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How to check for palindrome excluding the non-alphanumeric characters?

Here's the code that I attempted
public String isPalindrome(String s) {
String trimmed = s.replaceAll("[^A-Za-z0-9]", "");
String reversed = "";
int len = trimmed.length();
for (int i = len - 1; i >= 0; i--) {
char[] allChars = trimmed.toCharArray();
reversed += allChars[i];
}
if (trimmed.equalsIgnoreCase(reversed)) {
return "true";
} else {
return "false";
}
}
Sample Input 1
A man, a plan, a canal: Panama
Sample Output 1
true
Explanation 1
The given string is palindrome when considering only alphanumeric characters.
Sample Input 2
race a car
Sample Output 2
false
Explanation 2
The given string is not a palindrome when considering alphanumeric characters.

Your variable len comes from the length of the String s. But you use the value on the array coming from trimmed.
So if you want to remove the IndexOutOfBoundsException you should change your len declaration to:
int len = trimmed.length();

You can return boolean instead of String:
public static boolean isPalindrome(String s) {
String trimmed = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
int from = 0, to = trimmed.length() - 1;
while (from < to) {
if (trimmed.charAt(from) != trimmed.charAt(to)) {
return false;
}
from++;
to--;
}
return true;
}

You can use StringBuilder to reverse a String:
public static void main(String[] args) {
String input = "a#b!b^a";
String clean = input.replaceAll("[^A-Za-z0-9]", "");
String reverse = new StringBuilder(clean).reverse().toString();
boolean isPalindrome = reverse.equals(clean);
System.out.println(isPalindrome);
}

You can do like this in linear time as the loops are driven by the presence of non-alphabetic/digit characters. Also, no trimming or reversal of the string is required.
String[] test = {"A man, a plan, a canal: Panama",
"race a car","foobar", "ABC2CEc2cba"};
for (String s : test) {
System.out.printf("%5b -> %s%n", isPalindrome(s), s);
}
prints
true -> A man, a plan, a canal: Panama
false -> race a car
false -> foobar
true -> ABC2CEc2cba
The outer while loop drives then entire process until the indices cross or are equal. The inner loops simply skip over non-alphabetic/digit characters.
public static boolean isPalindrome(String s) {
int k = s.length() - 1;
int i = 0;
char c1 = '#';
char c2 = '#';
while (i <= k) {
while (!Character.isLetterOrDigit(c1 = s.charAt(i++)));
while (!Character.isLetterOrDigit(c2 = s.charAt(k--)));
if (Character.toLowerCase(c1) != Character.toLowerCase(c2)) {
return false;
}
}
return true;
}

LeetCode 14. longest common prefix

Question:
Write a function to find the longest common prefix string among an array of strings. If there is no common prefix, return an empty string "".
Example 1:
Input: ["flower","flow","flight"]
Output: "fl"
Example 2:
Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Code:
public class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs==null || strs.length==0)
return "";
for(int i=0;i<strs[0].length();i++) {
char x = strs[0].charAt(i);
for(int j=0;j<strs.length;j++) {
if((strs[j].length()==i)||(strs[j].charAt(i)!=x)) {
return strs[0].substring(0,i);
}
}
}
return strs[0];
}
}
This is the second solution, but I don't understand the inner loop.
I think if the second element in strs returns a string and ends the for loop, the third element will not have a chance to be compared.

You have to check same position in all of the words and just compare it.
positions
word 0 1 2 3 4 5
=====================
w[0] F L O W E R
w[1] F L O W
w[2] F L I G H T
In Java:
class Main {
public static void main(String[] args) {
String[] words = {"dog","racecar","car"};
String prefix = commonPrefix(words);
System.out.println(prefix);
// return empty string
String[] words2 = {"dog","racecar","car"};
String prefix2 = commonPrefix(words2);
System.out.println(prefix2);
// Return "fl" (2 letters)
}
private static String commonPrefix(String[] words) {
// Common letter counter
int counter = 0;
external:
for (int i = 0; i < words[0].length(); i++) {
// Get letter from first word
char letter = words[0].charAt(i);
// Check rest of the words on that same positions
for (int j = 1; j < words.length; j++) {
// Break when word is shorter or letter is different
if (words[j].length() <= i || letter != words[j].charAt(i)) {
break external;
}
}
// Increase counter, because all of words
// has the same letter (e.g. "E") on the same position (e.g. position "5")
counter++;
}
// Return proper substring
return words[0].substring(0, counter);
}
}

Your first loop is itterating over all chars in the first string of array. Second loop is checking char at i posistion of all strings of array. If characters do not match, or length of string is the same as i it returns substring result.
I think the best way to understand is debug this example.

If the char in the second string is different than the char in the first one, then it is correct to return, since it means that the common prefix ends there. Checking the third and following strings is not necessary.
Basically it returns as soon as it finds a mismatch char.

If we first sort them then it would be very easy we have to only go and compare the first and the last element in the vector present there so,
the code would be like,This is C++ code for the implementation.
class Solution {
public:
string longestCommonPrefix(vector<string>& str) {
int n = str.size();
if(n==0) return "";
string ans = "";
sort(begin(str), end(str));
string a = str[0];
string b = str[n-1];
for(int i=0; i<a.size(); i++){
if(a[i]==b[i]){
ans = ans + a[i];
}
else{
break;
}
}
return ans;
}
};

public class Solution {
public string LongestCommonPrefix(string[] strs) {
if(strs.Length == 0)
{
return string.Empty;
}
var prefix = strs[0];
for(int i=1; i<strs.Length; i++) //always start from 1.index
{
while(!strs[i].StartsWith(prefix))
{
prefix = prefix.Substring(0, prefix.Length-1);
}
}
return prefix;
}
}

unable to pass test case in autocomplete

Question
Autocomptete
Doug was using Google and was amazed to see the autocomptete feature How autocomptete works it search the database for all the possible words that can be formed using the characters that are provided by user (as input)
For ex If a user type 'cis' in the search bar then suggestions would be
• cisco
• cist
• cissp
• cism
• cisa
He thought about applying the same feature in his search engine. In his prototype he took a string as domain which contained all the words he could search.
As his designer you have to tell him how many autocomptete options will be provided to him if something is entered in the input field.
This is my code for the following problem.
import java.util.ArrayList;
import java.util.List;
public class Test {
public static void main(String[] args) {
String input1 = "Hello world with warm welcome from Mr.kajezevu";
String input2 = "w";
//output should be any word starting with w i.e {world,warm,welcome}
List < String > l = new ArrayList < String > ();
String[] str = input1.split("\\s+");//splits a given string at spaces
for (int i = 0; i < str.length; i++) {
if (str[i].length() >= input2.length()) { // checks if the length of input2 is not greater than the str[i] selected
if (input2.equals(str[i].substring(0, input2.length()))) { //comparing the two string if they are equal
l.add(str[i]);
}
}
}
String[] result = l.toArray(new String[l.size()]);
for (int i = 0; i < result.length; i++) {
System.out.println(result[i]);
}
}
}
But my solution is passing only one test case and also its failing complexity case.
i can't figure out whats wrong with it.

It seems you missed boundary conditions.
Below is code.
public static String[] autoComplete(String input1, String input2){
List<String> listOfPredictions = new ArrayList<String>();
String[] emptyArr = new String[0];
if(isEmpty(input1) || isEmpty(input2)){
return emptyArr;
}
input1 = input1.trim();
input2 = input2.trim();
String tokenizer = " " + input2;
int fromIdx = 1;
if(input1.startsWith(input2)){
fromIdx = input1.indexOf(" ");
listOfPredictions.add(input1.substring(0, fromIdx));
}
while(fromIdx > 0){
fromIdx = input1.indexOf(tokenizer, fromIdx) + 1;
if(fromIdx > 0){
listOfPredictions.add(input1.substring(fromIdx, input1.indexOf(" ", fromIdx)));
}
}
return listOfPredictions.toArray(emptyArr);
}
private static boolean isEmpty(String str){
return str == null || str.trim().length() == 0;
}

We also need to remove all duplicate words from the resulting array.
So first we break the string into words using the string.split() function.
Then push all those words that start with input2 string.
Then from the resulting array, we remove all duplicates by creating a Set and then converting it back into an Array.
function autoComplete(input1, input2) {
let results = [];
if(!input1 || !input1.length || !input2 || !input2.length) return results;
input1 = input1.trim();
input2 = input2.trim();
let allWords = input1.split(/\s+/);
allWords.forEach(word => {
if(word.startsWith(input2)) {
results.push(word);
}
})
results = [...[...new Set(results)]];
return results;
}

Decompressing String method

I need help with decompressing method. I have a working Compress method. Any suggestions as far as what I need to consider? Do I need parseInt or else....? Appreciate the advice. Here is what I have so far. If s = "ab3cca4bc", then it should return "abbbccaaaabc", for example of decompress.
class RunLengthCode {
private String pText, cText;
public RunLengthCode () {
pText = "";
cText = "";
}
public void setPText (String newPText) {
pText = newPText;
}
public void setCText (String newCText) {
cText = newCText;
}
public String getPText () {
return pText;
}
public String getCText () {
return cText;
}
public void compress () { // compresses pText to cText
String ans = "";
for (int i = 0; i < pText.length(); i++) {
char current = pText.charAt(i);
int cnt = 1;
String temp = "";
temp = temp + current;
while (i < pText.length() - 1 && (current == pText.charAt(i + 1))) {
cnt++;
i++;
temp = temp + current;
}
if (cnt > 2) {
ans = ans + current;
ans = ans + cnt;
}
else
ans = ans + temp;
setCText(ans);
}
}
public void decompress () {
}
}
public class {
public static void main(String [] args) {
Scanner in = new Scanner(System.in);
RunLengthCode myC = new RunLengthCode();
String pText, cText;
System.out.print("Enter a plain text consisting of only lower-case alphabets and spaces:");
pText = in.nextLine();
myC.setPText(pText);
myC.compress();
System.out.println(pText+" => "+myC.getCText());
System.out.print("Enter a compressed text consisting of only lower-case alphabets, spaces and digits:");
cText = in.nextLine();
myC.setCText(cText);
myC.decompress();
System.out.println(cText+" => "+myC.getPText());
}
}

You could create break the string into regx groups and combine them.
The following pattern works
(([A-Za-z]+[\d]*))
This will break your string "ab3cca4bc" into groups of
"ab3", "cca4", "bc"
So in a loop if the last character is a digit, you could multiply the character before it that many times.

Ok, so you've got an input string that looks like ab3cca4bc
1.) Loop over the length of the input String
2.) During each loop iteration, use the String.charAt(int) method to pick up the individual character
3.) The Character class has an isDigit(char) function that you can use to determine if a character is a number or not. You can then safely use Integer.parseInt(String) (you can use myChar+"" to convert a char into a String)
4.) If the char in question is a number, then you'll need to have an inner loop to repeat the previous character the correct number of times. How will you know what the last character was? Maybe have a variable that's instantiated outside the loop that you update each time you add a character on the end?

Trouble breaking from a method

I am having difficulties with my method returning true. It is a boolean method that takes two words and tries to see if one can be turned into the other by transposing two neighboring letters. I have had no troubles getting the false boolean. When the code gets to the for loop with an if statement in it it runs fine but does not return true when the if statement is satisfied. For some reason it continues through the for loop. For example, when comparing "teh" and "the" when the loop hits 1 the if statement is satisfied but does not return true, the for lo
public static boolean transposable(String word1, String word2)
{
ArrayList<Character> word1char = new ArrayList<Character>();
ArrayList<Character> word2char = new ArrayList<Character>();
int word1length = word1.length();
int word2length = word2.length();
int count = 0;
String w1 = word1.toUpperCase();
String w2 = word2.toUpperCase();
if(word1length != word2length)
{
return false;
}
for(int i = 0; i < word1length; i++)
{
char letter1 = w1.charAt(i);
word1char.add(letter1);
char letter2 = w2.charAt(i);
word2char.add(letter2);
}
for(int i = 0; i < word1length; i++)
{
char w1c = word1char.get(i);
char w2c = word2char.get(i);
if(w1c == w2c)
{
count++;
}
}
if(count < word1length - 2)
{
return false;
}
for(int i = 0; i < word1length; i++)
{
char w1c = word1char.get(i);
char w2c = word2char.get(i+1);
if(w1c == w2c)
{
return true;
}
}
return false;
}
op just keeps running. What am I doing wrong?

As pointed out in the comments this doesn't seem to be the easiest way around this problem. Here is a solution which tries to follow your logic and includes the use of toUpperCase() and ArrayLists.
Going over your code it looks like you were getting a bit lost in your logic. This is because you had one method trying to do everything. Break things down into smaller methods and you also will benefit by not having to repeat code and it keeps things much cleaner. The code below is tested with Java8 (although there is no reason why this should not work with Java 7).
public static void main(String args[]) {
String word1 = "Hello";
String word2 = "Hlelo";
transposable(word1, word2);
}
private static boolean transposable(String word1, String word2) {
// Get an ArrayList of characters for both words.
ArrayList<Character> word1CharacterList = listOfCharacters(word1);
ArrayList<Character> word2CharacterList = listOfCharacters(word2);
boolean areWordsEqual;
// Check that the size of the CharacterLists is the same
if (word1CharacterList.size() != word2CharacterList.size()) {
return false;
}
// check to see if words are equal to start with
areWordsEqual = checkIfTwoWordsAreTheSame(word1CharacterList, word2CharacterList);
System.out.print("\n" + "Words are equal to be begin with = " + areWordsEqual);
if (!areWordsEqual) {
/*
This loop i must start at 1 because you can't shift an ArrayList index of 0 to the left!
Loops through all the possible combinations and checks if there is a match.
*/
for (int i = 1; i < word1CharacterList.size(); i++) {
ArrayList<Character> adjustedArrayList = shiftNeighbouringCharacter(word2CharacterList, i);
areWordsEqual = checkIfTwoWordsAreTheSame(word1CharacterList, adjustedArrayList);
System.out.print("\n" + "Loop count " + i + " words are equal " + areWordsEqual + word1CharacterList + adjustedArrayList.toString());
if (areWordsEqual) {
break;
}
}
}
return areWordsEqual;
}
// takes in a String as a parameter and returns an ArrayList of Characters in the order of the String parameter.
private static ArrayList<Character> listOfCharacters(String word) {
ArrayList<Character> wordCharacters = new ArrayList<Character>();
String tempWord = word.toUpperCase();
for (int wordLength = 0; wordLength < tempWord.length(); wordLength++) {
Character currentCharacter = tempWord.charAt(wordLength);
wordCharacters.add(currentCharacter);
}
return wordCharacters;
}
// takes in two character arrayLists, and compares each index character.
private static boolean checkIfTwoWordsAreTheSame(ArrayList<Character> characterList1, ArrayList<Character> characterList2) {
// compare list1 against list two
for (int i = 0; i < characterList1.size(); i++) {
Character currentCharacterList1 = characterList1.get(i);
Character currentCharacterList2 = characterList2.get(i);
if (!currentCharacterList1.equals(currentCharacterList2)) {
return false;
}
}
return true;
}
// this method takes in an ArrayList of characters and the initial index that we want to shift one place to the left.
private static ArrayList<Character> shiftNeighbouringCharacter(ArrayList<Character> characterListToShift, int indexToShiftLeft) {
ArrayList<Character> tempCharacterList = new ArrayList<Character>();
int indexAtLeft = indexToShiftLeft - 1;
// fill the new arrayList full of nulls. We will have to remove these nulls later before we can add proper values in their place.
for (int i = 0; i < characterListToShift.size(); i++) {
tempCharacterList.add(null);
}
//get the current index of indexToShift
Character characterOfIndexToShift = characterListToShift.get(indexToShiftLeft);
Character currentCharacterInThePositionToShiftTo = characterListToShift.get(indexAtLeft);
tempCharacterList.remove(indexAtLeft);
tempCharacterList.add(indexAtLeft, characterOfIndexToShift);
tempCharacterList.remove(indexToShiftLeft);
tempCharacterList.add(indexToShiftLeft, currentCharacterInThePositionToShiftTo);
for (int i = 0; i < characterListToShift.size(); i++) {
if (tempCharacterList.get(i) == null) {
Character character = characterListToShift.get(i);
tempCharacterList.remove(i);
tempCharacterList.add(i, character);
}
}
return tempCharacterList;
}
Hope this helps. If you are still struggling then follow along in your debugger. :)