Develop Reference

How to compare two 2D Arrays in Java?

I am a beginner trying to write a function in Java that returns true if two passed 2D arrays of int type are the same size in every dimension, and false otherwise. Requirements are that if both arrays are null you should return true. If one is null and the other is not you should return false.
Somehow getting an error for my code:
public static boolean arraySameSize(int[][] a, int[][] b) {
if (a == null && b == null) {
return true;
}
if (a == null || b == null) {
return false;
}
if (a.length == b.length) {
for (int i = 0; i < a.length; i++) {
if (a[i].length == b[i].length) {
return true;
}
}
}
return false;
}
Any help would be appreciated!
Edit: Problem is "Runtime Error: null"

Your logic looks almost spot-on already. The only issue I see is in the logic handling the case where both arrays are not null and have the same first dimension. You should be returning false if any index does not have matching lengths:
public static boolean arraySameSize(int[][] a, int[][] b) {
if (a == null && b == null) {
return true;
}
if (a == null || b == null) {
return false;
}
if (a.length != b.length) {
return false;
}
// if the code reaches this point, it means that both arrays are not
// null AND both have the same length in the first dimension
for (int i=0; i < a.length; i++) {
if (a[i] == null && b[i] == null) {
continue;
}
if (a[i] == null || b[i] == null) {
return false;
}
if (a[i].length != b[i].length) {
return false;
}
}
return true;
}
Follow the demo link below to see some of examples of this method working correctly.
Demo

What am I doing wrong in this loop?

So for I am trying to make a boolean method that checks if an inputed string is a positive integer. It will return true if it is a positive integer and false if it is anything else. Here is my code:
public static boolean isPositiveInteger(String input) {
int stringLength = input.length();
int index = stringLength-1;
while(index>0) {
index--;
if(input.charAt(0) != '-' && input.charAt(index) >= '0' &&
input.charAt(index) <= '9') {
return true;
}
return false;
}
return false;
}
When the input is the string "fish33" the method will return true instead of false. Why is that?

Your while loop executes only once - return will stop execution. Moreover, you start with second to last, not with last character. Replace your code with this one:
public static boolean isPositiveInteger(String input) {
int stringLength = input.length();
int index = stringLength;
// special case when input is empty string
if (index == 0) {
return false;
}
while(index > 0) {
index--;
// if some of the characters is not digit, return false
if !(input.charAt(index) >= '0' &&
input.charAt(index) <= '9') {
return false;
}
}
// if the while loop does not find any other character, return true
return true;
}

There is no point to make so many manipulations, it can be solved in few lines
public static void main(String[] args) {
System.out.println(isPositiveInteger("1"));
System.out.println(isPositiveInteger("abc"));
}
public static boolean isPositiveInteger(String input) {
try {
Integer i = Integer.parseInt(input);
return i > 0;
}
catch(NumberFormatException nfe){
return false;
}
}

Nested if statement with multiple condition returns

What I'm trying to do is return "fizz" when a number is divisible by fizzNumber and return "fizzbuzz" when a number is divisible by fizzNumber AND buzzNumber. I know I can just make an && statement but I'm trying to use the least amount of code possible to achieve what I need. The problem here is that the nested if statement is unreachable, I'm assuming because the if statement ends after the return statement. How can I fix this so that this mimcks an && statement using a nested if while also returning both strings if the condition is true?
public String getValue(int n) {
if (n % fizzNumber == 0) {
return "fizz";
if (n % buzzNumber == 0) {
return "buzz";
}
return Integer.toString(n); // return the number itself as a String
}
}

Solution:
public String getValue(int n) {
if (n%fizzNumber==0) {
if (n%buzzNumber==0) {
// dividable by both fizzNumber and buzzNumber
return "fizzbuzz";
}
// dividable by fizzNumber only
return "fizz";
}
return Integer.toString(n);
}
or shorter (using terniary operator):
public String getValue(int n) {
return (n%fizzNumber==0) ? (n%buzzNumber==0 ? "fizzbuzz" : "fizz") : Integer.toString(n);
}

You could use this:
public String getValue(int n) {
String ret = "";
if(n % fizzNumber == 0) {
ret += "fizz";
}
if(n % buzzNumber == 0) {
ret += "buzz";
}
if(ret.equals("")) {
ret = Integer.toString(n);
}
return ret;
}
This can be optimized to:
public String getValue(int n) {
StringBuilder sb = new StringBuilder();
if(n % fizzNumber == 0) sb.append("fizz");
if(n % buzzNumber == 0) sb.append("buzz");
return sb.length() == 0 ? Integer.toString(n) : sb.toString();
}
As you can see, this solution doesn't use nested if-statements.

indexOf method using recursion

I need to write a recursive method called indexOf that accepts two Strings as parameters and that returns the starting index of the first occurrence of the second String inside the first String (or -1 if not found). I have to solve this problem using recursion.
These are some example results:
indexOf("Barack Obama", "Bar") 0
indexOf("Barack Obama", "ck") 4
indexOf("Barack Obama", "a") 1
indexOf("Barack Obama", "McCain") -1
indexOf("Barack Obama", "BAR") -1
This is my solution but it gives me 6 for indexOf("Barack Obama", "McCain")instead of -1.
public static int indexOf(String s1, String s2) {
if(s1.equals(s2))
return 0;
else
return indexOfHelper(s1, s2, 0);
}
private static int indexOfHelper(String s1, String s2, int ctr) {
if(s2.length() > s1.length())
return -1;
if(s2.length() == 0 || s1.length() == 0) //base case
return ctr;
else //recursive case
if(s1.charAt(0) == s2.charAt(0)){ //if there is a matching character
if(s1.substring(0, s2.length()).equals(s2))
return ctr; //if there is a matching character and the rest of the strings match as well
else
return -1; //if there is a matching character but the rest of the strings don't match
}
else
return 1 + indexOfHelper(s1.substring(1), s2, ctr);
}

I must say that recursive step was quite tricky. It passes all test cases. I hope following helps. Java and C++ code here
JAVA
int indexOf (String s1, String s2) {
if (s1.length() == 0 && s2.length() == 0) {
return 0;
} else if (s1.length() == 0) {
return -1;
} else if (s2.length() == 0) {
return 0;
} else if (s2.length()>s1.length()) {
return -1;
}
else if (s1.charAt(0) == s2.charAt(0)) {
int subIndex = indexOf(s1.substring(1),s2.substring(1));
return subIndex == -1 ? -1: subIndex;
} else {
int subIndex = indexOf(s1.substring(1),s2);
return subIndex == -1 ? -1: subIndex+1;
}
}
C++ code here
int indexOf (string s1, string s2) {
if (s1 == "" && s2 == "") {
return 0;
} else if (s1 == "") {
return -1;
} else if (s2=="") {
return 0;
} else if (s2.size()>s1.size()) {
return -1;
}
else if (s1 [0] == s2[0]) {
int subIndex = indexOf(s1.substr(1,s1.length()),s2.substr(1,s2.length()));
return subIndex == -1 ? -1: subIndex;
} else {
int subIndex = indexOf(s1.substr(1,s1.length()),s2);
return subIndex == -1 ? -1: subIndex+1;
}
}

Here is my solution
public static void main(String[] args) {
System.out.println(indexOf("Ninja!", "i"));
System.out.println(indexOf("ninja2", "ja2"));
System.out.println(indexOf("ninja2", "hae"));
}
public static int indexOf(String s, String contains) {
if (contains.length() > s.length()) {
return -1;
}
return indexOf(s, contains, 0);
}
private static int indexOf(String s, String contains, int index) {
if ((s.length() - contains.length()) < index) {
return -1;
}
if (s.substring(index, index + contains.length()).equals(contains)) {
return index;
} else {
return indexOf(s, contains, index + 1);
}
}

Creating a recursive method for Palindrome

I am trying to create a Palindrome program using recursion within Java but I am stuck, this is what I have so far:
public static void main (String[] args){
System.out.println(isPalindrome("noon"));
System.out.println(isPalindrome("Madam I'm Adam"));
System.out.println(isPalindrome("A man, a plan, a canal, Panama"));
System.out.println(isPalindrome("A Toyota"));
System.out.println(isPalindrome("Not a Palindrome"));
System.out.println(isPalindrome("asdfghfdsa"));
}
public static boolean isPalindrome(String in){
if(in.equals(" ") || in.length() == 1 ) return true;
in= in.toUpperCase();
if(Character.isLetter(in.charAt(0))
}
public static boolean isPalindromeHelper(String in){
if(in.equals("") || in.length()==1){
return true;
}
}
}
Can anyone supply a solution to my problem?

Here I am pasting code for you:
But, I would strongly suggest you to know how it works,
from your question , you are totally unreadable.
Try understanding this code. Read the comments from code
import java.util.Scanner;
public class Palindromes
{
public static boolean isPal(String s)
{
if(s.length() == 0 || s.length() == 1)
// if length =0 OR 1 then it is
return true;
if(s.charAt(0) == s.charAt(s.length()-1))
// check for first and last char of String:
// if they are same then do the same thing for a substring
// with first and last char removed. and carry on this
// until you string completes or condition fails
return isPal(s.substring(1, s.length()-1));
// if its not the case than string is not.
return false;
}
public static void main(String[]args)
{
Scanner sc = new Scanner(System.in);
System.out.println("type a word to check if its a palindrome or not");
String x = sc.nextLine();
if(isPal(x))
System.out.println(x + " is a palindrome");
else
System.out.println(x + " is not a palindrome");
}
}

Well:
It's not clear why you've got two methods with the same signature. What are they meant to accomplish?
In the first method, why are you testing for testing for a single space or any single character?
You might want to consider generalizing your termination condition to "if the length is less than two"
Consider how you want to recurse. One option:
Check that the first letter is equal to the last letter. If not, return false
Now take a substring to effectively remove the first and last letters, and recurse
Is this meant to be an exercise in recursion? That's certainly one way of doing it, but it's far from the only way.
I'm not going to spell it out any more clearly than that for the moment, because I suspect this is homework - indeed some may consider the help above as too much (I'm certainly slightly hesitant myself). If you have any problems with the above hints, update your question to show how far you've got.

public static boolean isPalindrome(String in){
if(in.equals(" ") || in.length() < 2 ) return true;
if(in.charAt(0).equalsIgnoreCase(in.charAt(in.length-1))
return isPalindrome(in.substring(1,in.length-2));
else
return false;
}
Maybe you need something like this. Not tested, I'm not sure about string indexes, but it's a start point.

I think, recursion isn't the best way to solve this problem, but one recursive way I see here is shown below:
String str = prepareString(originalString); //make upper case, remove some characters
isPalindrome(str);
public boolean isPalindrome(String str) {
return str.length() == 1 || isPalindrome(str, 0);
}
private boolean isPalindrome(String str, int i) {
if (i > str.length / 2) {
return true;
}
if (!str.charAt(i).equals(str.charAt(str.length() - 1 - i))) {
return false;
}
return isPalindrome(str, i+1);
}

Here is my go at it:
public class Test {
public static boolean isPalindrome(String s) {
return s.length() <= 1 ||
(s.charAt(0) == s.charAt(s.length() - 1) &&
isPalindrome(s.substring(1, s.length() - 1)));
}
public static boolean isPalindromeForgiving(String s) {
return isPalindrome(s.toLowerCase().replaceAll("[\\s\\pP]", ""));
}
public static void main(String[] args) {
// True (odd length)
System.out.println(isPalindrome("asdfghgfdsa"));
// True (even length)
System.out.println(isPalindrome("asdfggfdsa"));
// False
System.out.println(isPalindrome("not palindrome"));
// True (but very forgiving :)
System.out.println(isPalindromeForgiving("madam I'm Adam"));
}
}

public class palin
{
static boolean isPalin(String s, int i, int j)
{
boolean b=true;
if(s.charAt(i)==s.charAt(j))
{
if(i<=j)
isPalin(s,(i+1),(j-1));
}
else
{
b=false;
}
return b;
}
public static void main()
{
String s1="madam";
if(isPalin(s1, 0, s1.length()-1)==true)
System.out.println(s1+" is palindrome");
else
System.out.println(s1+" is not palindrome");
}
}

Some of the codes are string heavy. Instead of creating substring which creates new object, we can just pass on indexes in recursive calls like below:
private static boolean isPalindrome(String str, int left, int right) {
if(left >= right) {
return true;
}
else {
if(str.charAt(left) == str.charAt(right)) {
return isPalindrome(str, ++left, --right);
}
else {
return false;
}
}
}
public static void main(String []args){
String str = "abcdcbb";
System.out.println(isPalindrome(str, 0, str.length()-1));
}

Here are three simple implementations, first the oneliner:
public static boolean oneLinerPalin(String str){
return str.equals(new StringBuffer(str).reverse().toString());
}
This is ofcourse quite slow since it creates a stringbuffer and reverses it, and the whole string is always checked nomatter if it is a palindrome or not, so here is an implementation that only checks the required amount of chars and does it in place, so no extra stringBuffers:
public static boolean isPalindrome(String str){
if(str.isEmpty()) return true;
int last = str.length() - 1;
for(int i = 0; i <= last / 2;i++)
if(str.charAt(i) != str.charAt(last - i))
return false;
return true;
}
And recursively:
public static boolean recursivePalin(String str){
return check(str, 0, str.length() - 1);
}
private static boolean check (String str,int start,int stop){
return stop - start < 2 ||
str.charAt(start) == str.charAt(stop) &&
check(str, start + 1, stop - 1);
}

public static boolean isPalindrome(String str)
{
int len = str.length();
int i, j;
j = len - 1;
for (i = 0; i <= (len - 1)/2; i++)
{
if (str.charAt(i) != str.charAt(j))
return false;
j--;
}
return true;
}

Try this:
package javaapplicationtest;
public class Main {
public static void main(String[] args) {
String source = "mango";
boolean isPalindrome = true;
//looping through the string and checking char by char from reverse
for(int loop = 0; loop < source.length(); loop++){
if( source.charAt(loop) != source.charAt(source.length()-loop-1)){
isPalindrome = false;
break;
}
}
if(isPalindrome == false){
System.out.println("Not a palindrome");
}
else
System.out.println("Pailndrome");
}
}

String source = "liril";
StringBuffer sb = new StringBuffer(source);
String r = sb.reverse().toString();
if (source.equals(r)) {
System.out.println("Palindrome ...");
} else {
System.out.println("Not a palindrome...");
}

public class chkPalindrome{
public static String isPalindrome(String pal){
if(pal.length() == 1){
return pal;
}
else{
String tmp= "";
tmp = tmp + pal.charAt(pal.length()-1)+isPalindrome(pal.substring(0,pal.length()-1));
return tmp;
}
}
public static void main(String []args){
chkPalindrome hwObj = new chkPalindrome();
String palind = "MADAM";
String retVal= hwObj.isPalindrome(palind);
if(retVal.equals(palind))
System.out.println(palind+" is Palindrome");
else
System.out.println(palind+" is Not Palindrome");
}
}

Here is a recursive method that will ignore specified characters:
public static boolean isPal(String rest, String ignore) {
int rLen = rest.length();
if (rLen < 2)
return true;
char first = rest.charAt(0)
char last = rest.charAt(rLen-1);
boolean skip = ignore.indexOf(first) != -1 || ignore.indexOf(last) != -1;
return skip || first == last && isPal(rest.substring(1, rLen-1), ignore);
}
Use it like this:
isPal("Madam I'm Adam".toLowerCase(), " ,'");
isPal("A man, a plan, a canal, Panama".toLowerCase(), " ,'");
It does not make sense to include case insensitivity in the recursive method since it only needs to be done once, unless you are not allowed to use the .toLowerCase() method.

there's no code smaller than this:
public static boolean palindrome(String x){
return (x.charAt(0) == x.charAt(x.length()-1)) &&
(x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}
if you want to check something:
public static boolean palindrome(String x){
if(x==null || x.length()==0){
throw new IllegalArgumentException("Not a valid string.");
}
return (x.charAt(0) == x.charAt(x.length()-1)) &&
(x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}
LOL B-]

public static boolean isPalindrome(String p)
{
if(p.length() == 0 || p.length() == 1)
// if length =0 OR 1 then it is
return true;
if(p.substring(0,1).equalsIgnoreCase(p.substring(p.length()-1)))
return isPalindrome(p.substring(1, p.length()-1));
return false;
}
This solution is not case sensitive. Hence, for example, if you have the following word : "adinida", then you will get true if you do "Adninida" or "adninida" or "adinidA", which is what we want.
I like #JigarJoshi answer, but the only problem with his approach is that it will give you false for words which contains caps.

Palindrome example:
static boolean isPalindrome(String sentence) {
/*If the length of the string is 0 or 1(no more string to check),
*return true, as the base case. Then compare to see if the first
*and last letters are equal, by cutting off the first and last
*letters each time the function is recursively called.*/
int length = sentence.length();
if (length >= 1)
return true;
else {
char first = Character.toLowerCase(sentence.charAt(0));
char last = Character.toLowerCase(sentence.charAt(length-1));
if (Character.isLetter(first) && Character.isLetter(last)) {
if (first == last) {
String shorter = sentence.substring(1, length-1);
return isPalindrome(shorter);
} else {
return false;
}
} else if (!Character.isLetter(last)) {
String shorter = sentence.substring(0, length-1);
return isPalindrome(shorter);
} else {
String shorter = sentence.substring(1);
return isPalindrome(shorter);
}
}
}
Called by:
System.out.println(r.isPalindrome("Madam, I'm Adam"));
Will print true if palindrome, will print false if not.
If the length of the string is 0 or 1(no more string to check), return true, as the base case. This base case will be referred to by function call right before this. Then compare to see if the first and last letters are equal, by cutting off the first and last letters each time the function is recursively called.

Here is the code for palindrome check without creating many strings
public static boolean isPalindrome(String str){
return isPalindrome(str,0,str.length()-1);
}
public static boolean isPalindrome(String str, int start, int end){
if(start >= end)
return true;
else
return (str.charAt(start) == str.charAt(end)) && isPalindrome(str, start+1, end-1);
}

public class PlaindromeNumbers {
int func1(int n)
{
if(n==1)
return 1;
return n*func1(n-1);
}
static boolean check=false;
int func(int no)
{
String a=""+no;
String reverse = new StringBuffer(a).reverse().toString();
if(a.equals(reverse))
{
if(!a.contains("0"))
{
System.out.println("hey");
check=true;
return Integer.parseInt(a);
}
}
// else
// {
func(no++);
if(check==true)
{
return 0;
}
return 0;
}
public static void main(String[] args) {
// TODO code application logic here
Scanner in=new Scanner(System.in);
System.out.println("Enter testcase");
int testcase=in.nextInt();
while(testcase>0)
{
int a=in.nextInt();
PlaindromeNumbers obj=new PlaindromeNumbers();
System.out.println(obj.func(a));
testcase--;
}
}
}

/**
* Function to check a String is palindrome or not
* #param s input String
* #return true if Palindrome
*/
public boolean checkPalindrome(String s) {
if (s.length() == 1 || s.isEmpty())
return true;
boolean palindrome = checkPalindrome(s.substring(1, s.length() - 1));
return palindrome && s.charAt(0) == s.charAt(s.length() - 1);
}

Simple Solution
2 Scenario --(Odd or Even length String)
Base condition& Algo recursive(ch, i, j)
i==j //even len
if i< j recurve call (ch, i +1,j-1)
else return ch[i] ==ch[j]// Extra base condition for old length
public class HelloWorld {
static boolean ispalindrome(char ch[], int i, int j) {
if (i == j) return true;
if (i < j) {
if (ch[i] != ch[j])
return false;
else
return ispalindrome(ch, i + 1, j - 1);
}
if (ch[i] != ch[j])
return false;
else
return true;
}
public static void main(String[] args) {
System.out.println(ispalindrome("jatin".toCharArray(), 0, 4));
System.out.println(ispalindrome("nitin".toCharArray(), 0, 4));
System.out.println(ispalindrome("jatinn".toCharArray(), 0, 5));
System.out.println(ispalindrome("nittin".toCharArray(), 0, 5));
}
}

for you to achieve that, you not only need to know how recursion works but you also need to understand the String method.
here is a sample code that I used to achieve it: -
class PalindromeRecursive {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter a string");
String input=sc.next();
System.out.println("is "+ input + "a palindrome : " + isPalindrome(input));
}
public static boolean isPalindrome(String s)
{
int low=0;
int high=s.length()-1;
while(low<high)
{
if(s.charAt(low)!=s.charAt(high))
return false;
isPalindrome(s.substring(low++,high--));
}
return true;
}
}