We Keep Coding

How to check if there exists an element in a list less than equal to element in another list

Consider two Array list
a = [1,4,2,4]
b = [3,5]
I wanted to return the count of numbers for each number in B less than equal to number in A.
So the answer would be [2,4] since 3 in B has 2 numbers in A that are <= 3 i.e. [1,2]
And 5 in B has 4 numbers in A that are <= 5 . i.e [1,4,2,4]
Note: i tried using 2 loops but testcases time out.
this is the code
static ArrayList<Integer> counting(ArrayList<Integer>A, ArrayList<Integer>B)
{
ArrayList<Integer> finallist = new ArrayList<>();
for(int i : B)
{
int count = 0;
for(int j: A)
{
if(j<=i)
count++;
}
finallist.add(count);
}
return finallist;
}
we were supposed to edit the function and return arraylist

When I see an array and I need find smth. in it, so this is a magic key to use binary search. This gives you O(n log n) time complexity.
public static int[] findLessNumbers(int[] a, int[] b) {
TreeMap<Integer, Integer> numCount = new TreeMap<>();
for (int aa : a)
numCount.put(aa, numCount.getOrDefault(aa, 0) + aa);
int total = 0;
for (Map.Entry<Integer, Integer> entry : numCount.entrySet())
numCount.put(entry.getKey(), total += entry.getValue());
int[] res = new int[b.length];
for (int i = 0; i < res.length; i++)
res[i] = Optional.ofNullable(numCount.floorKey(b[i])).orElse(0);
return res;
}

What I understood from Your question was You weren't able to find the answer.
Hope this can help you.
Your code had a problem in counting function parameter, you should declare what is the data type that you are expecting.
static ArrayList counting(List A, List B) this isn't a way
instead you should have wrote static ArrayList counting(ArrayList<Integer> A, ArrayList<Integer> B)
and that will fix your problem.
here's my solution:
Counting function.
private static ArrayList counting(ArrayList<Integer>A, ArrayList<Integer>B ){
ArrayList<Integer> finallist = new ArrayList<>();
for(int i : B) {
int count = 0;
for(int j : A) {
if(j <= i) {
count++;
}
}
finallist.add(count);
}
return finallist;
}
Main from where am calling counting:
ArrayList<Integer> ans = counting(A,B);
for(int i : ans) {
System.out.print(i+" ");
}
Output:
2 4
This could resolve the issue you are facing.

Specific element permutation within an array of characters in JAVA?

How can I list all uppercase/lowercase permutations for any letter specified in a character array?
So, say I have an array of characters like so: ['h','e','l','l','o']
and I wanted print out possible combinations for say the letter 'l' so it would print out [hello,heLlo,heLLo,helLo].
This is what I have so far(the only problem is that I can print the permutations however I'm not able to print them inside the actual word. so my code prints [ll,lL,Ll,LL] instead of the example above.
my code:
import java.util.ArrayList;
import java.util.HashSet;
public class Main {
public static void main(String[] args) {
//Sample Word
String word = "Tomorrow-Today";
//Sample Letters for permutation
String rule_char_set = "tw";
ArrayList<Character> test1 = lettersFound(word, rule_char_set);
printPermutations(test1);
}
public static void printPermutations(ArrayList<Character> arrayList) {
char[] chars = new char[arrayList.size()];
int charIterator = 0;
for(int i=0; i<arrayList.size(); i++){
chars[i] = arrayList.get(i);
}
for (int i = 0, n = (int) Math.pow(2, chars.length); i < n; i++) {
char[] permutation = new char[chars.length];
for (int j =0; j < chars.length; j++) {
permutation[j] = (isBitSet(i, j)) ? Character.toUpperCase(chars[j]) : chars[j];
}
System.out.println(permutation);
}
}
public static boolean isBitSet(int n, int offset) {
return (n >> offset & 1) != 0;
}
public static ArrayList<Character> lettersFound(String word, String rule_char_set) {
//Convert the two parameter strings to two character arrays
char[] wordArray = word.toLowerCase().toCharArray();
char[] rule_char_setArray = rule_char_set.toLowerCase().toCharArray();
//ArrayList to hold found characters;
ArrayList<Character> found = new ArrayList<Character>();
//Increments the found ArrayList that stores the existent values.
int foundCounter = 0;
for (int i = 0; i < rule_char_setArray.length; i++) {
for (int k = 0; k < wordArray.length; k++) {
if (rule_char_setArray[i] == wordArray[k]) {
found.add(foundCounter, rule_char_setArray[i]);
foundCounter++;
}
}
}
//Convert to a HashSet to get rid of duplicates
HashSet<Character> uniqueSet = new HashSet<>(found);
//Convert back to an ArrayList(to be returned) after filtration of duplicates.
ArrayList<Character> filtered = new ArrayList<>(uniqueSet);
return filtered;
}
}

You need to make few changes in your program. Your logic is perfect that you need to find first the characters to be changed in the given word. After finding them, find powerset of characters to print all the permutation but this will only print permuatation of the characters of rule-char-set which are present in the given word.
Few changes you need to make is that first find all the indexes of word which contains characters of rule-char-set. Then find all subsets of indexes stored in an ArrayList and then for each element of each of the subsets, make the character present on that index to uppercase letter which will give you all permutation you require.
Consider an example that word = "Hello" and rule-char-set="hl" Then here first you need to find all indexes of h and l in the String word.
So here indexes are 0,2,3. Store it in ArrayList and then find its powerset.Then for each subset ,make the character present on that index to the uppercase letter.
Word[] = {'h','e','l','l','o'}
indexes = 0 , 1 , 2 , 3 , 4
index[]= { 0 , 2 ,3} //Store the indexes of characters which are to be changed
BITSET | SUBSET | word
000 | - | hello
001 | {3} | helLo
010 | {2} | heLlo
011 | {2,3} | heLLo
100 | {0} | Hello
101 | {0,3} | HelLo
110 | {0,2} | HeLlo
111 | {0,2,3} | HeLLo
Code :
import java.util.ArrayList;
import java.util.HashSet;
public class Main {
public static void main(String[] args) {
//Sample Word
String word = "Tomorrow-Today";
//Sample Letters for permutation
String rule_char_set = "tw";
ArrayList<Integer> test1 = lettersFound(word, rule_char_set); //To store the indexes of the characters
printPermutations(word,test1);
}
public static void printPermutations(String word,ArrayList<Integer> arrayList) {
char word_array[]=word.toLowerCase().toCharArray();
int length=word_array.length;
int index[]=new int[arrayList.size()];
for(int i=0; i<arrayList.size(); i++){
index[i] = arrayList.get(i);
}
for (int i = 0, n = (int) Math.pow(2, index.length); i < n; i++) {
char[] permutation = new char[length];
System.arraycopy(word_array,0,permutation,0,length);
//First copy the original array and change
//only those character whose indexes are present in subset
for (int j =0; j < index.length; j++) {
permutation[index[j]] = (isBitSet(i, j)) ? Character.toUpperCase(permutation[index[j]]) : permutation[index[j]];
}
System.out.println(permutation);
}
}
public static boolean isBitSet(int n, int offset) {
return (n >> offset & 1) != 0;
}
public static ArrayList<Integer> lettersFound(String word, String rule_char_set) {
//Convert the two parameter strings to two character arrays
char[] wordArray = word.toLowerCase().toCharArray();
char[] rule_char_setArray = rule_char_set.toLowerCase().toCharArray();
//ArrayList to hold found characters;
ArrayList<Integer> found = new ArrayList<Integer>();
//Increments the found ArrayList that stores the existent values.
int foundCounter = 0;
for (int i = 0; i < rule_char_setArray.length; i++) {
for (int k = 0; k < wordArray.length; k++) {
if (rule_char_setArray[i] == wordArray[k]) {
found.add(foundCounter, k); //Store the index of the character that matches
foundCounter++;
}
}
}
return found;
}
}
Output :
tomorrow-today
Tomorrow-today
tomorrow-Today
Tomorrow-Today
tomorroW-today
TomorroW-today
tomorroW-Today
TomorroW-Today

Sanket Makani answer is perfect.
I may offer a more objective approach of this problem.
As an input you have a string to modify, and characters, which should be replaced with the modified case ( upper or lower ).
As an output you will have all permutated strings.
I would create a structure which contains index, and possible values to change with:
class Change {
int index;
char values[];
}
We will need to make all possible combinations, so lets include field which will tell which character is currently used in to our structure, and add some methods:
class Change {
int index;
char values[];
int cur;
void reset() {cur=0;}
boolen isMax(){return cur==values.length-1;}
void next(){cur++;}
char getValue(){ return values[cur]; }
}
We will have a list or array of these classes then, which we will put in to a separate class
class Combination {
Change changes[];
void reset() { for (Change c: changes) c.reset();}
boolean next() {
for ( int i=0; i<changes.length; i++)
if ( changes[i].isMax())
changes[i].reset(); // next change will be taken in cycle, with "next()"
else {changes[i].next(); return true;}
return false; // all changes are max
}
}
So when you initialize your "Combination" class by your input data, you may use it in cycle then.
Combination c = new Combination();
.... // initialization here
c.reset();
do {
... // update and print your string
} while ( c.next() );
The initialization of "Combination" and using of values for updating the input string I leave after you :)

For the permutation case, I think recursion is the best fit in terms of readability, taking into account that maybe is not best in terms of performance.
My approach would be this:
public static void main(String[] args) {
generateCombinations("hello", "l", "");
}
public static void generateCombinations(String text, String changingLetters, String current) {
if (0 == text.length()) {
System.out.println(current);
return;
}
String currentLetter = text.substring(0, 1);
if (changingLetters.contains(currentLetter)) {
generateCombinations(text.substring(1), changingLetters, current + currentLetter.toUpperCase());
}
generateCombinations(text.substring(1), changingLetters, current + currentLetter);
}
The output for the main execution will be:
heLLo
heLlo
helLo
hello

Java, combination algorithm with arrays

I am trying to implement an algorithm to calculate all combinations of an Array where one character is replaced by '*' without changing the order of the Arrays entries.
For example the following Array with two entries:
{"A", "B"}
Should reproduce this Output:
[A, B]
[*, B]
[A, *]
[*, *]
My current code is:
public class TestCombination {
public static void combinations(List<String[]> values, String[] attr, String all, int iteration) {
String[] val = new String[attr.length];
for (int i = 0; i < attr.length; i++) {
val[i] = attr[i];
}
if (iteration < attr.length) {
val[iteration] = all;
}
values.add(val);
iteration = iteration + 1;
if (Math.pow(attr.length, 2) != iteration) {
combinations(values, attr, all, iteration);
}
}
public static void main() {
String[] values = new String[] {"A", "B"};
List<String[]> resultValues = new ArrayList<String[]>();
combinations(resultValues, values, "*", 0);
for (String[] res : resultValues) {
System.out.println(Arrays.deepToString(res));
}
}
}
The Output i get is:
[*, B]
[A, *]
[A, B]
[A, B]
This is especially because of this not correct code:
if (iteration < attr.length) {
val[iteration] = all;
}
I do not have any idea, how the next possible index can be calculated to replace the Array value at that index by '*'.
Can you give me please some hints on that?

One simple approach is to use a bit mask of length n. Iterate all n-digit binary numbers, and then for each of the n positions do the following:
If position i has one, output an asterisk *
If position i has zero, output the original value.
This will cover all combinations.
String[] a = new String[] {"A", "B", "C"};
for (int mask = 0 ; mask != 1<<a.length ; mask++) {
for (int i = 0 ; i != a.length ; i++) {
if ((mask & 1<<i) != 0) {
System.out.print("* ");
} else {
System.out.print(a[i]+" ");
}
}
System.out.println();
}
Demo.

My solution is a modification of #dasblinkenlight solution where I'm using recursive function and not mask bit. My solution is in javascript.
var arr = ['A', 'B', 'C'],
len = arr.length,
pattern = function(startIndex, arr) {
var newArr = [].concat(arr),
i;
newArr[startIndex] = '*';
console.log(newArr.toString());
for (i = startIndex + 1; i < len; i++) {
pattern(i, newArr);
}
};
console.log(arr.toString())
for (i = 0; i < len; i++) {
pattern(i, arr);
}

You can use a recursive function to get the current string and one index and one time change the character at that index to * and another time call the function without changing the character at that index. Print the result when index reaches end of the string:
public class Main{
public static void main(String args[]){
f(new StringBuilder("ABC"),0);
}
public static void f(StringBuilder str, int index){
if (index == str.length()){
System.out.println(str);
return;
}
f(str, index+1);
char c = str.charAt(index);
str.setCharAt(index, '*');
f(str, index+1);
str.setCharAt(index, c);
}
}
output
ABC
AB*
A*C
A**
*BC
*B*
**C
***

Pair Palindrome

I have this code to find all pairs of string to form a palindrome. e.g) D: { AB, DEEDBA } => AB + DEEDBA -> YES and will be returned. Another example, { NONE, XENON } => NONE + XENON = > YES.
What would be running time of this ?
public static List<List<String>> pairPalindrome(List<String> D) {
List<List<String>> pairs = new LinkedList<>();
Set<String> set = new HashSet<>();
for (String s : D) {
set.add(s);
}
for (String s : D) {
String r = reverse(s);
for (int i = 0; i <= r.length(); i++) {
String prefix = r.substring(0, i);
if (set.contains(prefix)) {
String suffix = r.substring(i);
if (isPalindrom(suffix)) {
pairs.add(Arrays.asList(s, prefix));
}
}
}
}
return pairs;
}
private static boolean isPalindrom(String s) {
int i = 0;
int j = s.length() - 1;
char[] c = s.toCharArray();
while (i < j) {
if (c[i] != c[j]) {
return false;
}
i++;
j--;
}
return true;
}
private static String reverse(String s) {
char[] c = s.toCharArray();
int i = 0;
int j = c.length - 1;
while (i < j) {
char temp = c[i];
c[i] = c[j];
c[j] = temp;
i++;
j--;
}
return new String(c);
}

I'm going to take a few guesses here as I don't have much experience with Java.
First, isPalindrome is O(N) with the size of suffix string. Add operation to 'pairs' would probably be O(1).
Then, we have the for loop, it's O(N) with the length of r. Getting a substring I'd think is O(M) with the size of the substring. Checking if a hashmap contains a certain key, with a perfect hash function would be (IIRC) O(1), in your case we can assume O(lgN) (possibly). So, first for loop has O(NMlgK), where K is your hash table size, N is r's length and M is substring's length.
Finally we have the outmost for loop, it runs for each string in the string list, so that's O(N). Then, we reverse each of them. So for each of these strings we have another O(N) operation inside, with the other loop being O(NMlgK). So, overall complexity is O(L(N + NMlgK)), where L is the amount of strings you have. But, it'd reduce to O(LNMlgK). I'd like if someone verified or corrected my mistakes.
EDIT: Actually, substring length will at most be N, as the length of the entire string, so M is actually N. Now I'd probably say it's O(LNlgK).

Find all substrings that are palindromes

If the input is 'abba' then the possible palindromes are a, b, b, a, bb, abba.
I understand that determining if string is palindrome is easy. It would be like:
public static boolean isPalindrome(String str) {
int len = str.length();
for(int i=0; i<len/2; i++) {
if(str.charAt(i)!=str.charAt(len-i-1) {
return false;
}
return true;
}
But what is the efficient way of finding palindrome substrings?

This can be done in O(n), using Manacher's algorithm.
The main idea is a combination of dynamic programming and (as others have said already) computing maximum length of palindrome with center in a given letter.
What we really want to calculate is radius of the longest palindrome, not the length.
The radius is simply length/2 or (length - 1)/2 (for odd-length palindromes).
After computing palindrome radius pr at given position i we use already computed radiuses to find palindromes in range [i - pr ; i]. This lets us (because palindromes are, well, palindromes) skip further computation of radiuses for range [i ; i + pr].
While we search in range [i - pr ; i], there are four basic cases for each position i - k (where k is in 1,2,... pr):
no palindrome (radius = 0) at i - k
(this means radius = 0 at i + k, too)
inner palindrome, which means it fits in range
(this means radius at i + k is the same as at i - k)
outer palindrome, which means it doesn't fit in range
(this means radius at i + k is cut down to fit in range, i.e because i + k + radius > i + pr we reduce radius to pr - k)
sticky palindrome, which means i + k + radius = i + pr
(in that case we need to search for potentially bigger radius at i + k)
Full, detailed explanation would be rather long. What about some code samples? :)
I've found C++ implementation of this algorithm by Polish teacher, mgr Jerzy Wałaszek.
I've translated comments to english, added some other comments and simplified it a bit to be easier to catch the main part.
Take a look here.
Note: in case of problems understanding why this is O(n), try to look this way:
after finding radius (let's call it r) at some position, we need to iterate over r elements back, but as a result we can skip computation for r elements forward. Therefore, total number of iterated elements stays the same.

Perhaps you could iterate across potential middle character (odd length palindromes) and middle points between characters (even length palindromes) and extend each until you cannot get any further (next left and right characters don't match).
That would save a lot of computation when there are no many palidromes in the string. In such case the cost would be O(n) for sparse palidrome strings.
For palindrome dense inputs it would be O(n^2) as each position cannot be extended more than the length of the array / 2. Obviously this is even less towards the ends of the array.
public Set<String> palindromes(final String input) {
final Set<String> result = new HashSet<>();
for (int i = 0; i < input.length(); i++) {
// expanding even length palindromes:
expandPalindromes(result,input,i,i+1);
// expanding odd length palindromes:
expandPalindromes(result,input,i,i);
}
return result;
}
public void expandPalindromes(final Set<String> result, final String s, int i, int j) {
while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {
result.add(s.substring(i,j+1));
i--; j++;
}
}

So, each distinct letter is already a palindrome - so you already have N + 1 palindromes, where N is the number of distinct letters (plus empty string). You can do that in single run - O(N).
Now, for non-trivial palindromes, you can test each point of your string to be a center of potential palindrome - grow in both directions - something that Valentin Ruano suggested.
This solution will take O(N^2) since each test is O(N) and number of possible "centers" is also O(N) - the center is either a letter or space between two letters, again as in Valentin's solution.
Note, there is also O(N) solution to your problem, based on Manacher's algoritm (article describes "longest palindrome", but algorithm could be used to count all of them)

I just came up with my own logic which helps to solve this problem.
Happy coding.. :-)
System.out.println("Finding all palindromes in a given string : ");
subPal("abcacbbbca");
private static void subPal(String str) {
String s1 = "";
int N = str.length(), count = 0;
Set<String> palindromeArray = new HashSet<String>();
System.out.println("Given string : " + str);
System.out.println("******** Ignoring single character as substring palindrome");
for (int i = 2; i <= N; i++) {
for (int j = 0; j <= N; j++) {
int k = i + j - 1;
if (k >= N)
continue;
s1 = str.substring(j, i + j);
if (s1.equals(new StringBuilder(s1).reverse().toString())) {
palindromeArray.add(s1);
}
}
}
System.out.println(palindromeArray);
for (String s : palindromeArray)
System.out.println(s + " - is a palindrome string.");
System.out.println("The no.of substring that are palindrome : "
+ palindromeArray.size());
}
Output:-
Finding all palindromes in a given string :
Given string : abcacbbbca
******** Ignoring single character as substring palindrome ********
[cac, acbbbca, cbbbc, bb, bcacb, bbb]
cac - is a palindrome string.
acbbbca - is a palindrome string.
cbbbc - is a palindrome string.
bb - is a palindrome string.
bcacb - is a palindrome string.
bbb - is a palindrome string.
The no.of substring that are palindrome : 6

I suggest building up from a base case and expanding until you have all of the palindomes.
There are two types of palindromes: even numbered and odd-numbered. I haven't figured out how to handle both in the same way so I'll break it up.
1) Add all single letters
2) With this list you have all of the starting points for your palindromes. Run each both of these for each index in the string (or 1 -> length-1 because you need at least 2 length):
findAllEvenFrom(int index){
int i=0;
while(true) {
//check if index-i and index+i+1 is within string bounds
if(str.charAt(index-i) != str.charAt(index+i+1))
return; // Here we found out that this index isn't a center for palindromes of >=i size, so we can give up
outputList.add(str.substring(index-i, index+i+1));
i++;
}
}
//Odd looks about the same, but with a change in the bounds.
findAllOddFrom(int index){
int i=0;
while(true) {
//check if index-i and index+i+1 is within string bounds
if(str.charAt(index-i-1) != str.charAt(index+i+1))
return;
outputList.add(str.substring(index-i-1, index+i+1));
i++;
}
}
I'm not sure if this helps the Big-O for your runtime, but it should be much more efficient than trying each substring. Worst case would be a string of all the same letter which may be worse than the "find every substring" plan, but with most inputs it will cut out most substrings because you can stop looking at one once you realize it's not the center of a palindrome.

I tried the following code and its working well for the cases
Also it handles individual characters too
Few of the cases which passed:
abaaa --> [aba, aaa, b, a, aa]
geek --> [g, e, ee, k]
abbaca --> [b, c, a, abba, bb, aca]
abaaba -->[aba, b, abaaba, a, baab, aa]
abababa -->[aba, babab, b, a, ababa, abababa, bab]
forgeeksskeegfor --> [f, g, e, ee, s, r, eksske, geeksskeeg,
o, eeksskee, ss, k, kssk]
Code
static Set<String> set = new HashSet<String>();
static String DIV = "|";
public static void main(String[] args) {
String str = "abababa";
String ext = getExtendedString(str);
// will check for even length palindromes
for(int i=2; i<ext.length()-1; i+=2) {
addPalindromes(i, 1, ext);
}
// will check for odd length palindromes including individual characters
for(int i=1; i<=ext.length()-2; i+=2) {
addPalindromes(i, 0, ext);
}
System.out.println(set);
}
/*
* Generates extended string, with dividors applied
* eg: input = abca
* output = |a|b|c|a|
*/
static String getExtendedString(String str) {
StringBuilder builder = new StringBuilder();
builder.append(DIV);
for(int i=0; i< str.length(); i++) {
builder.append(str.charAt(i));
builder.append(DIV);
}
String ext = builder.toString();
return ext;
}
/*
* Recursive matcher
* If match is found for palindrome ie char[mid-offset] = char[mid+ offset]
* Calculate further with offset+=2
*
*
*/
static void addPalindromes(int mid, int offset, String ext) {
// boundary checks
if(mid - offset <0 || mid + offset > ext.length()-1) {
return;
}
if (ext.charAt(mid-offset) == ext.charAt(mid+offset)) {
set.add(ext.substring(mid-offset, mid+offset+1).replace(DIV, ""));
addPalindromes(mid, offset+2, ext);
}
}
Hope its fine

public class PolindromeMyLogic {
static int polindromeCount = 0;
private static HashMap<Character, List<Integer>> findCharAndOccurance(
char[] charArray) {
HashMap<Character, List<Integer>> map = new HashMap<Character, List<Integer>>();
for (int i = 0; i < charArray.length; i++) {
char c = charArray[i];
if (map.containsKey(c)) {
List list = map.get(c);
list.add(i);
} else {
List list = new ArrayList<Integer>();
list.add(i);
map.put(c, list);
}
}
return map;
}
private static void countPolindromeByPositions(char[] charArray,
HashMap<Character, List<Integer>> map) {
map.forEach((character, list) -> {
int n = list.size();
if (n > 1) {
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (list.get(i) + 1 == list.get(j)
|| list.get(i) + 2 == list.get(j)) {
polindromeCount++;
} else {
char[] temp = new char[(list.get(j) - list.get(i))
+ 1];
int jj = 0;
for (int ii = list.get(i); ii <= list
.get(j); ii++) {
temp[jj] = charArray[ii];
jj++;
}
if (isPolindrome(temp))
polindromeCount++;
}
}
}
}
});
}
private static boolean isPolindrome(char[] charArray) {
int n = charArray.length;
char[] temp = new char[n];
int j = 0;
for (int i = (n - 1); i >= 0; i--) {
temp[j] = charArray[i];
j++;
}
if (Arrays.equals(charArray, temp))
return true;
else
return false;
}
public static void main(String[] args) {
String str = "MADAM";
char[] charArray = str.toCharArray();
countPolindromeByPositions(charArray, findCharAndOccurance(charArray));
System.out.println(polindromeCount);
}
}
Try out this. Its my own solution.

// Maintain an Set of palindromes so that we get distinct elements at the end
// Add each char to set. Also treat that char as middle point and traverse through string to check equality of left and right char
static int palindrome(String str) {
Set<String> distinctPln = new HashSet<String>();
for (int i=0; i<str.length();i++) {
distinctPln.add(String.valueOf(str.charAt(i)));
for (int j=i-1, k=i+1; j>=0 && k<str.length(); j--, k++) {
// String of lenght 2 as palindrome
if ( (new Character(str.charAt(i))).equals(new Character(str.charAt(j)))) {
distinctPln.add(str.substring(j,i+1));
}
// String of lenght 2 as palindrome
if ( (new Character(str.charAt(i))).equals(new Character(str.charAt(k)))) {
distinctPln.add(str.substring(i,k+1));
}
if ( (new Character(str.charAt(j))).equals(new Character(str.charAt(k)))) {
distinctPln.add(str.substring(j,k+1));
} else {
continue;
}
}
}
Iterator<String> distinctPlnItr = distinctPln.iterator();
while ( distinctPlnItr.hasNext()) {
System.out.print(distinctPlnItr.next()+ ",");
}
return distinctPln.size();
}

Code is to find all distinct substrings which are palindrome.
Here is the code I tried. It is working fine.
import java.util.HashSet;
import java.util.Set;
public class SubstringPalindrome {
public static void main(String[] args) {
String s = "abba";
checkPalindrome(s);
}
public static int checkPalindrome(String s) {
int L = s.length();
int counter =0;
long startTime = System.currentTimeMillis();
Set<String> hs = new HashSet<String>();
// add elements to the hash set
System.out.println("Possible substrings: ");
for (int i = 0; i < L; ++i) {
for (int j = 0; j < (L - i); ++j) {
String subs = s.substring(j, i + j + 1);
counter++;
System.out.println(subs);
if(isPalindrome(subs))
hs.add(subs);
}
}
System.out.println("Total possible substrings are "+counter);
System.out.println("Total palindromic substrings are "+hs.size());
System.out.println("Possible palindromic substrings: "+hs.toString());
long endTime = System.currentTimeMillis();
System.out.println("It took " + (endTime - startTime) + " milliseconds");
return hs.size();
}
public static boolean isPalindrome(String s) {
if(s.length() == 0 || s.length() ==1)
return true;
if(s.charAt(0) == s.charAt(s.length()-1))
return isPalindrome(s.substring(1, s.length()-1));
return false;
}
}
OUTPUT:
Possible substrings:
a
b
b
a
ab
bb
ba
abb
bba
abba
Total possible substrings are 10
Total palindromic substrings are 4
Possible palindromic substrings: [bb, a, b, abba]
It took 1 milliseconds