
int remLen = remove.length();
remove = remove.toLowerCase();
while (true) {
remIdx = base.toLowerCase().indexOf(remove);
if (remIdx == -1)
break;
base = base.substring(0, remIdx) + base.substring(remIdx + remLen);
}
return base;
}
I just ran your code in an IDE. It compiles correctly and matches all tests shown on codingbat. There must be some bug with codingbat's test cases.
If you are curious, this problem can be solved with a single line of code:
public String withoutString(String base, String remove) {
return base.replaceAll("(?i)" + remove, ""); //String#replaceAll(String, String) with case insensitive regex.
}
Regex explaination:
The first argument taken by String#replaceAll(String, String) is what is known as a Regular Expression or "regex" for short.
Regex is a powerful tool to perform pattern matching within Strings. In this case, the regular expression being used is (assuming that remove is equal to IS):
(?i)IS
This particular expression has two parts: (?i) and IS.
IS matches the string "IS" exactly, nothing more, nothing less.
(?i) is simply a flag to tell the regex engine to ignore case.
With (?i)IS, all of: IS, Is, iS and is will be matched.
As an addition, this is (almost) equivalent to the regular expressions: (IS|Is|iS|is), (I|i)(S|s) and [Ii][Ss].
EDIT
Turns out that your output is not correct and is failing as expected. See: dansalmo's answer.
public String withoutString(String base, String remove) {
String temp = base.replaceAll(remove, "");
String temp2 = temp.replaceAll(remove.toLowerCase(), "");
return temp2.replaceAll(remove.toUpperCase(), "");
}
Please find below my solution
public String withoutString(String base, String remove) {
final int rLen=remove.length();
final int bLen=base.length();
String op="";
for(int i = 0; i < bLen;)
{
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
i +=rLen;
continue;
}
op += base.substring(i, i + 1);
i++;
}
return op;
}
Something things go really weird on codingBat this is just one of them.
I am adding to a previous solution, but using a StringBuilder for better practice. Most credit goes to Anirudh.
public String withoutString(String base, String remove) {
//create a constant integer the size of remove.length();
final int rLen=remove.length();
//create a constant integer the size of base.length();
final int bLen=base.length();
//Create an empty string;
StringBuilder op = new StringBuilder();
//Create the for loop.
for(int i = 0; i < bLen;)
{
//if the remove string lenght we are looking for is not less than the base length
// and the base substring equals the remove string.
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
//Increment by the remove length, and skip adding it to the string.
i +=rLen;
continue;
}
//else, we add the character at i to the string builder.
op.append(base.charAt(i));
//and increment by one.
i++;
}
//We return the string.
return op.toString();
}
Taylor's solution is the most efficient one, however I have another solution that is a naive one and it works.
public String withoutString(String base, String remove) {
String returnString = base;
while(returnString.toLowerCase().indexOf(remove.toLowerCase())!=-1){
int start = returnString.toLowerCase().indexOf(remove.toLowerCase());
int end = remove.length();
returnString = returnString.substring(0, start) + returnString.substring(start+end);
}
return returnString;
}
#Daemon
your code works. Thanks for the regex explanation. Though dansalmo pointed out that codingbat is displaying the intended output incorrectly, I through in some extra lines to your code to unnecessarily account for the double spaces with the following:
public String withoutString(String base, String remove){
String result = base.replaceAll("(?i)" + remove, "");
for(int i = 0; i < result.length()-1;){
if(result.substring(i,i+2).equals(" ")){
result = result.replace(result.substring(i,i+2), " ");
}
else i++;
}
if(result.startsWith(" ")) result = result.substring(1);
return result;
}
public String withoutString(String base, String remove){
return base.replace(remove,"");
}