Develop Reference

Palindrome in java

Here what I tried
sample input is "aabaa"
eg: in if condition val[0] = a[4]
if it is equal i stored it in counter variable if it is half of the length it original string it is palindrome
if it is not it is not a palindrome
I tried with my basic knowledge in java if there is any errors let me know
boolean solution(String inputString) {
int val = inputString.length();
int count = 0;
for (int i = 0; i<inputString.length(); i++) {
if(inputString.charAt(i) == inputString.charAt(val-i)) {
count = count++;
if (count>0) {
return true;
}
}
}
return true;
}

How about
public boolean isPalindrome(String text) {
String clean = text.replaceAll("\\s+", "").toLowerCase();
int length = clean.length();
int forward = 0;
int backward = length - 1;
while (backward > forward) {
char forwardChar = clean.charAt(forward++);
char backwardChar = clean.charAt(backward--);
if (forwardChar != backwardChar)
return false;
}
return true;
}
From here

In your version you compare first element with last, second with second last etc.
last element in this case is inputString.length()-1(so need to use 'inputString.charAt(val-i-1)' . If you iterate till end, then the count should be equal to length of the string.
for(int i = 0; i<inputString.length(); i++){
if(inputString.charAt(i) == inputString.charAt(val-i-1)){
count ++;
}
}
return (count==val); //true when count=val
Or alternatlively iterate till the mid point of the array, then count value is val/2.
for(int i = 0; i<inputString.length()/2; i++){
if(inputString.charAt(i) == inputString.charAt(val-i-1)){
count ++;
}
}
return (count==val/2); //true when count=val/2

There's no constraints in the question so let me throw in a more cheesy solution.
boolean isPalindrome(String in)
final String inl = in.toLowerCase();
return new StringBuilder(inl).reverse().toString().equals(inl);
}

A palindrome is a word, sentence, verse, or even a number that reads the same forward and backward. In this java solution, we’ll see how to figure out whether the number or the string is palindrome in nature or not.
Method - 1
class Main {
public static void main(String[] args) {
String str = "Nitin", revStr = "";
int strLen = str.length();
for (int i = (strLen - 1); i >=0; --i) {
revStr = revStr + str.charAt(i);
}
if (str.toLowerCase().equals(revStr.toLowerCase())) {
System.out.println(str + " is a Palindrome String.");
}
else {
System.out.println(str + " is not a Palindrome String.");
}
Method - 2
class Main {
public static void main(String[] args) {
int n = 3553, revNum = 0, rem;
// store the number to the original number
int orgNum = n;
/* get the reverse of original number
store it in variable */
while (n != 0) {
remainder = n % 10;
revNum = revNum * 10 + rem;
n /= 10;
}
// check if reversed number and original number are equal
if (orgNum == revNum) {
System.out.println(orgNum + " is Palindrome.");
}
else {
System.out.println(orgNum + " is not Palindrome.");
}

How would I use printf() and a String array of inputed words to print out within a box of asterisks

In this format where 'Nivla' would be within a String List, and input is available to be formatted by the user.
************
* Nivla *
************
Also, the box would increase in size with the lines of the string array. E.g. ["Nivla is" , "not intelligent"];
**********************
* Nivla is *
* not intelligent *
**********************
Also, I'm having trouble centering the string around the center like above.
The code I'm current using is:
public void run() {
Scanner s = new Scanner(System.in);
ArrayList<String> inputs = new ArrayList<>();
String input;
do {
input = s.nextLine();
inputs.add(input);
} while(!input.equals(""));
printBox(inputs);
}
public void printBox(ArrayList inputs) {
for (int i = 0; i < inputs.size(); i++) {
System.out.printf("*\t%s\t*\n", inputs.get(i));
}
}
Is there any way to solve my issue?

You need to make the amount of spaces dependent on the length of the longest input instead of just printing tabs on each side. You should also add a padding parameter in your printBox method to specify how many spaces to put on each side of the word.
Here is one possible solution that works pretty decently:
int max = 0;
int padding = 10;
public void run() {
Scanner s = new Scanner(System.in);
ArrayList<String> inputs = new ArrayList<>();
String input;
do {
input = s.nextLine();
inputs.add(input);
max = Math.max(max, input.length());
} while(!input.equals(""));
printBox(inputs, padding);
}
public void printBox(ArrayList inputs, int padding) {
printStars();
// go to size - 1 because the last input is always ""
for (int i = 0; i < inputs.size() - 1; i++) {
int len = ((String)inputs.get(i)).length();
int frontPad = (max + len)/2 + padding;
// need to round or else sometimes the padding will be one too short
int backPad = padding + (int)Math.round(((max - len)/2.0));
System.out.printf("*%" + frontPad + "s%" + backPad + "s\n", inputs.get(i), "*");
}
printStars();
}
private void printStars() {
for (int i = 0; i <= max + padding*2; i++) {
System.out.print("*");
}
System.out.println();
}

New to programming, getting a run time error don't know why

Basically summarized in the title. https://ideone.com/E2BMS8 <-- that's a link to the code. I understand if you don't want to click it though so I'll paste it here as well. will just be disorganized. The code is supposed to flip the letters but keep words in the same position. I would like to figure that part out on my own though. Just need help with the run time error.
import java.util.*;
class Ideone {
public static void main (String[] args) throws java.lang.Exception {
Scanner input = new Scanner(System.in);
String sent, accum = "";
char check, get;
int len, count = 0;
System.out.print("Please enter the sentance you want reversed: ");
sent = input.nextLine();
len = sent.length();
for (int i = 0; i < len; i++) {
check = sent.charAt(len - i);
count += 1;
if (check == ' ') {
for (int p = 0; p < count; p++) {
while (p < count) {
get = sent.charAt(len - p);
accum += (get + ' ');
}
}
}
}
System.out.println("Reversed: " + accum);
}
}

The error String index out of range is cause because of the len is one more than the index range. Remove one on the index such I did below:
import java.util.*;
public class Ideone {
public static void main (String[] args) throws java.lang.Exception {
Scanner input = new Scanner(System.in);
String sent, accum = "";
char check, get;
int len, count = 0;
System.out.print("Please enter the sentance you want reversed: ");
sent = input.nextLine();
len = sent.length();
for (int i = 0; i < len; i++) {
check = sent.charAt(len - i - 1);
count += 1;
if (check == ' ') {
for (int p = 0; p < count; p++) {
get = sent.charAt(len - p - 1);
accum += (get + ' ');
}
}
}
System.out.println("Reversed: " + accum);
}
}

This is a classical "off by one" error -- something you will run into a lot as you find your programming feet. The issue in this case is the 0-based indexing. That is, the first character of a string is at index 0, and the last is at index "string length - 1". If we use sent = "Test"; as an example, then:
sent.charAt(0) == 'T'
sent.charAt(1) == 'e'
sent.charAt(2) == 's'
sent.charAt(3) == 't'
sent.charAt(4) == ??? // "That's an error, Jim!"
Note that index 4 -- which perhaps confusingly is also the length of the string -- is out of bounds. So, what happens during the first iteration of the loop, when i == 0:
check = sent.charAt(len - i); // ERROR! Because ...
==> = sent.charAt((4) - (0));
==> = sent.charAt( 4 ); // Doh!
I leave it to you to figure out how you might fix it.

How to find a number with no digit occurring more than once (J3 CCC 2013 from 1987 to 2013)

I'm a grade 11 student and have been given the task to complete this question for homework:
Problem J3/S1: From 1987 to 2013
You might be surprised to know that 2013 is the first year since 1987
with distinct digits.
The years 2014, 2015, 2016, 2017, 2018, 2019
each have distinct digits.
2012 does not have distinct digits,
since the digit 2 is repeated.
Given a year, what is the next year with distinct digits?
Input
The input consists of one integer Y (0 ≤ Y ≤ 10000),
representing the starting year.
Output
The output will be the single integer D,
which is the next year after Y with distinct digits.
Sample Input 1
1987
Sample Output 1
2013
Sample Input 2
999
Sample Output 2
1023
I usually answer these types of questions rather quickly but I am stumped when it comes to this one. I have spent several hours and cannot figure it out. I found out How to identify if a number is distinct or not, but I can't figure out how to add on years and check again, I keep getting errors. I would really appreciate someone's help.
Please keep in mind that I am in grade 11 and this is my first year of working with Java, so please do not use advanced coding, and methods because I won't understand. If you can, please answer it in a class and not the main method.
here is what I tried:
import java.util.*;
import java.io.*;
public class Leavemealone
{
public static void main(String[] args) throws IOException
{
BufferedReader objReader = new BufferedReader(new InputStreamReader(System.in));
int ctr = 0;
String inputStr = "";
int input = 0;
int inputCheck = 0;
System.out.println("Enter somthin: ");
input = Integer.parseInt (objReader.readLine ());
while(ctr == 0)
{
inputStr += input;
Scanner sc = new Scanner(inputStr);
int n = sc.nextInt(); // get year
String s = String.valueOf(n);
int[] num = new int[4];
for (int i = 0; i < s.length(); i++)
{
int x = Integer.parseInt(s.substring(i, i + 1)); // integer at this part in the string
num[i] += x;
}
String apple = (num[0] + "" + num[1] + "" + num[2] + "" + num[3]);
if (num[0] != num[1] &&
num[1] != num[2] &&
num[2] != num[3] &&
num[0] != num[2] &&
num[0] != num[3] &&
num[1] != num[3])
{
ctr++;
//distinct
}
else
{
input++;
//not distinct
}
}
}
}
Thanks in advance!
this is the other code I found online, I just don't know how to put it in a class
import java.util.Scanner;
import java.io.*;
public class Thegoodone
{
public static void main(String[] args) throws IOException
{
BufferedReader objReader = new BufferedReader(new InputStreamReader (System.in));
int ctr = 0;
String input = "";
int inputCheck = 0;
while (ctr == 0)
{
System.out.println("Enter somthin: ");
inputCheck = Integer.parseInt (objReader.readLine ());
if (inputCheck > 0 && inputCheck < 10000)
{
input += inputCheck;
ctr += 1;
}
else
{
System.out.println("invalid input ");
}
}
Scanner sc = new Scanner(input);
int n = sc.nextInt(); // get year
n++; // start from the next year
while (!hasDistinctDidgets(n)) //if there is repeating digits
{
n++;// next year
}
System.out.println(n);// prints year
}
public static boolean hasDistinctDidgets(int n)
{
//System.out.println("a" + n);
String s = String.valueOf(n); // converts the year from int to String
int[] numbers = new int[10]; // index position represents number, element value represents occurrence of that number
for (int i = 0; i < s.length(); i++)
{
int x = Integer.parseInt(s.substring(i, i + 1)); // integer at this part in the string
numbers[x]++; //increase occurrence of this integer in the array
}
//check if any digit occurred more than once in the array
for (int i = 0; i < numbers.length; i ++)
{
if (numbers[i] > 1) //digit occurred more than once
{
return false; //not distinct
}
}
return true; // hasn't returned false yet, so the integer has distinct digits
}
}
so this is how I tried to put it in a class:
import java.util.Scanner;
import java.io.*;
public class Danny3
{
public static void main(String[] args) throws IOException
{
BufferedReader objReader = new BufferedReader(new InputStreamReader (System.in));
int ctr = 0;
String input = "";
int inputCheck = 0;
while (ctr == 0)
{
System.out.println("Enter somthin: ");
inputCheck = Integer.parseInt (objReader.readLine ());
if (inputCheck > 0 && inputCheck < 10000)
{
input += inputCheck;
ctr += 1;
}
else
{
System.out.println("invalid input ");
}
}
Scanner sc = new Scanner(input);
// System.out.println(output);
int n = sc.nextInt(); // get year
n++; // start from the next year
DistinctCheck processing = new DistinctCheck(n);
int output = processing.getSum();
System.out.println(output);
}
}
class DistinctCheck
{
//private int year = 0;
private boolean hasDistinctDidgets;
private int b = 0;
DistinctCheck(int temp)
{
hasDistinctDidgets(temp);
}
private void yearAdd(int b)
{
while(!hasDistinctDidgets(b)) //if there is repeating digits
{
b++;// next year
}
}
private boolean hasDistinctDidgets(int year)
{
String s = String.valueOf(year); // converts the year from int to String
int[] numbers = new int[10]; // index position represents number, element value represents occurrence of that number
for (int i = 0; i < s.length(); i++)
{
int x = Integer.parseInt(s.substring(i, i + 1)); // integer at this part in the string
numbers[x]++; //increase occurrence of this integer in the array
}
//check if any digit occurred more than once in the array
for (int i = 0; i < numbers.length; i ++)
{
if (numbers[i] > 1) //digit occurred more than once
{
return false; //not distinct
}
}
return true; // hasn't returned false yet, so the integer has distinct digits
}
int getSum()
{
return b;// prints year
}
}

I would start with a method to determine if a given int consists of distinct digits. You could use a Set<Character> and add each character from the String to the Set. You will get false on a duplicate. Like,
static boolean distinctDigits(int i) {
String s = String.valueOf(i);
Set<Character> set = new HashSet<>();
for (char c : s.toCharArray()) {
if (!set.add(c)) {
return false;
}
}
return true;
}
Then your main just needs to invoke that. Like,
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int v = s.nextInt();
while (v < 10000) {
v++;
if (distinctDigits(v)) {
break;
}
}
System.out.println(v);
}

i figured it out:
import java.util.*;
public class Apple
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int num = input.nextInt();
Distinct findDistinct = new Distinct(num); // objecct
String output = findDistinct.getDistinctYear();
System.out.println(output);
}
} // end of main
class Distinct
{
private int ctr = 0;
private String yearStr = "";
private String distinctYear = "";
private int year = 0;
Distinct(int n)
{
year = n;
makeDistinct();
}
private void makeDistinct()
{
while(ctr == 0)
{
year += 1; // year will keep increasing until it is distinct
yearStr = Integer.toString(year);
if(isDistinct(yearStr) == true) // if the number is distinct
{
distinctYear = yearStr;
ctr++;
}
}
}
private boolean isDistinct(String yearStr)
{
String eachNum[] = yearStr.split(""); // breaks up each number (char) of yearStr
for(int i = 0; i < eachNum.length; i++)
{
for(int j = 0; j < i; j++)
{
if (eachNum[i].equals(eachNum[j])) // not distinct
{
return false;
}
}
}
return true; // is distinct
}
String getDistinctYear()
{
return distinctYear;
}
}

interviewstreet.com - String similarity

I'm trying to solve the string similarity question on interviewstreet.com. My code is working for 7/10 cases (and it is exceeding the time limit for the other 3).
Here's my code -
public class Solution {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
String v1 = user_input.next();
int number_cases = Integer.parseInt(v1);
String[] cases = new String[number_cases];
for(int i=0;i<number_cases;i++)
cases[i] = user_input.next();
for(int k=0;k<number_cases;k++){
int similarity = solve(cases[k]);
System.out.println(similarity);
}
}
static int solve(String sample){
int len=sample.length();
int sim=0;
for(int i=0;i<len;i++){
for(int j=i;j<len;j++){
if(sample.charAt(j-i)==sample.charAt(j))
sim++;
else
break;
}
}
return sim;
}
}
Here's the question -
For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.
Calculate the sum of similarities of a string S with each of it's suffixes.
Input:
The first line contains the number of test cases T. Each of the next T lines contains a string each.
Output:
Output T lines containing the answer for the corresponding test case.
Constraints:
1 <= T <= 10
The length of each string is at most 100000 and contains only lower case characters.
Sample Input:
2
ababaa
aa
Sample Output:
11
3
Explanation:
For the first case, the suffixes of the string are "ababaa", "babaa", "abaa", "baa", "aa" and "a". The similarities of each of these strings with the string "ababaa" are 6,0,3,0,1,1 respectively. Thus the answer is 6 + 0 + 3 + 0 + 1 + 1 = 11.
For the second case, the answer is 2 + 1 = 3.
How can I improve the running speed of the code. It becomes harder since the website does not provide a list of test cases it uses.

I used char[] instead of strings. It reduced the running time from 5.3 seconds to 4.7 seconds and for the test cases and it worked. Here's the code -
static int solve(String sample){
int len=sample.length();
char[] letters = sample.toCharArray();
int sim=0;
for(int i=0;i<len;i++){
for(int j=i;j<len;j++){
if(letters[j-i]==letters[j])
sim++;
else
break;
}
}
return sim;
}

used a different algorithm. run a loop for n times where n is equals to length the main string. for each loop generate all the suffix of the string starting for ith string and match it with the second string. when you find unmatched character break the loop add j's value to counter integer c.
import java.io.BufferedReader;
import java.io.InputStreamReader;
class Solution {
public static void main(String args[]) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(in.readLine());
for (int i = 0; i < T; i++) {
String line = in.readLine();
System.out.println(count(line));
}
}
private static int count(String input) {
int c = 0, j;
char[] array = input.toCharArray();
int n = array.length;
for (int i = 0; i < n; i++) {
for (j = 0; j < n - i && i + j < n; j++)
if (array[i + j] != array[j])
break;
c+=j;
}
return c;
}
}

I spent some time to resolve this question, and here is an example of my code (it works for me, and pass thru all the test-cases):
static long stringSimilarity(String a) {
int len=a.length();
char[] letters = a.toCharArray();
char localChar = letters[0];
long sim=0;
int sameCharsRow = 0;
boolean isFirstTime = true;
for(int i=0;i<len;i++){
if (localChar == letters[i]) {
for(int j = i + sameCharsRow;j<len;j++){
if (isFirstTime && letters[j] == localChar) {
sameCharsRow++;
} else {
isFirstTime = false;
}
if(letters[j-i]==letters[j])
sim++;
else
break;
}
if (sameCharsRow > 0) {
sameCharsRow--;
sim += sameCharsRow;
}
isFirstTime = true;
}
}
return sim;
}
The point is that we need to speed up strings with the same content, and then we will have better performance with test cases 10 and 11.

Initialize sim with the length of the sample string and start the outer loop with 1 because we now in advance that the comparison of the sample string with itself will add its own length value to the result.

import java.util.Scanner;
public class StringSimilarity
{
public static void main(String args[])
{
Scanner user_input = new Scanner(System.in);
int count = Integer.parseInt(user_input.next());
char[] nextLine = user_input.next().toCharArray();
try
{
while(nextLine!= null )
{
int length = nextLine.length;
int suffixCount =length;
for(int i=1;i<length;i++)
{
int j =0;
int k=i;
for(;k<length && nextLine[k++] == nextLine[j++]; suffixCount++);
}
System.out.println(suffixCount);
if(--count < 0)
{
System.exit(0);
}
nextLine = user_input.next().toCharArray();
}
}
catch (Exception e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}