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Longest Most Common Substring Based on Whole-Word Phrases

I've been doing a lot of research around this topic and can't quite crack this one easily. There are a lot of valuable solutions I've come across online for solving this problem based on characters, but how would you solve this problem based on whole-word phrases to avoid the result returning a phrase that contains a partial word at the start or end of the phrase?
For example, given an Array of Strings, the output would be the most common whole-word phrase that is contained in most (not all) of the Strings within the Array.
This example below is the closest I've found so far but it only works about half of the time and includes partial word results which isn't quite what I'm after. I'm sure someone has solved this one before.
// function to find the stem (longest common
// substring) from the string array
public static String findstem(String arr[])
{
// Determine size of the array
int n = arr.length;
// Take first word from array as reference
String s = arr[0];
int len = s.length();
String res = "";
for (int i = 0; i < len; i++) {
for (int j = i + 1; j <= len; j++) {
// generating all possible substrings
// of our reference string arr[0] i.e s
String stem = s.substring(i, j);
int k = 1;
for (k = 1; k < n; k++)
// Check if the generated stem is
// common to all words
if (!arr[k].contains(stem))
break;
// If current substring is present in
// all strings and its length is greater
// than current result
if (k == n && res.length() < stem.length())
res = stem;
}
}
return res;
}
// Driver Code
public static void main(String args[])
{
String arr[] = { "grace", "graceful", "disgraceful",
"gracefully" };
String stems = findstem(arr);
System.out.println(stems);
}

Does this do what you intended. It simply checks to see if any word is a substring of itself and others.
If you want to check for real word substrings you would need to reference some dictionary which would be very time consuming.
String arr[] = { "grace", "graceful", "disgraceful",
"gracefully" };
String save = "";
int count = 0;
for (int i = 0; i < arr.length && count != arr.length; i++) {
count = 0;
for (int k = 0; k < arr.length; k++) {
if (arr[k].contains(arr[i])) {
count++;
save = arr[i];
}
}
}
System.out.println(save);

replacing all consecutive duplicates from the string with allowed no of occurrences [duplicate]

This question already has answers here:
How to replace multiple consecutive occurrences of a character with a maximum allowed number of occurences?
(2 answers)
Closed 3 years ago.
I need to write a method that takes a String as a parameter and returns a new String obtained by replacing every instance of repeated adjacent letters with a 'n' instances of that string.
For example, if "aaabcccd" as an input String and n =2, it returns "aabccd". I already tried the following code, but not getting expected output
String in = "aaadbbb";
char[] s = in.toCharArray();
int len = s.length;
int n = 2;
StringBuffer new_s = new StringBuffer("");
int count = 1;
char prev='\0';
for (int i = 0; i < len - 1; i++) {
if (s[i] == s[i + 1]) {
if(count <= n){
new_s.append(s[i]);
count++;
}else{
count=1;
}
} else {
new_s.append(s[i]);
}
}
System.out.println(new_s);
output-aaadb
expected-aadbb

Can be done with regexp magic using backreferences.
String in = "aaaaddbbbbc";
int n = 2;
String pattern = String.format("(([a-z])\\2{%d})\\2+", n - 1);
System.out.println(in.replaceAll(pattern, "$1"));
Outputs:
aaddbbc
Explanation:
The number inside {} is n-1.
([a-z]) is a capture group, matching any single lowercase letter from a to z. Since it's a second group of parentheses in the expression, it can be referenced as 2.
(([a-z])\\2{n}) means "match n+1 repetitions of same letter". It makes up a first capture group, and we'll use that as replacement
\\2+ matches all the extra repetitions of the same letter. They are discarded after replacement.

public static String test(String input, int repetitions) {
String flag = "";
String replacement = "";
String output = input;
ArrayList<Character> prevLetters = new ArrayList<Character>();
for(int x = 0; x < input.length(); x++) {
if(!prevLetters.contains(input.charAt(x))) {
for(int y = 0; y <= repetitions ; y++) {
flag += String.valueOf(input.charAt(x));
}
if(input.contains(flag)) {
replacement = flag.substring(0, flag.length()-1);
while(output.contains(flag)){
output = output.replace(flag, replacement);
}
}
flag = "";
prevLetters.add(input.charAt(x));
}
}
return output;
}
That is my solution, which follows a similar idea as yours. Rather than comparing each character value however, I thought it would be easier to simply check for a break in the rules (character appearing n+1 times in a row) and 'fix' it.
If you are interested in using your method, one potential issue that I noticed is that you aren't assigning count to 1 in your last else. You also won't have the chance to add the final character due to you only adding the character at index 'i' when the duration for the loop is len - 1.

To add one more alternative:
String in = "aaadbbbjjkllllllopp";
int n = 2;
StringBuilder sb = new StringBuilder();
char temp = in.charAt(0);
for(int i = 0; i < in.length()-1;){ // note that the incrementation of i is moved to the while loop
temp = in.charAt(i); // save current char in temp variable
int count = 0;
while (i < in.length() && in.charAt(i) == temp) { ///iterate as long as you find same chars or hit the end of the string
i++;
count++;
}
if (count > n){ // if and only if count is greater than max allowed set it to max allowed
count = n;
}
for(int j = 0; j < count; j++){ // append count chars
sb.append(temp);
}
}
System.out.println(sb.toString());

Look at this solution. You should take care of the last char in your input string, as you iterate only to the last but one.
private void replaceConsecutiveDuplicates() {
String input = "aaadbbb";
int n = 2;
StringBuffer sb = new StringBuffer();
int count = 1;
char current;
for( int i = 0; i < input.length(); ++i){
current = input.charAt(i);
if (i + 1 < input.length() && current == input.charAt(i + 1)) {
++count;
} else if (count > 1) {
for(int j = 0; j < n; ++j) {
sb.append(current);
}
count = 1;
}
else {
sb.append(current);
}
}
System.out.println(sb.toString());
}

I think you're on the right track. I'm not sure whether this is an assignment, so I don't want to just straight up give you an answer, but here are some hints that might help:
You're already iterating over the string. This is great! However, I think you want to compare the current character with the previous character, and not the next character.
You don't need to convert your input to a char array to iterate over it, just use charAt(idx)
You never seem to use prev, but I think you had the right idea in mind when you declared it!
Break your problem into two parts: When to update count and when to append a character. You can tackle both in your for loop, but instead of trying to do both things in the same if statements, break it up into multiple ifs.
The 3 things to do are:
Update Prev Value
Update Count
Update new String
Getting the right order for these and the exact implementation I'll leave to you (again, because I'm not sure if this is an assignment or not)
Update: Since others posted, here is my solution (with single for loop):
private String replaceConsecutiveDuplicates(String input, int n) {
if (input == null || input.length() < n) return input;
if (n == 0) return "";
StringBuffer sb = new StringBuffer();
int count = 1;
char prev = input.charAt(0);
sb.append(prev);
char current;
for( int i = 1; i < input.length(); i++) {
current = input.charAt(i);
if (prev == current) {
if (++count > n) continue;
} else {
count = 1;
}
prev = current;
sb.append(current);
}
return sb.toString();
}

First repeated string java

OK, i know this question was asked many times here, but i still don't get it, how to find the first repeated character in a string ?
I did something which was close, but it gave me all repeated characters instead of only the first one.
Here's what i did :
private static void stringChar(){
String s = "sababa";
int count = 0;
char c[] = s.toCharArray();
System.out.println("Duplicate characters are :");
for(int i = 0; i < s.length(); i++){
for(int j = i + 1; j < s.length(); j++){
if(c[i] == c[j]) {
System.out.println(c[j]);
count++;
break;
}
}
}
}

One quick and dirty (yet effective) approach which comes to mind is to maintain a map whose keys are the characters. In this case, we don't even need a formal map, because we can use an integer array which maps to the underlying ASCII values of the characters.
The basic idea here is to walk down the string, and check the array if we have seen it before. If so, then print that character and exit.
int[] nums = new int[128]; // assuming only basic ASCII characters
String str = "stuff";
for (int i=0; i < str.length(); ++i) {
int index = str.charAt(i);
if (nums[index] != 0) {
System.out.println("The first repeated character is: " + str.charAt(i));
break;
}
++nums[index];
}
Demo

You need to break the outer loop if you have already found the repeating character.
boolean found = false;
for(int i = 0; i < s.length(); i++){
for(int j = i + 1; j < s.length(); j++){
if(c[i] == c[j]) {
System.out.println(c[j]);
found = true;
break;
}
}
if (found) {
break;
}
}
If you want the count of it
int count = 0;
char repeatingChar = '0'; //dummy to overcome compiler warning
for(int i = 0; i < s.length(); i++){
for(int j = i + 1; j < s.length(); j++){
if(c[i] == c[j]) {
repeatingChar = c[j];
count++;
}
}
if (count > 0) {
System.out.println("Repeating char is " + repeatingChar + ". Occurred " + count + " times");
break;
}
}

The first repeated char is one you have seen before as you iterate the chars in the string. You can keep track of the first time you see a char with a Boolean array. The array is indexed with the char value. Java automatically "widens" char to int. So, for each char, check if it's been seen before and if not, mark that it has been seen.
String s = "sababa";
boolean[] seenChars = new boolean[Character.MAX_VALUE + 1]; // initialized to all false.
for (char c : s.toCharArray()) {
if (Character.isSurrogate(c)) throw new IllegalArgumentException("No first char seen before seeing a surrogate. This algorithm doesn't work for codepoints needing surrogates in UTF-16.");
if (seenChars[c]) {
System.out.println("First repeated char is: " + c);
break;
}
seenChars[c] = true;
}
You might notice that I've been saying char instead of character. Character is not a well-defined term. Java's text datatypes use the UTF-16 encoding of the Unicode character set. Some Unicode codepoints require two UTF-16 code units (char), which are sometimes called surrogate pairs. Since they are not all unique as individual chars, the algorithm doesn't work if they are present. For example, try String s = "sab🚲aba"; (You can also write it as "sab\uD83D\uDEB2aba".) The answer would be "a", again. But now try String s = "sab🚲🚶aba"; (You can also write it as "sab\uD83D\uDEB2\uD83D\uDEB6aba".) The answer should still be "a" but the algorithm would say "\uD83D" (which of course can't be displayed because it is only part of a codepoint).

Short and Simple >
void findFirstDupChar() {
String s = "google";
Integer dupIndex = s.length() - 1;
boolean isDuplicateExists = false;
Map<Character, Integer> map = new LinkedHashMap<>();
char[] words = s.toCharArray();
for (int i = 0; i < words.length; i++) {
Integer index = map.put(words[i], i);
if (index != null) {
if (index < dupIndex) {
dupIndex = index;
isDuplicateExists = true;
}
}
}
System.out.println(isDuplicateExists ? words[dupIndex] + ", index=" + dupIndex : "No duplicateds found");
}
This will print output as following >
g, index=0

Sorting two dimensional String array as per LastName without any Collection API

As I attended interview yesterday for an Organisation as I done every thing is best but I was unable to build logic for a Question called :
String [][]Names={{"John","Pepper"},
{"Smith","Adams"},
{"Katpiller","RhodSon"},
{"BillMark","pearson"}
};
As per the Last Name in the above array need to be sorted without any Collection.sort or CompareTo() or any other Collection API.
My Implementation is :
String str[]=new String[3];
for(int j=0; j<Name.length;j++)
{
for (int i=0 ; i<2; i++)
{
str[i]=Name[j][i];
}
for(int i=0;i<str.length;i++)
{
for(int k=i+1;k<str.length;k++)
{
if(str[i].compareTo(str[k])>0)
{
String temp= str[i];
str[i]=str[k];
str[k]=temp;
}
}
System.out.print(str[i]+ " ");
}
System.out.println();
}
Please Help me out this Hurdle Thanks in Advance

for(int j = Names.length - 2; j >= 0; j--)
{
String name = Names[j][0];
String lastName = Names[j][1];
int i = j + 1;
while(i < Names.length && strCompare(Names[i][1],lastName) < 0)
{
Names[i - 1][0] = Names[i][0];
Names[i - 1][1] = Names[i][1];
i = i + 1;
}
Names[i - 1][0] = name;
Names[i - 1][1] = lastName;
}
for(String[] name : Names)
System.out.print(Arrays.toString(name));
This is an implementation of Insertion Sort. Which sorts in place without the need of other using additional space and it works good on small inputs.
And my own reallyyyyyy bad implementation of a string comparation (String#compareTo) haha.. just to give you an idea on how to do it. (dont know if it works on all cases)
int strCompare(String str1, String str2){
//gets char arrays for both strings.
char[] strChar1 = str1.toUpperCase().toCharArray();
char[] strChar2 = str2.toUpperCase().toCharArray();
//gets max length of the strings
int maxLength = strChar1.length > strChar2.length? strChar1.length : strChar2.length;
//loops character by character to compare them.
for(int i = 0; i < maxLength; i++)
{
//if the length of the char array 1 is less than the index, then its smaller than str2.
if(strChar1.length <= i)
return -1;
//if the length of the char array 2 is less than the index, then its smaller than str1.
if(strChar2.length <= i)
return 1;
//compare characters and return -1 or 1 depending which one is larger.
if(strChar1[i] < strChar2[i]){
return -1;
} else if (strChar1[i] > strChar2[i]){
return 1;
}
}
//return 0 to say that they are equal if the loop doesnt return anything.
return 0;
}
Hope it helps.
Output : [Smith, Adams][John, Pepper][Katpiller, RhodSon][BillMark, pearson]

IMHO there are few questions to ask before trying to solve this problem :
should the sort be case sensitive (I suppose not)
should the sort be optimized for a great number of names (I suppose not)
should the array be sorted in place (I suppose yes)
If the answer to the second question is true, a correct algorithm must be implemented (quick sort, merge sort), and the lowercased name must be pre-calculated, if not a direct bubble sort is enough (O(n2/2) but simpler to implement)
int k;
String[] temp;
for (int i=0; i<len(Names) - 1; i++) {
k = i;
for (j=i + 1; i<len(Names); j++) {
if (Names[k][1].toLowerCase().compareTo(Names[i][1].toLowerCase()) < 0) {
k = j;
}
}
if (k != i) {
temp = Names[i];
Names[i] = Names[k];
Names[k] = temp;
}
}
If String.compareTo is not allowed, it simply has to be re-implemented and I suppose there are no accented chars in last name (é, è, ö, ü, etc.)
int compare(String a, String b) {
int n = (a.size() < b.size()) ? a.size(), b.size();
int delta;
for (int i =0; i<n, i++) {
delta = a.codePointAt(i) < b.codePointAt(j);
if (delta != 0) {
return delta;
}
}
return len(b) - len(a);
}
and the line if (Names[k][1]..compareTo(Names[i][1]...) { has to be rewritten as :
if (compare(Names[k][1].toLowerCase(), Names[i][1].toLowerCase()) < 0) {
This algorithm is really badly optimized, but it was simple to write (and would be simple to test). It is then possible to add as needed :
better sort algo (quick sort)
pre calculation of lowercase last name
management of accented characters.

How Do I make matrix 16 x 16 from String?

I have a string
String word = "FrenciusLeonardusNaibaho";
while I'm trying to make matrix like this:
char matriks[][] = new char[16][16];
int k = 0;
for (int i = 1; i < 16; i++) {
for (int j = 1; j < 16; j++) {
matriks[i][j] = word.charAt(k);
k++;
}
}
I got this error
String index out of range: 24
How can I achieve this?
Thanks..

You are overflowing beyond the end of word at word.charAt(k);. Basically you dont have enough alphabets to fill your matrix.
You can do something like this
if(k >= word.length())
break;
Below the inner loop. Or you can init the element to some default value with this condition.
Additionally as others have mentioned, i,j should start at 0, unless you have a good reason to start at 1.

char matriks[][] = new char[16][16];
int k = 0;
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
matriks[i][j] = word.charAt(k%word.length());
k++;
}
}
So it can go from start to end,then restart.

try adding
if(k >= word.length())
k = 0;
to your inner for loop, this will continue filling the array from the beginning of the word.

'Out of bounds' or 'out of range' occures when you try to read or write in an array, list, string or whatever with a range beyond it's boundary. You can't read a a character at index 8 when your string contains only 7 character. It's not your string's RAM and it would cause RAM corruption like it is happening sometimes in C-arrays.
When you set up your array and your for-loop try to check if you are still in bounds of your string with a size or length function of your container. In special case of string it is length.
I think you are trying to split a list of names stored in a string. In such a case it is easier to create a dynamic container, something like list (http://www.easywayserver.com/blog/java-list-example/).
Here I have a little example. For those purposes I prefer a while-loop. In cases I know the length of a list at least at runtime without interpreting data a for-loop is a good choice, but not in this:
String names = "Foo Bar";
List<String> seperatedNames = new List<String>();
String name = "";
int i = 0;
while (i < names.length()) {
if (names.charAt(i) == ' ') { // you can check for upper case char too
seperatedNames.add(name); // add name to list
name = ""; // clear name-buffer
i++; // increment i, else it would produce an infinite loop
}
name += names.charAt(i++); // add current char to name-buffer and increment current char
}
I hope I could help a bit.

of course, you will get this error surely because the character in your word are only 24 character.
to avoid this your need to check the length of your word and need to break the all looping.
Try this code.
char matriks[][] = new char[16][16];
int k = 0;
int lenght = word.length();
outerloop:
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
matriks[i][j] = word.charAt(k);
k++;
if(k >= lenght){
break outerloop;
}
}
}

You are filling 16x16 array and iterating the loop 16x16 times but your word size is less than 16x16. So put a check when k becomes equal to the word length then terminate the loop.Change your code like this.
char matriks[][] = new char[16][16];
int k = 0;
for (int i = 1; i < 16; i++) {
for (int j = 1; j < 16; j++) {
if(k >=word.length)
break;
matriks[i][j] = word.charAt(k);
k++;
}
}