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1900 Discussions

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3 days ago+ 0 comments

By comparing both ends characters.

boolean isPalindrome = false; for(int i=0;i

    if(isPalindrome)
        System.out.println("Yes");
    else
        System.out.println("No");

5 days ago+ 0 comments

    System.out.println(A.contentEquals(new StringBuilder(A).reverse())?"Yes":"No");

2 weeks ago+ 0 comments

 public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String str = input.next();
        System.out.println(str.length() > 50 ? "String is out of range"
        : str.equals(new StringBuilder(str).reverse().toString()) ? "Yes" : "No");
        input.close();

    }

3 weeks ago+ 0 comments

Nice challenge! Palindrome checks are always fun and a great way to test string manipulation skills. bet bricks 7

1 month ago+ 0 comments

Language: Java 15

Using Array:

public class Solution {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        String A=sc.next();
        sc.close();
        
        String[] str = A.split("");
        int i=0;
        int j=(str.length)-1;
        boolean flag= true;
        while(i<j){
            if (str[i].compareTo(str[j]) !=0){
                flag=false;
                break;
            }
            
                i++;
                j--;  
        }
        if(flag==true){
            System.out.println("Yes");
        }
        else if(flag==false){System.out.println("No");}
    }
}

Using CharAt Method:

public class Solution {

public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    String A=sc.next();
    sc.close();


    int i=0;
    int j=(A.length())-1;
    boolean flag= true;
    while(i<j){
        if (A.charAt(i) != A.charAt(j)){
            flag=false;
            break;
        }

            i++;
            j--;  
    }
    if(flag==true){
        System.out.println("Yes");
    }
    else if(flag==false){System.out.println("No");}
}

}

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1900 Discussions

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Language: Java 15

Using CharAt Method:

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