how to find upside down of a 2digit number - java

I am trying to write program to find upside down of all double digit number. Actually I have written this code and it works pretty well too. Any other alternatives to find upside down of an integer. Say 18=81 and 96=69....
List < Integer > al = new ArrayList < Integer > ();
al.add(10);al.add(11);al.add(16);al.add(18);al.add(19);al.add(60);al.add(61);
al.add(66);al.add(69);al.add(80);al.add(81);al.add(86);al.add(88);al.add(91);
al.add(96);al.add(98);al.add(99);
Scanner s = new Scanner(System.in);
Integer n = s.nextInt();
if (((String)"" + n).length() == 2) {
if (al.contains(n))
System.out.println("yes");
else
System.out.println("no");
}

This is a simple, general algorithm analysing an integer w.r.t. digits with vertical symmetry. (It depends on a certain style of writing digits; note that "continental" '1' is not symmetrical.)
private static int[] other = new int[]{0, 1, -1, -1, -1, -1, 9, -1, 8, 6};
public static int invert( int n ){
int units = n % 10;
n /= 10;
int inv = other[units];
if( n == 0 ) return inv;
return inv < 0 ? -1 : invert( n )*10 + inv;
}
If a negative value is returned, the number is not symmetrical.
public static void main(String[] args) throws Exception {
for( int i = 100; i <= 199; ++i ){
int invi = invert( i );
if( invi > 0 ){
System.out.println( i + ": " + invi );
}
}
}

This is a better way of doing it.
It not only prints yes or no, but also the up side down number
public static void main(String[] args) {
List<String> al = new ArrayList<String>();
al.add("0");
al.add("1");
al.add("6");
al.add("8");
al.add("9");
Scanner s = new Scanner(System.in);
Integer n = s.nextInt();
String num = n + "";
if (num.length() == 2) {
if(al.contains(num.charAt(0) + "") && al.contains(num.charAt(1) + "")) {
System.out.println("Yes");
String upSideDownNumber = "";
if(num.charAt(1) == 6) {
upSideDownNumber += 9;
} else if(num.charAt(1) == 9) {
upSideDownNumber += 6;
} else {
upSideDownNumber += num.charAt(1);
}
if(num.charAt(0) == 6) {
upSideDownNumber += 9;
} else if(num.charAt(0) == 9) {
upSideDownNumber += 6;
} else {
upSideDownNumber += num.charAt(0);
}
System.out.println("The up side down number is " + upSideDownNumber);
} else {
System.out.println("No");
}
} else {
System.out.println("No");
}
}

Since only 0,1,6,8,9 look like digits when inverted. You can so below:
public static void main(String[] args) throws IOException {
Set<Integer> upsideDownProperDigits = new TreeSet<>(Arrays.asList(0,1,8,9,6));
Scanner s = new Scanner(System.in);
Integer n = s.nextInt();
boolean found = true;
while( n != 0){
int digit = n % 10;
if(!upsideDownProperDigits.contains(digit)){
found = false;
break;
}
n = n/10;
}
if(found){
System.out.println("yes");
}
else{
System.out.println("no");
}
}

Related

Give amount of happy numbers in a given (input) range

As an exercise I need to make a code that will give the amount of happy numbers* in a given range. As a test I need to insert the code in a program that checks 10 different outcomes, for every good outcome you get 4 points. I want the full 40 points and my teacher said that I need to change only a little bit in my code. So far I have this:
*A happy number is a number which eventually reaches 1 when replaced by the sum of the square of each digit. For instance, 13 is a happy number because 1^2 + 3^2 = 10 and 1^2 + 0^2 = 1
Important to know: When you end up with a 4, it is definitely not a happy number.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String st1 = scan.nextLine().trim();
String st2 = scan.nextLine().trim();
int min = Integer.parseInt(st1);
int max = Integer.parseInt(st2);
Set<Integer> happyNumbers = getHappyNumbers(min, max);
System.out.println(happyNumbers.size());
}
public static Set<Integer> getHappyNumbers(int min, int max) {
Set<Integer> out = new HashSet<>();
for (int i = min; i < max; i++) {
if (isHappy(i)) {
out.add(i);
}
}
return out;
}
private static boolean isHappy(int i) {
int sum = 0;
while (i != 0) {
sum += Math.pow((i % 10), 2);
i /= 10;
}
if (sum == 4) return false;
else if(sum == 1) return true;
else if (sum !=1) {return isHappy(sum);}
else return true;
}
}
My teacher also said the mistake is in the following part:
if (sum == 4) return false;
else if(sum == 1) return true;
else if (sum !=1) {return isHappy(sum);}
else return true;
Please help :)
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
public class HappyNumbersMain {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("Lower limit (>0): ");
String st1 = scan.nextLine().trim();
System.out.print("Upper limit: ");
String st2 = scan.nextLine().trim();
int min = Integer.parseInt(st1);
int max = Integer.parseInt(st2);
Set<Integer> happyNumbers = getHappyNumbers(min, max);
System.out.println("Happy numbers between " + min + " and " + max + ":" + happyNumbers);
}
public static Set<Integer> getHappyNumbers(int min, int max) {
Set<Integer> out = new TreeSet<>();
for (int i = min; i <= max; i++) {
if (isHappy(i)) {
out.add(i);
}
}
return out;
}
private static boolean isHappy(int i) {
// Stopping conditions
if (i == 4) {
return false;
} else if (i == 1) {
return true;
}
int sum = 0;
while (i > 0) {
sum += Math.pow((i % 10), 2);
i /= 10;
}
return isHappy(sum);
}
}
Above works as expected, some notes:
When doing recursive functions, put stopping conditions as the first thing it checks
Adding some feedback to the user helps in understanding how to use your program
TreeSet will sort values, HashSet does not, better for display purposes
You should always verify user input (I did not do that, but you should)
if (sum == 4)
return false;
else
if(sum == 1)
return true;
else
return isHappy(sum);
Alternative:
switch ( sum ): {
case 4: return false; break;
case 1: return true; break;
default: return isHappy(sum); break;
}
The break isn't needed, since the return in all control paths.

Return the sum of all integers from a random String without using regex

I was asked this question in an interview recently (Java programming que)
Return the sum of all integers from a random String.
Just iterate over the string, handle one digit at a time. This is pretty much exactly what the regex would do anyway:
String testStrings[] = { "-1a2b3c", "123ab!45c", "abcdef", "0123.4",
"dFD$#23+++12##T1234;/.,10" };
for (String testString : testStrings) {
String currentNumber = "";
int sum = 0;
for (int i = 0; i < testString.length(); i++) {
char currentChar = testString.charAt(i);
// Add digits or a leading minus to "currentNumber"
if (Character.isDigit(currentChar)
|| (currentNumber.equals("") && currentChar == '-')) {
currentNumber += currentChar;
} else {
// We've stumbled across a non-digit char.
//Try to parse the "currentNumber" we have so far
if (!currentNumber.equals("") && !currentNumber.equals("-"))
sum += Integer.parseInt(currentNumber);
currentNumber = "";
}
}
// Add the last "currentNumber" in case the string ends with a
// number
if (!currentNumber.equals("") && !currentNumber.equals("-"))
sum += Integer.parseInt(currentNumber);
System.out.println(sum);
}
Output:
4
168
0
127
1279
public class Random {
public int SumofNumbers(String s){
char[] str = s.toCharArray();
String answer="";
int sum = 0;
List<String> al = new ArrayList();
for (int i=0;i< str.length;i++){
if (checkNumber(str[i])){
answer=answer+str[i];
}
else
{
if(!answer.isEmpty()){
al.add(answer);
answer = "";
}
}
if (i == str.length -1 && !answer.isEmpty()) {
al.add(answer);
}
}
for (String a1 : al){
sum = sum + Integer.valueOf(a1);
}
return sum;
}
private boolean checkNumber(char c) {
if ((int)c > 47 && (int)c < 58){
return true;
}else if ((int)c == 45){
return true;
}
return false;
}
public static void main(String [] args){
Random r = new Random();
String test = "123ab!45c";
System.out.println(r.SumofNumbers(test));
}
}
public class StringToIntAddition {
public static void main(String[] args) throws Exception {
String str = "2e40 ssdf 23-9", number="";
int sum=0;
for(int i=0; i<str.length() ;i++){
if(Character.isDigit(str.charAt(i))){
number += str.charAt(i);
}
else if(!number.isEmpty()){
sum += Integer.parseInt(number);
number= "";
}
if (str.charAt(i) == '-'){
number = "-" ;
}
}
if(!number.isEmpty()){
sum += Integer.parseInt(number);
}
System.out.println("number= " + sum);
}
}
I've got a slightly 'cute' way to do this in Java 8: implement it as a Collector
public DigitCollector {
private boolean negative = false;
private int current = 0;
private int total = 0;
public int getTotal() {
if (negative) {
total -= current;
} else {
total += current;
}
current = 0;
negative = false;
return total;
}
public void accept(Character ch) {
if (Character.isDigit(ch)) {
current = 10 * current + Integer.parseInt(ch.toString());
} else if (ch.equals('-')) {
negative = true;
} else {
getTotal();
}
}
}
Now you can collect a stream of characters:
text.chars().map(ch -> new Character((char)ch))
.collect(DigitCollector::new, DigitCollector::accept, null)
.getTotal();
I realise the mapping ch -> new Character((char)ch)) looks strange but .chars() returns a stream of integers instead of characters. See here for reasons why (though pretty much everyone agrees it was a mistake).
This is a slightly longwinded way of doing it but it's pretty flexible: you could take a stream of Character from anywhere and do any sort of manipulation you wanted before collecting them. It seems to me to be a natural representation of the problem and, mostly, I just reckon streams are cooler than traditional iteration :-)
There's already quite a few answers, but this one seemed fun. I have a different solution that should be pretty efficient:
public static int countString(String input) {
if (input == null) return 0;
int sum = 0;
int accumulator = 0;
boolean lastCharWasDigit = false;
for (int i = 0, len = input.length(); ++i) {
char c = input.charAt(i);
// If a non-digit character is found, clear the
// accumulator and add it to the sum.
if (c < '0' || c > '9') {
sum += accumulator;
accumulator = 0;
lastCharWasDigit = false;
continue;
}
// If the previous character was a digit, that means
// this is a continuation. Multiply by ten to shift
// it over one power of ten before adding the new value
if (lastCharWasDigit) {
accumulator *= 10;
}
// Add the integer value of the character
int charValue = c - '0';
accumulator += charValue;
lastCharWasDigit = true;
}
// Finally, clear the accumulator for any ending digits,
// and return the sum
sum += accumulator;
return sum;
}

Roman number to decimal in Java

I have to make a program that converts Roman numbers to decimal. I am confused about how to write the conditions for the Roman numbers, such as IV (4), IX (9), XL (40) and CM(900). The code that I wrote works for all the other numbers.
public static void main(String[] args) {
System.out.print("Enter a roman numeral: ");
Scanner in = new Scanner(System.in);
String Roman = in.next();
int largo = Roman.length();
char Roman2[] = new char[largo];
int Roman3[] = new int[largo];
for (int i = 0; i < largo; i++) {
Roman2[i] = Roman.charAt(i);
}
for (int i = 0; i < largo; i++) {
if (Roman2[i] == 'I') {
Roman3[i] = 1;
} else if (Roman2[i] == 'V') {
Roman3[i] = 5;
} else if (Roman2[i] == 'X') {
Roman3[i] = 10;
} else if (Roman2[i] == 'L') {
Roman3[i] = 50;
} else if (Roman2[i] == 'C') {
Roman3[i] = 100;
} else if (Roman2[i] == 'M') {
Roman3[i] = 1000;
}
}
int total = 0;
for (int m = 0; m < Roman3.length; m++) {
total += Roman3[m];
}
System.out.println("The Roman is equal to " + total);
}
You can check the previous digit.
For example, I added the condition that detects IV :
if (Roman2[i]=='I'){
Roman3[i]=1;
} else if (Roman2[i]=='V'){
Roman3[i]=5;
if (i>0 && Roman2[i-1]=='I') { // check for IV
Roman3[i]=4;
Roman3[i-1]=0;
}
} else if (Roman2[i]=='X'){
Roman3[i]=10;
} else if (Roman2[i]=='L'){
Roman3[i]=50;
} else if (Roman2[i]=='C'){
Roman3[i]=100;
} else if (Roman2[i]=='M'){
Roman3[i]=1000;
}
Define enum like below:
public enum RomanSymbol {
I(1), V(5), X(10), L(50), C(100), D(500), M(1000);
private final int value;
private RomanSymbol(final int value) {
this.value = value;
}
public int getValue() {
return this.value;
}
public int calculateIntEquivalent(final int lastArabicNumber, final int totalArabicResult) {
if (lastArabicNumber > this.value) {
return totalArabicResult - this.value;
} else {
return totalArabicResult + this.value;
}
}
}
And use it like RomanSymbol.I.getValue() which will return 1 and similarly for other.
So if you accept character from user, you can get the values as:
char symbol = 'I';//lets assume this is what user has entered.
RomanSymbol rSymbol = RomanSymbol.valueOf(String.valueOf(symbol));
int invalue = rSymbol.getValue();
And if you have string like IV, then you could calculate on something like for example:
int lastValue = rSymbol.calculateIntEquivalent(intValue, 0);
lastValue = rSymbol.calculateIntEquivalent(intValue, lastValue); //and so on

not getting correct numbers in for loop

Hy there. Below you can see sagment of my code. So lets go to the problem.
int i is not returning correct values and i cannot figure it out why.
LIST: [AGRFT, AGRFT, ARNES, ASCII, ASEAN, Aaron, Abdul, Abdul]
So for example. User inputs AS***, the program should return i is at 2. However i am getting i is at 0.
If i remember right it should go like this:
User_input= AS***
User_input.lenght() should be 5
first it should be user_input.charAt(0)=='*' NO
second it should be user_input.charAt(1)=='*' NO
third it should be user_input.charAt(2)=='*' YES
BREAK
i is at 2.
SO what am i missing?
I am getting 0.
Oh and also at
for(i=0; i < user_input.length();i++){
i am getting warning that i++ is Dead code?
if (dolzina == 5) {
for(i=0; i < user_input.length();i++){
if (user_input.charAt(i)=='*');
break;
}
System.out.println("i is at "+ i);
this is my full code for refrence. What it does it reads from txt file add wor
public class proba {
public static void main(String[] args) {
String izbira;
int dolzina=0;
int i=0;
Scanner in = new Scanner(System.in);
String user_input;
Scanner input = new Scanner(System.in);
String regex;
List<String> list5 = new ArrayList<String>();
int beseda;
String prefix = null;
try {
File file = new File("sort.txt");
FileReader fileReader = new FileReader(file);
BufferedReader bufferedReader = new BufferedReader(fileReader);
String vrstica;
while ((vrstica = bufferedReader.readLine()) != null) {
if (vrstica.length() == 5) {
list5.add(vrstica);
}
}
System.out.println(list5);
do{
do {
System.out.println("Enter lenght of word:");
if (in.hasNextInt()) {
dolzina = in.nextInt();
} else if (in.hasNextLine()) {
System.out.printf("Wrong entry!%n ",
in.nextLine());
}
} while (dolzina <= 0);
Collections.sort(list5);
System.out.println("Enter a word for unknown character enter * :");
user_input = input.nextLine();
System.out.println("Sorted list: [length: " + list5.size() + "]");
if (dolzina == 5) {
for(i=0; i < user_input.length();i++){
if (user_input.charAt(i)=='*');
break;
}
System.out.println("i je"+ i);
prefix=user_input.substring(0,i);
System.out.println(prefix);
int start=binarySearchfirst(list5,prefix);
int end=binarySearchlast(list5,prefix);
System.out.println(start);
System.out.println(end);
for (int b=start;b<=end;b++)
{
user_input = user_input.replace("*", ".");
String s = (String) list5.get(b);
if (s.matches(user_input))
System.out.println(s);
}
}
dolzina=-1;
System.out.println("Ponovni vnos (da/ne):");
Scanner inn= new Scanner (System.in);
izbira = inn.next();
}while (izbira.equalsIgnoreCase("da"));
bufferedReader.close();
} catch (IOException e) {
e.printStackTrace();
}}
public static int binarySearchfirst(List<String> integerList, String prefix) {
int low = 0;
int high = integerList.size() - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (integerList.get(mid).startsWith(prefix)) {
if (mid == 0 || !integerList.get(mid - 1).startsWith(prefix)) {
return mid;
} else {
high = mid - 1;
}
} else if (prefix.compareTo(integerList.get(mid)) > 0) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return low;
}
public static int binarySearchlast(List<String> integerList, String prefix) {
int low = 0;
int high = integerList.size()-1;
while (low <= high) {
int mid = (low+high)/2;
if (integerList.get(mid).startsWith(prefix)) {
if (mid == integerList.size()-1 || !integerList.get(mid+1).startsWith(prefix)) {
return mid;
}
else {
low = mid+1;
}
}
else if (prefix.compareTo(integerList.get(mid)) > 0) {
low = mid+1;
}
else {
high = mid-1;
}
}
return high;
}
}
You have an extra semi-colon after your if statement:
for(i=0; i < user_input.length();i++)
{ if (user_input.charAt(i)=='*');
break;
}
So the break is executed the first time through the loop no matter what. This is also why i++ is being reported as dead code...it's never being executed.

How to find Largest Prime Number, Smallest Factor except 1, Sum of Numbers

Herro, Um I'm not sure if this is how you use this site but uh lets get to it... So I need help on this project and I have to do this -
Input [][] Output
A - F -> Multiply by 3, Divide by 4
G - J -> Divide by 3 + 25
K - N -> Find Greatest Factor * 2
O - Q -> Find largest prime inclusive * 3
R - W -> Find Smallest Factor Except 1
X - Z -> Sum Numbers
So I was wondering if my first couple is correct, and need help on the empty space. So the Letter represents the number of the as in "Z" is the last letter so its 26, and "A" is the first so its 1. So if an responses... Thanks ! package Fun;
import java.util.Scanner;
public class Fun {
public static void main(String[] args) {
// TODO Auto-generated method stub
run();
}
public static void run()
{
input();
evaluateAlphabet();
evaluate();
}
public static void input()
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter a Letter, please");
x = sc.next();
}
public static String x = " ";
public static int temp = 0;
public static int answer = 0;
public static void evaluateAlphabet()
{
if(x.equals("A"))
{
temp = 1;
}
else if (x.equals("B"))
{
temp = 2;
}
else if (x.equals("C"))
{
temp = 3;
}
else if (x.equals("D"))
{
temp = 4;
}
else if (x.equals("E"))
{
temp = 5;
}
else if (x.equals("F"))
{
temp = 6;
}
else if (x.equals("G"))
{
temp = 7;
}
else if (x.equals("H"))
{
temp = 8;
}
else if (x.equals("I"))
{
temp = 9;
}
else if (x.equals("J"))
{
temp = 10;
}
else if (x.equals("K"))
{
temp = 11;
}
else if (x.equals("L"))
{
temp = 12;
}
else if (x.equals("M"))
{
temp = 13;
}
else if (x.equals("N"))
{
temp = 14;
}
else if (x.equals("O"))
{
temp = 15;
}
else if (x.equals("P"))
{
temp = 16;
}
else if (x.equals("Q"))
{
temp = 17;
}
else if (x.equals("R"))
{
temp = 18;
}
else if (x.equals("S"))
{
temp = 19;
}
else if (x.equals("T"))
{
temp = 20;
}
else if (x.equals("U"))
{
temp = 21;
}
else if (x.equals("V"))
{
temp = 22;
}
else if (x.equals("W"))
{
temp = 23;
}
else if (x.equals("X"))
{
temp = 24;
}
else if (x.equals("Y"))
{
temp = 25;
}
else if (x.equals("Z"))
{
temp = 26;
}
else if (x.equals("Qwerty"))
{
temp = 27;
}
}
public static void evaluate()
{
if(temp>=1 && temp<= 6)
{
answer = (temp * 3)/4;
System.out.println("Answer is " + answer);
}
else if(temp >= 7 && temp<= 10)
{
answer = (temp/3) + 25;
System.out.println("Answer is " + answer);
}
else if(temp >= 11 && temp<= 14)
{
}
else if(temp>=15 && temp<= 17)
{
for(int i = temp; i>0; i--)
{
for(int j = 2; j <=i/2 + 1; j++)
{
if(i%j==0)
{
break;
}
if(j==i/2 + 1)
{
answer = i * 3;
}
}
}
System.out.println("Answer is " + answer);
}
else if(temp>=18 && temp<= 23)
{
answer = temp;
}
else if(temp>= 24 && temp<=26)
answer = (answer * 12)%26;
System.out.println("Answer is " + answer);
}
}
-Corruption
Here is a better approach for char to int conversion:
Assuming the string has at least 1 character(check for its length), you can get the temp by doing:
temp = x.getCharAt(0) - 'A' + 1;
or, safety first:
temp = 0;
if (x.matches("^[A-Z]{1}$") {
temp = x.getCharAt(0) - 'A' + 1;
}
What's happening here? Every character has an ASCII code, an integer. So, when you have 2 chars and you try to get the distance between them, the result is an int. 'A' - 'A' = 0(that's why i added a + 1), 'B' - 'A' = 1 and so on.
For the if condition, I am using a RegExp. ^ means start of the input, [A-Z]{1} means one of A-Z, $ means the end of the input. So, if it's an A-Z, temp will get a value, anything else won't make it in the if and your temp will remain 0 so you can easily test if you've got an A-Z or not.
That's all for code review, I won't give you solutions, you must work harder, use Google. You won't enjoy and learn if I give you everything ready for a copy paste.

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