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Pancake Sorting with a Twist (Java)

I am attempting to solve a version of a pancake sorting algorithm. In this problem I am given a string that is composed of any combination of characters A-F and has a maximum length of 6. For instance I may receive the String 'ACFE'. In this problem I am trying to use pancake sorting to fix the string to be in Alphabetical Order. So the above example would become 'ACEF'.
That is pretty simple and straightforward. Here is the catch: the characters in the input string can be Uppercase OR Lowercase. Whenever you flip characters in the string, the flipped characters switch case. So an uppercase A would become 'a'. The goal at the end is to flip the string into order and also have all of the characters in uppercase as well.
I have had no problem putting together the algorithm to solve the sorting part of the algorithm, but it is the part where I am trying to make sure that we aren't done flipping the characters until they are all uppercase that I am having trouble with and can't seem to solve.
To make things easier on myself, I have made a HashMap of Characters to Integers so that it is easier to sort the characters (we can just use an equivalent Integer value). I also break the string apart at the beginning into a char[] and put it in reverse order to make the algorithm easier for myself.
Here is the code I use to do everything:
private static final HashMap<Character, Integer> numericalEquivalent = new HashMap<>();
static {
numericalEquivalent.put('A', 6);
numericalEquivalent.put('B', 5);
numericalEquivalent.put('C', 4);
numericalEquivalent.put('D', 3);
numericalEquivalent.put('E', 2);
numericalEquivalent.put('F', 1);
numericalEquivalent.put('a', 6);
numericalEquivalent.put('b', 5);
numericalEquivalent.put('c', 4);
numericalEquivalent.put('d', 3);
numericalEquivalent.put('e', 2);
numericalEquivalent.put('f', 1);
}
private static int flip(char[] arr, int i, int numFlips) {
char temp;
int start = 0;
if (start < i) {
while (start < i) {
temp = (Character.isUpperCase(arr[start]) ? Character.toLowerCase(arr[start]) : Character.toUpperCase(arr[start]));
arr[start] = (Character.isUpperCase(arr[i]) ? Character.toLowerCase(arr[i]) : Character.toUpperCase(arr[i]));
arr[i] = temp;
start++;
i--;
}
numFlips++;
}
return numFlips;
}
private static int findMax(char[] arr, int n) {
int mi, i;
for (mi = 0, i = 0; i < n; ++i)
if (numericalEquivalent.get(arr[i]) > numericalEquivalent.get(arr[mi]))
mi = i;
return mi;
}
private static int getFlips (char[] pancakes) {
int n = pancakes.length;
int numFlips = 0;
for (int curr_size = n; curr_size > 1; --curr_size) {
int mi = findMax(pancakes, curr_size);
if (mi != curr_size - 1) {
numFlips = flip(pancakes, mi, numFlips);
if (!isSorted(pancakes))
numFlips = flip(pancakes, curr_size - 1, numFlips);
}
}
return numFlips;
}
private static boolean isSorted(char[] arr) {
for (int i = 0; i < arr.length - 1; i++) {
if (numericalEquivalent.get(arr[i]) > numericalEquivalent.get(arr[i + 1]))
return false;
}
return true;
}
public static void main(String[] args) {
while(true) {
String input = scanner.nextLine();
if (input.equals("0")) break;
else System.out.println(getFlips(new StringBuilder(input).reverse().toString().toCharArray()));
}
}
My goal is to get back the minimum number of flips that it will take to flip the characters into order. How can I modify this code, though, to make sure it accounts for characters being lowercase and the need to make sure they all end up in Uppercase?

You can change the stop condition from if (!isSorted(pancakes)) to if (!isSortedAndUppercase(pancakes)) where isSortedAndUppercase(pancakes) is defined as:
private static boolean isSortedAndUppercase(char[] arr){
return isUpperCase(arr) && isSorted(arr);
}
private static boolean isUpperCase(char[] arr) {
String s = String.valueOf(arr);
return s.equals(s.toUpperCase());
}
and stop the search only when the stop condition is met.
Consider using Breadth-First Search for this task.
Side notes:
There is no need to map chars into integers. Try the following:
char[] chars = "ABCDEF".toCharArray();
for (int i = 0; i < chars.length; i++) {
System.out.println(chars[i] +" int value: "+(int)chars[i]);
System.out.println(Character.toLowerCase(chars[i]) +" int value: "+(int)Character.toLowerCase(chars[i]));
}

Algorithm to create all permutations and lengths

I am looking to create an algorithm preferably in Java. I would like to go through following char array and create every possible permutations and lengths out of it.
For example, loop and print the following:
a
aa
aaaa
aaaaa
.... keep going ....
aaaaaaaaaaaaaaaaa ....
ab
aba
abaa .............
Till I hit all possible lengths and permutations from my array.
private void method(){
char[] data = "abcdefghiABCDEFGHI0123456789".toCharArray();
// loop and print each time
}
I think it would be silly to come up with 10s of for loops for this. I am guessing some form of recursion would help here but can't get my head around to even start with. Could I get some help with this please? Even if pointing me to a start or a blog or something. Been Googling and looking around and many permutations examples exists but keeps to fixed max length. None seems to have examples on multiple length + permutations. Please advice. Thanks.

Another way to do it is this:
public class HelloWorld{
public static String[] method(char[] arr, int length) {
if(length == arr.length - 1) {
String[] strArr = new String[arr.length];
for(int i = 0; i < arr.length; i ++) {
strArr[i] = String.valueOf(arr[i]);
}
return strArr;
}
String[] before = method(arr, length + 1);
String[] newArr = new String[arr.length * before.length];
for(int i = 0; i < arr.length; i ++) {
for(int j = 0; j < before.length; j ++) {
if(i == 0)
System.out.println(before[j]);
newArr[i * before.length + j] = (arr[i] + before[j]);
}
}
return newArr;
}
public static void main(String []args){
String[] all = method("abcde".toCharArray(), 0);
for(int i = 0; i < all.length; i ++) {
System.out.println(all[i]);
}
}
}
However be careful you'll probably run out of memory or the program will take a looooong time to compile/run if it does at all. You are trying to print 3.437313508041091e+40 strings, that's 3 followed by 40 zeroes.
Here's the solution also in javascript because it starts running but it needs 4 seconds to get to 4 character permutations, for it to reach 5 character permutations it will need about 28 times that time, for 6 characters it's 4 * 28 * 28 and so on.
const method = (arr, length) => {
if(length === arr.length - 1)
return arr;
const hm = [];
const before = method(arr, length + 1);
for(let i = 0; i < arr.length; i ++) {
for(let j = 0; j < before.length; j ++) {
if(i === 0)
console.log(before[j]);
hm.push(arr[i] + before[j]);
}
}
return hm;
};
method('abcdefghiABCDEFGHI0123456789'.split(''), 0).forEach(a => console.log(a));

private void method(){
char[] data = "abcdefghiABCDEFGHI0123456789".toCharArray();
// loop and print each time
}
With your given input there are 3.43731350804×10E40 combinations. (Spelled result in words is eighteen quadrillion fourteen trillion three hundred ninety-eight billion five hundred nine million four hundred eighty-one thousand nine hundred eighty-four. ) If I remember it correctly the maths is some how
1 + x + x^2 + x^3 + x^4 + ... + x^n = (1 - x^n+1) / (1 - x)
in your case
28 + 28^2 + 28^3 + .... 28^28
cause you will have
28 combinations for strings with length one
28*28 combinations for strings with length two
28*28*28 combinations for strings with length three
...
28^28 combinations for strings with length 28
It will take a while to print them all.
One way I can think of is to use the Generex library, a Java library for generating String that match a given regular expression.
Generex github. Look at their page for more info.
Generex maven repo. Download the jar or add dependency.
Using generex is straight forward if you are somehow familiar with regex.
Example using only the first 5 chars which will have 3905 possible combinations
public static void main(String[] args) {
Generex generex = new Generex("[a-e]{1,5}");
System.out.println(generex.getAllMatchedStrings().size());
Iterator iterator = generex.iterator();
while (iterator.hasNext()) {
System.out.println(iterator.next());
}
}
Meaning of [a-e]{1,5} any combination of the chars a,b,c,d,e wit a min length of 1 and max length of 5
output
a
aa
aaa
aaaa
aaaaa
aaaab
aaaac
aaaad
aaaae
aaab
aaaba
aaabb
aaabc
aaabd
aaabe
aaac
....
eeee
eeeea
eeeeb
eeeec
eeeed
eeeee

You can have a for loop that starts from 1 and ends at array.length and in each iteration call a function that prints all the permutations for that length.
public void printPermutations(char[] array, int length) {
/*
* Create all permutations with length = length and print them
*/
}
public void method() {
char data = "abcdefghiABCDEFGHI0123456789".toCharArray();
for(int i = 1; i <= data.length; i ++) {
printPermutations(data, i);
}
}

I think the following recursion could solve your problem:
public static void main(String[] args) {
final String[] data = {"a", "b", "c"};
sampleWithReplacement(data, "", 1, 5);
}
private static void sampleWithReplacement(
final String[] letters,
final String prefix,
final int currentLength,
final int maxLength
) {
if (currentLength <= maxLength) {
for (String letter : letters) {
final String newPrefix = prefix + letter;
System.out.println(newPrefix);
sampleWithReplacement(letters, newPrefix, currentLength + 1, maxLength);
}
}
}
where data specifies your possible characters to sample from.

Is this what you're talking about?
public class PrintPermutations
{
public static String stream = "";
public static void printPermutations (char[] set, int count, int length)
{
if (count < length)
for (int i = 0; i < set.length; ++i)
{
stream += set[i];
System.out.println (stream);
printPermutations (set, count + 1, length);
stream = stream.substring (0, stream.length() - 1);
}
}
public static void main (String[] args)
{
char[] set = "abcdefghiABCDEFGHI0123456789".toCharArray();
printPermutations (set, 0, set.length);
}
}
Test it using a smaller string first.

On an input string 28 characters long this method is never going to end, but for smaller inputs it will generate all permutations up to length n, where n is the number of characters. It first prints all permutations of length 1, then all of length 2 etc, which is different from your example, but hopefully order doesn't matter.
static void permutations(char[] arr)
{
int[] idx = new int[arr.length];
char[] perm = new char[arr.length];
Arrays.fill(perm, arr[0]);
for (int i = 1; i < arr.length; i++)
{
while (true)
{
System.out.println(new String(perm, 0, i));
int k = i - 1;
for (; k >= 0; k--)
{
idx[k] += 1;
if (idx[k] < arr.length)
{
perm[k] = arr[idx[k]];
break;
}
idx[k] = 0;
perm[k] = arr[idx[k]];
}
if (k < 0)
break;
}
}
}
Test:
permutations("abc".toCharArray());
Output:
a
b
c
aa
ab
ac
ba
bb
bc
ca
cb
cc

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

Same char in one string

I need to know how many chars of the same type are in one string.
I have tried this
String x ="(3+3)*(4-2)";
int a = x.indexOf( "(" );
But that only give me the first index

You can use a loop and use the other method indexOf(int, int):
String x ="(3+3)*(4-2)";
int a = x.indexOf( "(" );
while (a >= 0) {
System.out.println("Char '(' found at: "+a);
a = x.indexOf('(', a+1);
}

It seems like it would be better to put it in a separate function:
// accepts a string and a char to find the number of occurrences of
public static int get_count(String s, char c) {
int count = 0; // count initially 0
for (int i = 0; i < s.length(); i++) // loop through the whole string
if (s.charAt(i) == c)
count ++; // increment every time an occurrence happens
return count; // return the count in the end
}
You can call it like this:
System.out.println(get_count("(3+3)*(4-2)", '('));
// Output: 2

There's a few ways I could think of doing this, but one of the simplest would be to simply loop the through characters in the String....
String x ="(3+3)*(4-2)";
int count = 0;
for (char c : x.toCharArray()) {
if (c == '(') {
count++;
}
}
System.out.println(count);
And just because it can be done...you could use a little regexp...(I know, overkill)
Pattern p = Pattern.compile("\\(");
Matcher matcher = p.matcher(x);
while (matcher.find()) {
count++;
}
System.out.println(count);

The code below does what you want. If performance is critical you can make optimization with this. If you want more elegant solutions you may take a look at regex library of java.
int occurences = 0;
String x ="(3+3)*(4-2)";
char tolookfor = '(';
for(int i = 0; i < x.length() ; i++)
{
if(x.charAt(i) == tolookfor)
occurences++;
}

You can try this
String x ="(3+3)*(4-2)";
char[] arr=x.toCharArray();
Map<String,Integer> map=new HashMap<>();
for(int i=0;i<arr.length;i++){
Integer upTo=map.get(String.valueOf(arr[i]));
if (upTo==null) {
upTo=0;
}
map.put(String.valueOf(arr[i]),upTo+1) ;
}
for (Map.Entry<String,Integer> entry:map.entrySet()){
System.out.println("Number of "+entry.getKey()+" in this string is: "+entry.getValue());
}
out put
Number of 3 in this string is: 2
Number of 2 in this string is: 1
Number of 4 in this string is: 1
Number of * in this string is: 1
Number of + in this string is: 1
Number of ( in this string is: 2
Number of ) in this string is: 2
Number of - in this string is: 1

It’s unbelievable how complicated the answers to such a simple question can be.
x.indexOf( "(" );
But that only give me the first index
Use x.indexOf( "(", fromIndex ); to find more occurrences. Point.
By the way, if you are looking for a single char you can use x.indexOf( '('); and x.indexOf( '(', fromIndex ); to be more efficient.
So the most efficient way without reinventing the wheel would be:
int count=0;
for(int pos=s.indexOf('('); pos!=-1; pos=s.indexOf('(', pos+1)) count++;

Use StringUtils.countMatches
StringUtils.countMatches(value,"(");
or
public static int countMatches(String value, String valueToCount) {
if (value.isEmpty() || valueToCount.isEmpty()) {
return 0;
}
int count = 0;
int index = 0;
while ((index = value.indexOf(valueToCount, index)) != -1) {
count++;
index += valueToCount.length();
}
return count;
}

This will help you!
public static int counter(String x, char y) {
char[] array=x.toCharArray();
int count=0;
for(int i=0;i<x.length();i++)
{
if(y==array[i]) count++;
}
return (count>0)? count:0;
}