publicbooleanisPalindrome(String text){Stringclean=text.replaceAll("\\s+","").toLowerCase();intlength=clean.length();intforward=0;intbackward=length -1;while(backward > forward) {charforwardChar=clean.charAt(forward++);charbackwardChar=clean.charAt(backward--);if(forwardChar != backwardChar)r...
Example 2: Check if String with spaces is Empty or Null class Main { public static void main(String[] args) { // create a string with white spaces String str = " "; // check if str1 is null or empty System.out.println("str is " + isNullEmpty(str)); } // method check if ...
Learn how to check if two strings are anagrams in Java with step-by-step examples and code snippets.
values()) if __name__ == "__main__": from doctest import testmod testmod() input_a = input("Enter the first string ").strip() input_b = input("Enter the second string ").strip() status = check_anagrams(input_a, input_b) print(f"{input_a} and {input...
is anagram What is this? This package will check if two words / strings are anagrams.https://en.wikipedia.org/wiki/Anagram Installation It's available on npm. npm install --save is-anagram Usage constisAnagram=require('is-anagram');isAnagram('silent','listen')// trueisAnagram('rail safe...
: true Is string- 'my car color is red' is pangram?: false Here in the above code, we have two input strings. Now we create a function named CheckPangram() to check if the given string is pangram or not. In this function, we first convert the given string into lower case ...
Java Program to check if two strings are an anagram or not import java.util.*; class test{ public static void main(String args[]){ String str1,str2; System.out.println("Enter the two String value"); Scanner sc=new Scanner(System.in); ...
/*Java Program to check whether string is empty or not*/ public class JavaisEmptyPrg { public static void main(String args[]) { String str1="www.includehelp.com"; String str2=""; if(str1.isEmpty()==true) System.out.println("Str1 is an empty string."); el...
Here is source code of the Python Program to check whether a string is a palindrome or not using recursion. The program output is also shown below. def is_palindrome(s): if len(s) < 1: return True else: if s[0] == s[-1]: return is_palindrome(s[1:-1]) else: return False ...
c. if all checks for each position pass, return true. Both the runtime and space complexity is O(N). classSolution {publicbooleanisTransformable(String s, String t) {intn =s.length(); List<Integer>[] sIdx =newList[10];for(inti = 0; i < 10; i++) { ...