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recursion subtraction problem from CodingBat.com

I'm trying to solve this question from CodingBat.com. - https://codingbat.com/prob/p143900
I run the recursion twice, then I subtract their values to give the final answer. Individually, I and J are getting the correct values, but when I do i - j, the result doesn't make sense.
public int countHi2(String str) {
if(str.length() == 0)
return 0;
int i = count(str, "hi");
int j = count(str, "xhi");
return i-j;
}
public int count(String str, String match)
{
int i = str.indexOf(match);
if(i != -1)
return 1 + countHi2(str.substring(i+match.length()-1));
else
return 0 + countHi2(str.substring(1));
}

I think you treat rescursive illegally. You should not use external method or even split a string.
public int countHi2(String str) {
if (str == null || str.length() < 2)
return 0;
if (str.startsWith("hi"))
return 1 + countHi2(str.substring(2));
return countHi2(str.substring(str.startsWith("xhi") ? 3 : 1));
}

you might wanna use:
str.substring(i+match.length())
suppose your match was on index 3 and the length of the word to match is 2, then you want the substring from index 5 to the end.
thus,
str.substring(5);
which includes the index 5
though, i don't understand why you used
return 0 + countHi2(str.substring(1));
because this condition will hold if no match is found, so what is the point of checking for a match again in a substring !!!

Check if string is a palindrome after removing one character. My working solution is too slow

I cant seem to find a proper solution to an exercise. The exercise asks to create a method that returns true if a string can be a palindrome by removing one character. I have a solution that works but fails tests of large (100,000 character) strings because its exceeding the time limit of 1 second. Can somebody point me in the right direction?
I realize my approach is brute force and I'm sure there's a better way to solve it. I'm assuming my problem lies with the iteration.
public class Main {
public static boolean makePalindrome(String mjono) {
StringBuilder sb = new StringBuilder(mjono);
for (int i = 0; i < mjono.length(); i++) {
sb.deleteCharAt(i);
if(isPalindrome(sb.toString())){
return true;
} else {
sb.insert(i, mjono.charAt(i));
}
}
return false;
}
private static boolean isPalindrome(String string) {
return string.equals(new StringBuilder(string).reverse().toString());
}
public static void main(String[] args) {
System.out.println(makePalindrome("ABCBXA"));
System.out.println(makePalindrome("ABCBAX"));
System.out.println(makePalindrome("ABCXBA"));
System.out.println(makePalindrome("ABCDE"));
System.out.println(makePalindrome("BAAAAC"));
}
}
These are the tests it fails:
#Test(timeout=1000)
public void suuri2() {
int n = 100000;
char[] t = new char[n];
for (int i = 0; i < n; i++) t[i] = 'A';
t[12345] = 'B';
testaaSuuri(new String(t), true);
}
#Test(timeout=1000)
public void suuri3() {
int n = 100000;
char[] t = new char[n];
for (int i = 0; i < n; i++) t[i] = 'A';
t[12345] = 'B';
t[54321] = 'C';
testaaSuuri(new String(t), false);
}
Thanks in advance.

Well, there's of course the naive solution running in O(n ^ 2) by trying each possibility to remove one char.
But we can certainly do better:
We can define a palindrome recursively:
palindrome = x.palindrome.x | x | x.x , where x is an arbitrary token
So how does that help us? Pretty simple: we can derive a rules that allow checking whether the string is palindromic in O(n).
A palindrome consists of a char c, followed by a string that must be empty or palindromic, followed by another c, if it's longer than 1 char. If it's of length 1, it's automatically palindromic.
Thus, the last character must be equal to the first, the second to the second to the last, and so on. So basically:
boolean isPalindrome(String s){
for(int i = 0 ; i < s.length() / 2 ; i++)
if(s.charAt(i) != s.charAt(s.length() - i - 1))
return false;
return true;
}
We have to alter this rule a bit, since once we may remove a single character. This introduces splitting the whole problem into two parts, as we can see from a definition:
palindrome_1 = s.x.palindrome.reverse(s) | s.palindrome.x.reverse(s) | palindrome
As we can easily see, this contains the original palindrome-definition, but in addition allows introduction of one additional char x.
static boolean isPalindrome_1(String s){
for(int i = 0 ; i < s.length() / 2 ; i++)
if(s.charAt(i) != s.charAt(s.length() - i - 1))
return isPalindrome(s , i + 1 , s.length() - i - 1) ||
isPalindrome(s , i , s.length() - i - 2);
return true;
}
static boolean isPalindrome(String s , int lower , int upper){
while(lower < upper){
if(s.charAt(lower) != s.charAt(upper))
return false;
lower++;
upper--;
}
return true;
}
An explanation/or at least an attempt to explain this:
This piece of code:
if(s.charAt(i) != s.charAt(s.length() - i - 1))
return isPalindrome(s , i + 1 , s.length() - i - 1) ||
isPalindrome(s , i , s.length() - i - 2);
Is required, if the definition of palindrome doesn't apply to our input-string. In that case, we have to check two possibilities, how the code was built:
s.x.palindrome.reverse(s)
s.palindrome.x.reverse(s)
If the definition of palindrome doesn't apply, we have reached a point, were we have to ommit either the character of at the start of the remaining string (x.palindrome) or the end of the remaining string (palindrome.x) and see, if the rest matches the definition of a palindrome. This is done by calling isPalindrome(...) with two different substrings, that are cut by one character at either the start or the end of the remaining string.
An few examples of how this code works:
A B C D E F E D C B A
| | portion that runs inside isPalindrome_1
A B D E F E D C B A
| | | | portion that can be checked inside isPalindrome_1
| | isPalindrome(s , i , s.length() - i - 2)
| | isPalindrome(s , i + 1 , s.length() - i - 1)
As we can see in the second example, the code searched for the first pair of chars that isn't equal. At this point, we have two substrings to search further, which each ommit one character, either at the beginning or the end of the string.
Efficiency:
This code runs in-place - there are never made any copies of the input-string. Runtime is O(n) (O(2 * n) to be precise). Building a faster solution won't be possible - at least until we get quantum computers ;)

Hint 1: since this is an exercise, posting solutions is inappropriate. (It detracts from the learning experience of doing the exercise yourself.)
Hint 2: The following operations are all O(N) for an N character String or StringBuilder:
Adding or removing a character from a StringBuilder
Creating a new StringBuilder from an existing StringBuilder
Reversing a StringBuilder
Creating a String from a StringBuilder (toString())
Comparing two equal or "almost equal" String objects.
(In most cases you copy or compare N characters. For insertion and deletion, you copy on average 0.5 N characters assuming that the buffer does not need to grow, but that is still O(N). For equals ... it is complicated, but the worst-case is clearly O(N).)
So a fast palindrome tester for large strings needs to avoid these operations.
Hint 3: you can treat the string as an array of characters, either by converting it into a char[] or using charAt(...).
Hint 4: you don't have to physically remove the char from the string. You can just get your algorithm to pretend it isn't there.

We can solve this using LCS(Longest Common Sub-sequence).
LCS tells us the length of the longest sub-sequence in two strings.
boolean isPalindromAfterRemovingOneChar(String s) {
int lengthOfLCS = lcs(s, s.reverse(), s.length());
return (s.length()- lengthOfLCS) == 1;
}

function test(s) {
const check = isPalindrome(s)
if (!check) {
const arr = s.split('')
const arrCheck = []
arr.forEach((element, i) => {
if (element !== arr[arr.length - i - 1]) {
const news = Array.from(arr)
console.log(arr, news.splice(i, 1))
isPalindrome(news.join(''))
}
});
console.log('arrCheck', arrCheck)
}
function isPalindrome(s) {
var reversedString = s.split("").reverse().join("");
if (s === reversedString) {
console.log('this string is palindrome', s)
return true
} else {
console.log('no')
return false
}
test('aaab')

Only need to compare the first half to the second half. Don't waste time reversing the whole String either.
private boolean isPalindrome(String string) {
char[] values = string.toCharArray();
for (int i = 0; i < values.length / 2; i++) {
if (values[i] != values[values.length - 1 - i])
return false;
}
return true;
}

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).

Finding the Number of Times an Expression Occurs in a String Continuously and Non Continuously

I had a coding interview over the phone and was asked this question:
Given a String (for example):
"aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc"
and an expression (for example):
"a+b+c-"
where:
+: means the char before it is repeated 2 times
-: means the char before it is repeated 4 times
Find the number of times the given expression appears in the string with the operands occurring non continuously and continuously.
The above expression occurs 4 times:
1) aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc
^^ ^^ ^^^^
aa bb cccc
2) aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc
^^ ^^ ^^^^
aa bb cccc
3) aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc
^^ ^^ ^^^^
aa bb cccc
4) aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc
^^ ^^ ^^^^
aa bb cccc
I had no idea how to do it. I started doing an iterative brute force method with lots of marking of indices but realized how messy and hard that would to code half way through:
import java.util.*;
public class Main {
public static int count(String expression, String input) {
int count = 0;
ArrayList<char[]> list = new ArrayList<char[]>();
// Create an ArrayList of chars to iterate through the expression and match to string
for(int i = 1; i<expression.length(); i=i+2) {
StringBuilder exp = new StringBuilder();
char curr = expression.charAt(i-1);
if(expression.charAt(i) == '+') {
exp.append(curr).append(curr);
list.add(exp.toString().toCharArray());
}
else { // character is '-'
exp.append(curr).append(curr).append(curr).append(curr);
list.add(exp.toString().toCharArray());
}
}
char[] inputArray = input.toCharArray();
int i = 0; // outside pointer
int j = 0; // inside pointer
while(i <= inputArray.length) {
while(j <= inputArray.length) {
for(int k = 0; k< list.size(); k++) {
/* loop through
* all possible combinations in array list
* with multiple loops
*/
}
j++;
}
i++;
j=i;
}
return count;
}
public static void main(String[] args) {
String expression = "a+b+c-";
String input = "aaksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc";
System.out.println("The expression occurs: "+count(expression, input)+" times");
}
}
After spending a lot of time doing it iteratively he mentioned recursion and I still couldn't see a clear way doing it recursively and I wasn't able to solve the question. I am trying to solve it now post-interview and am still not sure how to go about this question. How should I go about solving this problem? Is the solution obvious? I thought this was a really hard question for a coding phone interview.

Non-recursion algorithm that requires O(m) space and operates in O(n*m), where m is number of tokens in query:
#Test
public void subequences() {
String input = "aabbccaacccccbbd";
String query = "a+b+";
// here to store tokens of a query: e.g. {a, +}, {b, +}
char[][] q = new char[query.length() / 2][];
// here to store counts of subsequences ending by j-th token found so far
int[] c = new int[query.length() / 2]; // main
int[] cc = new int[query.length() / 2]; // aux
// tokenize
for (int i = 0; i < query.length(); i += 2)
q[i / 2] = new char[] {query.charAt(i), query.charAt(i + 1)};
// init
char[] sub2 = {0, 0}; // accumulator capturing last 2 chars
char[] sub4 = {0, 0, 0, 0}; // accumulator capturing last 4 chars
// main loop
for (int i = 0; i < input.length(); i++) {
shift(sub2, input.charAt(i));
shift(sub4, input.charAt(i));
boolean all2 = sub2[1] != 0 && sub2[0] == sub2[1]; // true if all sub2 chars are same
boolean all4 = sub4[3] != 0 && sub4[0] == sub4[1] // true if all sub4 chars are same
&& sub4[0] == sub4[2] && sub4[0] == sub4[3];
// iterate tokens
for (int j = 0; j < c.length; j++) {
if (all2 && q[j][1] == '+' && q[j][0] == sub2[0]) // found match for "+" token
cc[j] = j == 0 // filling up aux array
? c[j] + 1 // first token, increment counter by 1
: c[j] + c[j - 1]; // add value of preceding token counter
if (all4 && q[j][1] == '-' && q[j][0] == sub4[0]) // found match for "-" token
cc[j] = j == 0
? c[j] + 1
: c[j] + c[j - 1];
}
if (all2) sub2[1] = 0; // clear, to make "aa" occur in "aaaa" 2, not 3 times
if (all4) sub4[3] = 0;
copy(cc, c); // copy aux array to main
}
}
System.out.println(c[c.length - 1]);
}
// shifts array 1 char left and puts c at the end
void shift(char[] cc, char c) {
for (int i = 1; i < cc.length; i++)
cc[i - 1] = cc[i];
cc[cc.length - 1] = c;
}
// copies array contents
void copy(int[] from, int[] to) {
for (int i = 0; i < from.length; i++)
to[i] = from[i];
}
The main idea is to catch chars from the input one by one, holding them in 2- and 4-char accumulators and check if any of them match some tokens of the query, remembering how many matches have we got for sub-queries ending by these tokens so far.
Query (a+b+c-) is splitted into tokens (a+, b+, c-). Then we collect chars in accumulators and check if they match some tokens. If we find match for first token, we increment its counter by 1. If we find match for another j-th token, we can create as many additional subsequences matching subquery composed of tokens [0...j], as many of them now exist for subquery composed of tokens [0... j-1], because this match can be appended to every of them.
For example, we have:
a+ : 3 (3 matches for a+)
b+ : 2 (2 matches for a+b+)
c- : 1 (1 match for a+b+c-)
when cccc arrives. Then c- counter should be increased by b+ counter value, because so far we have 2 a+b+ subsequences and cccc can be appended to both of them.

Let's call the length of the string n, and the length of the query expression (in terms of the number of "units", like a+ or b-) m.
It's not clear exactly what you mean by "continuously" and "non-continuously", but if "continuously" means that there can't be any gaps between query string units, then you can just use the KMP algorithm to find all instances in O(m+n) time.
We can solve the "non-continuous" version in O(nm) time and space with dynamic programming. Basically, what we want to compute is a function:
f(i, j) = the number of occurrences of the subquery consisting of the first i units
of the query expression, in the first j characters of the string.
So with your example, f(2, 41) = 2, since there are 2 separate occurrences of the subpattern a+b+ in the first 41 characters of your example string.
The final answer will then be f(n, m).
We can compute this recursively as follows:
f(0, j) = 0
f(i, 0) = 0
f(i > 0, j > 0) = f(i, j-1) + isMatch(i, j) * f(i-1, j-len(i))
where len(i) is the length of the ith unit in the expression (always 2 or 4) and isMatch(i, j) is a function that returns 1 if the ith unit in the expression matches the text ending at position j, and 0 otherwise. For example, isMatch(15, 2) = 1 in your example, because s[14..15] = bb. This function takes just constant time to run, because it never needs to check more than 4 characters.
The above recursion will already work as-is, but we can save time by making sure that we only solve each subproblem once. Because the function f() depends only on its 2 parameters i and j, which range between 0 and m, and between 0 and n, respectively, we can just compute all n*m possible answers and store them in a table.
[EDIT: As Sasha Salauyou points out, the space requirement can in fact be reduced to O(m). We never need to access values of f(i, k) with k < j-1, so instead of storing m columns in the table we can just store 2, and alternate between them by always accessing column m % 2.]

Wanted to try it for myself and figured I could then share my solution as well. The parse method obviously has issues when there is indeed a char 0 in the expression (although that would probably be the bigger issue itself), the find method will fail for an empty needles array and I wasn't sure if ab+c- should be considered a valid pattern (I treat it as such). Note that this covers only the non-continous part so far.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Matcher {
public static void main(String[] args) {
String haystack = "aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc";
String[] needles = parse("a+b+c-");
System.out.println("Needles: " + Arrays.toString(needles));
System.out.println("Found: " + find(haystack, needles, 0));
needles = parse("ab+c-");
System.out.println("Needles: " + Arrays.toString(needles));
System.out.println("Found: " + find(haystack, needles, 0));
}
private static int find(String haystack, String[] needles, int i) {
String currentNeedle = needles[i];
int pos = haystack.indexOf(currentNeedle);
if (pos < 0) {
// Abort: Current needle not found
return 0;
}
// Current needle found (also means that pos + currentNeedle.length() will always
// be <= haystack.length()
String remainingHaystack = haystack.substring(pos + currentNeedle.length());
// Last needle?
if (i == needles.length - 1) {
// +1: We found one match for all needles
// Try to find more matches of current needle in remaining haystack
return 1 + find(remainingHaystack, needles, i);
}
// Try to find more matches of current needle in remaining haystack
// Try to find next needle in remaining haystack
return find(remainingHaystack, needles, i) + find(remainingHaystack, needles, i + 1);
}
private static String[] parse(String expression) {
List<String> searchTokens = new ArrayList<String>();
char lastChar = 0;
for (int i = 0; i < expression.length(); i++) {
char c = expression.charAt(i);
char[] chars;
switch (c) {
case '+':
// last char is repeated 2 times
chars = new char[2];
Arrays.fill(chars, lastChar);
searchTokens.add(String.valueOf(chars));
lastChar = 0;
break;
case '-':
// last char is repeated 4 times
chars = new char[4];
Arrays.fill(chars, lastChar);
searchTokens.add(String.valueOf(chars));
lastChar = 0;
break;
default:
if (lastChar != 0) {
searchTokens.add(String.valueOf(lastChar));
}
lastChar = c;
}
}
return searchTokens.toArray(new String[searchTokens.size()]);
}
}
Output:
Needles: [aa, bb, cccc]
Found: 4
Needles: [a, bb, cccc]
Found: 18

How about preprocessing aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc?
This become a1k1s1d1b1a2l1a1s1k1d1h1f1b2l1a1j1d1f1h1a1c4a1o1u1d1g1a1l1s1a2b2l1i1s1d1f1h1c4
Now find occurrences of a2, b2, c4.

Tried it code below but right now it gives only first possible match based of depth first.
Need to be changed to do all possible combination instead of just first
import java.util.ArrayList;
import java.util.List;
public class Parsing {
public static void main(String[] args) {
String input = "aksdbaalaskdhfbblajdfhacccc aoudgalsaa bblisdfhcccc";
System.out.println(input);
for (int i = 0; i < input.length(); i++) {
System.out.print(i/10);
}
System.out.println();
for (int i = 0; i < input.length(); i++) {
System.out.print(i%10);
}
System.out.println();
List<String> tokenisedSearch = parseExp("a+b+c-");
System.out.println(tokenisedSearch);
parse(input, 0, tokenisedSearch, 0);
}
public static boolean parse(String input, int searchFromIndex, List<String> tokensToSeach, int currentTokenIndex) {
if(currentTokenIndex >= tokensToSeach.size())
return true;
String token = tokensToSeach.get(currentTokenIndex);
int found = input.indexOf(token, searchFromIndex);
if(found >= 0) {
System.out.println("Found at Index "+found+ " Token " +token);
return parse(input, searchFromIndex+1, tokensToSeach, currentTokenIndex+1);
}
return false;
}
public static List<String> parseExp(String exp) {
List<String> list = new ArrayList<String>();
String runningToken = "";
for (int i = 0; i < exp.length(); i++) {
char at = exp.charAt(i);
switch (at) {
case '+' :
runningToken += runningToken;
list.add(runningToken);
runningToken = "";
break;
case '-' :
runningToken += runningToken;
runningToken += runningToken;
list.add(runningToken);
runningToken = "";
break;
default :
runningToken += at;
}
}
return list;
}
}

Recursion may be the following (pseudocode):
int search(String s, String expression) {
if expression consists of only one token t /* e. g. "a+" */ {
search for t in s
return number of occurrences
} else {
int result = 0
divide expression into first token t and rest expression
// e. g. "a+a+b-" -> t = "a+", rest = "a+b-"
search for t in s
for each occurrence {
s1 = substring of s from the position of occurrence to the end
result += search(s1, rest) // search for rest of expression in rest of string
}
return result
}
}
Applying this to entire string, you'll get number of non-continuous occurrences. To get continuous occurrences, you don't need recursion at all--just transform expression into string and search by iteration.

If you convert the search string first with a simple parser/compiler so a+ becomes aa etc. then you can simply take this string and run a regular expression match against your hay stack. (Sorry, I'm no Java coder so can't deliver any real code but it is not really difficult)

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