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?
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):
radius = 0) at i - kradius = 0 at i + k, too)radius at i + k is the same as at i - k)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)i + k + radius = i + pri + 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
