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Java for loop with char arrays dis-function

I'm stuck with a loop issue here, I'm working on a script who will receive let's say the String "geij" or "abab" and will have to turn it into a double like "6478" or "0101". I do the conversion from letter to number thanks to a two-dimensional array :
String crypt = "geij";
char twoD[][] = {{'a','b','c','d','e','f','g','h','i','j'}, {'0','1','2','3','4','5','6','7','8','9'}};
First I pass the String into a char array :
char tab[] = crypt.toCharArray();
Then I use a loop to convert from letter to number :
for(int c=0;c<tab.length;c++) {
for(int z=0;z<twoD.length;z++) {
if(tab[c] == twoD[0][z]) {
tab[c] = twoD[1][z];
}
}
Then I create a new instance of String named 'second' to turn the array into a String
String second = new String(tab);
And I turned this String into a double
double finalC = Double.parseDouble(second);
The issue is with this loop, If the String crypt is "abab", the loop will return 0101 as it is supposed to, but if the String contains any letter after "a" or "b" from the first array of the two-dimensional array, like for example the String "geij" the program will simply return "geij".
I don't understand why the program doesn't go further than b and it is starting to give me an egghead. If anyone has an idea I'll be grateful !
Here is an example of the inside of the tab array after the loop for the String "abcd" :
Indice : 0 value: 0
Indice : 1 value: 1
Indice : 2 value: c
Indice : 3 value: d

Kevin Cruijssen resolves your problem but you can more:
Use HashMap to this problem. For now, your algorithm time complexity is O(n*m) (n-base string length, m - amount of letters in the table) because you must iterate through the whole array of letters for each letter.
Using HashMap you can find the right letter in O(1). A lot faster. So now your algorithm has O(n) time complexity.
Simple example:
Map<Character, Integer> encoding = new HashMap<>();
encoding.put('a', 0);
encoding.put('b', 1);
encoding.put('c', 2);
encoding.put('d', 3);
String toEncode = "abcd";
char[] chars = toEncode.toCharArray();
StringBuilder sb = new StringBuilder();
for(char c : chars){
int newInt = encoding.getOrDefault(c, -5); //-5 is just a flag that there is no char to encode
if(newInt == -5){
continue; //or do something else, e.g throw exception;
}
sb.append(newInt);
}
System.out.println(sb.toString());
//Parse double if you want, but remember that what *Nikolas* said in the comments under your post.
//Double.parseDouble(sb.toString());

The problem is in your inner loop: twoD.length is 2, because twoD contains your two inner array of characters.
You should use twoD[0].length instead:
for(int c=0; c<tab.length; c++) {
for(int z=0; z<twoD[0].length; z++) {
...
However, since you are using all ten digits, perhaps better to use that instead:
char twoD[][] = {{'a','b','c','d','e','f','g','h','i','j'}, {'0','1','2','3','4','5','6','7','8','9'}};
int amountOfDigitsUsed = 10; // Equal to `twoD[0].length` or `twoD[1].length`.
for(int c=0; c<tab.length; c++) {
for(int z=0; z<amountOfDigitsUsed; z++) {
...
Regardless whether you use a hard-coded twoD conversion and amountOfDigits used or not. In your current implementation your twoD.length is 2, causing the issues you have right now.

Length of your twoD array is 2. Your second loop should iterate from z = 0 to twoD[0].length.
Try naming your variables meaningfully so it will be easier to find bugs like this. Also check out foreach loops so you don't have to worry about indexes. Java Maps could be better for mapping characters to numbers.

Since it seems as though in your case the characters are incrementing along with their int values, you don't need a map at all. You can cast the character to an int, and then subtract a's int value. This is a slight variation of B_Osipiuk's answer:
String toEncode = "abcd";
char[] chars = toEncode.toCharArray();
StringBuilder sb = new StringBuilder();
for(char c : chars){
int newInt = c - 'a';
if (newInt < 0 || newInt > ('j'-'a')) {
continue; //or do something else, e.g throw exception;
}
sb.append(newInt);
}
System.out.println(sb.toString());

How to print all non null elements of an array using Array.toString()

So I need to print out an array of integers. The problem is that when user enters the numbers to be processed and sorted, I do not know how many numbers there will be entered by the user. The only rule to that is that user can enter only less than 10000 numbers.
So I made an array which can hold 10000 numbers but if user enters less than 10000 numbers into the array then Array.toString() function prints out everything, even the empty spaces.
Is there any way to bypass that or are there any other methods for outputting an array in one line that would format it the output to look like this: [1, 2, 3, 4, 5, 6]
Thanks a lot!

It would be much easier to store the user's input in an ArrayList, and then just print myArryList.toString().

An ArrayList<T> (optionally <T> for generics). Is an array that dynamically adds more memory if more input comes available. The amortized cost of adding an element to such list is O(1), but it offers a convenient way to process input.
To answer your question, as #Mureinik already answered you then can use ArrayList.toString() to convert the list to a textual representation.
To answer your real question, you can do the following:
public static<T> String toStringArrayNonNulls (T[] data) {
StringBuilder sb = new StringBuilder();
sb.append("[");
int n = data.length;
int i = 0;
for(; i < n; i++) {
if(data[i] != null) {
sb.append(data[i].toString());
break;
}
}
for(; i < n; i++) {
if(data[i] != null) {
sb.append(",");
sb.append(data[i].toString());
}
}
sb.append("]");
return sb.toString();
}
And call that method with any type of array you want.
Examples
String[] names = new String[] {"Alice",null,"Charly","David",null,null};
System.out.println(toStringArrayNonNulls(names));
Integer[] primes = new Integer[] {2,3,null,null,11};
System.out.println(toStringArrayNonNulls(primes));
Object[] namesAndPrimes = new Integer[] {"Alice",2,null,3,null,"Charly",null,11};
System.out.println(toStringArrayNonNulls(namesAndPrimes));

If you want a dynamic array in Java, the best way is to learn how to use lists :
List<Integer> integers = new ArrayList<Integer>();
It prints only as many numbers as you put into it.
If you dont want to rewrite your program, this is also solution :
Integer[]x = new Integer[50];
List<Integer> integers = new ArrayList<Integer>();
integers = new ArrayList<Integer>(Arrays.asList(x));

multiply string element per int element from two arrays Java

I am learning Java and looking for a comprehensive code of multiplying the elements from 2 arrays, possibly without importing anything to achieve it.
In Python it's quite easy:
a=['a','b','c','d']
b=[1,2,3,4]
[x*y for x,y in zip(a,b)]
['a', 'bb', 'ccc', 'dddd']
How can I achieve the same thing in Java, when the first array is an array of strings and the second is integers?

I'm afraid Java isn't going to support this kind of thing natively, and you'll need to perform some of your own logic to implement it. Let's say you've got your String[]..
String[] a = {"a", "b", "c", "d"};
And you've got your int[]..
int[] b = {1,2,3,4};
Next, you'll need to check that the arrays are the same size.
if(a.length == b.length) {
// Continue.
}
Then you need to implement a loop, to go through each item in the arrays.
for(int x = 0; x < a.length; x++)
{
// Some looping code.
}
And you're going to grab each item.
String value = a[x];
int multiplier = b[x];
If you're not importing anything, you declare the total value:
String total = "";
But if you're allowing for a StringBuilder, then you'll import it and declare..
StringBuilder total = new StringBuilder();
NOTE: StringBuilder is strongly recommended here.
And then you're looping multiplier amount of times..
for(int y = 0; y < multiplier; y++)
{
// If you use StringBuilder..
total.append(value);
// If you don't..
total += value;
}
// If you use StringBuilder..
a[x] = total.toString();
// If you don't...
a[x] = total;
This will set the value of a[x] to the repeated String.
NOTE: Something that's also important is leaning good practise. If you're using Java code, it's considered terrible practise to repeatedly concatenate String objects. StringBuilder is more efficient, and is the Java standard. I would strongly recommend using this.
Have fun putting it all together!!

To create string filled with multiple instances of same character like "ccc" you can firs create array of characters which will hold only 3 characters like
char[] myCharacters = new char[3];
Now this array is filled with zeroes ('\0'), so you need to fill it with desired character 'c'. You simply do it using for loop
for (int i = 0; i<myCharacters; i++){
myCharacters[i] = 'c';
}
After this your array will contain ['c', 'c', 'c'].
Now you can use this array to create string using characters from it. To do so you just need to pass this array to String constructor like
String myString = new String(myCharacters);
And there you go. Now you have "ccc" String. Repeat these steps for each pair of elements from a and b arrays.
You can also use shorter version which kinds of do the same
String myString = new String(new char[3]).replace('\0','c');//will produce "ccc"

Exporting specific pattern of string using split method in a most efficient way

I want to export pattern of bit stream in a String varilable. Assume our bit stream is something like bitStream="111000001010000100001111". I am looking for a Java code to save this bit stream in a specific array (assume bitArray) in a way that all continous "0"s or "1"s be saved in one array element. In this example output would be somethins like this:
bitArray[0]="111"
bitArray[1]="00000"
bitArray[2]="1"
bitArray[3]="0"
bitArray[4]="1"
bitArray[5]="0000"
bitArray[6]="1"
bitArray[7]="0000"
bitArray[8]="1111"
I want to using bitArray to calculate the number of bit which is stored in each continous stream. For example in this case the final output would be, "3,5,1,1,1,4,1,4,4". I figure it out that probably "split" method would solve this for me. But I dont know what splitting pattern would do that for me, if i Using bitStream.split("1+") it would split on contious "1" pattern, if i using bitStream.split("0+") it will do that base on continous"0" but how it could be based on both?
Mathew suggested this solution and it works:
var wholeString = "111000001010000100001111";
wholeString = wholeString.replace('10', '1,0');
wholeString = wholeString.replace('01', '0,1');
stringSplit = wholeString.split(',');
My question is "Is this solution the most efficient one?"

Try replacing any occurrence of "01" and "10" with "0,1" and "1,0" respectively. Then once you've injected the commas, split the string using the comma as the delimiting character.
String wholeString = "111000001010000100001111"
wholeString = wholeString.replace("10", "1,0");
wholeString = wholeString.replace("01", "0,1");
String stringSplit[] = wholeString.split(",");

You can do this with a simple regular expression. It matches 1s and 0s and will return each in the order they occur in the stream. How you store or manipulate the results is up to you. Here is some example code.
String testString = "111000001010000100001111";
Pattern pattern = Pattern.compile("1+|0+");
Matcher matcher = pattern.matcher(testString);
while (matcher.find())
{
System.out.print(matcher.group().length());
System.out.print(" ");
}
This will result in the following output:
3 5 1 1 1 4 1 4 4
One option for storing the results is to put them in an ArrayList<Integer>
Since the OP wanted most efficient, I did some tests to see how long each answer takes to iterate over a large stream 10000 times and came up with the following results. In each test the times were different but the order of fastest to slowest remained the same. I know tick performance testing has it's issues like not accounting for system load but I just wanted a quick test.
My answer completed in 1145 ms
Alessio's answer completed in 1202 ms
Matthew Lee Keith's answer completed in 2002 ms
Evgeniy Dorofeev's answer completed in 2556 ms
Hope this helps

I won't give you a code, but I'll guide you to a possible solution:
Construct an ArrayList<Integer>, iterate on the array of bits, as long as you have 1's, increment a counter and as soon as you have 0, add the counter to the ArrayList. After this procedure, you'll have an ArrayList that contain numbers, etc: [1,2,2,3,4] - Representing a serieses of 1's and 0's.
This will represent the sequences of 1's and 0's. Then you construct an array of the size of the ArrayList, and fill it accordingly.
The time complexity is O(n) because you need to iterate on the array only once.

This code works for any String and patterns, not only 1s and 0s. Iterate char by char, and if the current char is equal to the previous one, append the last char to the last element of the List, otherwise create a new element in the list.
public List<String> getArray(String input){
List<String> output = new ArrayList<String>();
if(input==null || input.length==0) return output;
int count = 0;
char [] inputA = input.toCharArray();
output.add(inputA[0]+"");
for(int i = 1; i <inputA.length;i++){
if(inputA[i]==inputA[i-1]){
String current = output.get(count)+inputA[i];
output.remove(count);
output.add(current);
}
else{
output.add(inputA[i]+"");
count++;
}
}
return output;
}

try this
String[] a = s.replaceAll("(.)(?!\\1)", "$1,").split(",");

I tried to implement #Maroun Maroun solution.
public static void main(String args[]){
long start = System.currentTimeMillis();
String bitStream ="0111000001010000100001111";
int length = bitStream.length();
char base = bitStream.charAt(0);
ArrayList<Integer> counts = new ArrayList<Integer>();
int count = -1;
char currChar = ' ';
for (int i=0;i<length;i++){
currChar = bitStream.charAt(i);
if (currChar == base){
count++;
}else {
base = currChar;
counts.add(count+1);
count = 0;
}
}
counts.add(count+1);
System.out.println("Time taken :" + (System.currentTimeMillis()-start ) +"ms");
System.out.println(counts.toString());
}
I believe it is more effecient way, as he said it is O(n) , you are iterating only once. Since the goal to get the count only not to store it as array. i woul recommen this. Even if we use Regular Expression ( internal it would have to iterate any way )
Result out put is
Time taken :0ms
[1, 3, 5, 1, 1, 1, 4, 1, 4, 4]

Try this one:
String[] parts = input.split("(?<=1)(?=0)|(?<=0)(?=1)");
See in action here: http://rubular.com/r/qyyfHNAo0T

Count of Each Alphabet using Array

I have a string as an input eg. Testerty. I want to find the count of each alphabet in the string. I have tried using a HashMap. But I want to implement this using array.
Can you please suggest some way.

You can use ASCII to assign the letters number values:
int[] letters = new int[128]; // There are 128 different possible characters.
for(int i = 0; i < input.length; i++) {
char current = input.charAt(i);
int index = Character.getNumericValue(char);
letters[index]++;
}

ArrayList<Character> ch = new ArrayList<Character>();
ArrayList<Integer> count = new ArrayList<Integer>();
someMethod(String input) {
for(char c : input.toCharArray()) {
if(ch.indexOf(c) != -1) {
i.set(ch.indexOf(c), i.get(ch.indexOf(c))+1);
} else {
ch.add(c);
i.add(1);
}
}
}

doing it with a Map is easier, where the letters are keys and the values are counts. Using an array is more tricky; you could assign each letter a number, and use that number as an index into the array, and store the counts in the array. So 'A' is 1, 'B' is 2, etc....The algorithm is
Get next letter of string.
Get the index for the letter.
Increment the value at that index in the array by 1.
Of course you need to do null checking and whatever.
Note that this is logically a Map. It's just when you use a Map, the Map does step 2 above for you.

You should use a collection implementing Multiset iunterface, i.e. HashMultiset (both taken from Google Guava library). Multiset is designed to hold counts for objects in collection:
Multiset<String> m = HashMultiset.create(Arrays.asList("a", "a", "b", "c", "b", "a"));
// m.toString() prints "a x 3, b x 2, c x 1"
// m.count() gives 6

One way could be, you first create an array, then traverse string using charAt(index) method ,match the current char against those in the array.If you find the match ,increment the value there,or add it as a new entry in the array.