Move all special char to the end of the String in Java

To add all special characters to the end of a string in Java, it's necessary to iterate through the input string, identify the alphanumeric letters, and then rearrange them so that the special characters are at the end. Java's built-in character classification methods can be used to distinguish between letters, numerals, and non-alphanumeric characters, which is often necessary for this operation. The outcome should relocate the special characters without changing their relative order while retaining the order of the alphanumeric letters.

Example 1:

Input:

String str = "Hello!@#World123$%"

Output:

The resulted string is given by HelloWorld123!@#$%

Explanation:

Special characters (!@#$%) are relocated to the end, maintaining their original order, while all alphanumeric characters (HelloWorld123) are pushed to the front.

Example 2:

Input:

String str = "2024Year!!!New#Goals"

Output:

The resulted string is given by 2024YearNewGoals!!!#

Explanation:

In the same arrangement, the special characters (!!!#) are moved to the end, and the alphanumeric characters (2024YearNewGoals) are positioned up front.

Example 3:

Input:

String str = "Java$Is#Fun*&"

Output:

The resulted string is given by JavaIsFun$#*&

Explanation:

After the special characters ($#*&), which remain in the same order as before, are the alphanumeric characters (JavaIsFun).

Naïve Approach

The goal is to iterate through the input string and preserve two strings: one with regular characters (a, A, 1, '', etc.) and another with special characters (@, $, etc.). Concatenate the two strings at the end and return.

Algorithm:

Step 1: Initialize the string s.

Step 2: Set up two blank result strings, result1 for alphanumeric characters and result2 for special characters.

Step 3: Go through every character in the string s in a loop.

Step 4: Verify that ch corresponds to the regular expression [a-zA-Z0-9\\s+].

Step 4.1: Add ch to result1 if that is the case.

Step 4.2: Add ch to result2 if it's false.

Step 5: Return the combination of result 1 (alphanumeric characters) and result 2 (special characters).

Step 6: Print the updated result after invoking the function with the input string.

Implementation:

FileName: CharactersTraversing.java

Output:

 
The resulted string is given by 2024YearNewGoals!!!#   

Complexity Analysis:

The time complexity of the above code is O(N), and the space complexity is O(N), where 'N' represents the length of the string.

Approach: Using Pointers

The approach provides a Java program that, given a string, splits special characters from alphanumeric letters and spaces, adding the special characters at the end. The moveCharToEnd method checks if each character in the input string str is a space or an alphanumeric character by iterating through each character. If it is, it's added to the red string. If not, it is appended to the Char string, which is used to store special characters. The method concatenates the res and Char strings after the loop and returns the outcome.

Algorithm:

Step 1: Create two blank strings, called char for special characters and res for alphanumeric characters and spaces.

Step 2: Every character in the input string str is iterated through.

Step 3: Determine whether each character at index I is a space or an alphanumeric character (inside A, A, 0-9).

Step 4: Add the character to res if it's a space or an alphanumeric character.

Step 5: If not, append the character (signifying that it is a unique character) to char.

Step 6: Concatenate res, which contains alphanumeric characters and spaces, with char, which includes special characters, once the loop is finished.

Step 7: Return the concatenated string. It starts with all alphanumeric characters and ends with special characters.

Step 8: Print the result that was returned. The sequence of spaces and alphanumeric letters in the original string will be preserved when the special characters are added at the end.

Implementation:

FileName: PointersCharacterTravaersing.java

Output:

 
The resulted string is given by 2024YearNewGoals!!!#   

Complexity Analysis:

The Time Complexity of the above code is O(N), where N represents the length of the input string and the Space Complexity is O(N).

Approach: Using 'isalnum()' method

The below approach provided defines a class called isalnumMethod, which has a moveAllSpecialChars method that divides special characters in an input string into different categories from regular characters (letters, numbers, and spaces). Using character.isLetterOrDigit() and Character.isWhitespace() to determine which characters are regular, the procedure loops through each character in the string. The normalCh string receives these regular characters as an appendix, whereas the specialCh string stores the special characters. Ultimately, all special characters are moved to the end of the two strings by concatenating them.

Algorithm:

Step 1: Start with an empty string called "special_ch," in order to store the special characters, and to store the regular characters, start with an empty string called "normal_ch."

Step 2: Go through every "ch" character in the input string one at a time.

Step 3: The steps given below should be followed for each character in the provided string.

Step 3.1: Use the isalnum() function to determine whether ch is an alphanumeric character, a space, or a whitespace character. Alternatively, use the isspace() method.

Step 3.2: Add ch to the normal_chars string if it's a space or an alphanumeric character.

Step 3.3: Add ch to the special_chars string if it's a unique character.

Step 4: Give back the concatenation of normal_ch and special_ch after going through every character in the string.

Implementation:

FileName: IsalnumMethod.java

Output:

 
The resulted string is given by 2024YearNewGoals!!!#   

Complexity Analysis:

The Time Complexity of the above code is O(N), where n represents the length of the input string and the Space Complexity is O(N).