I have opened an account for Ridit, one of 7-years-old students learning Java at SPOJ. The first task i gave to him was PALIN -The Next Palindrome. Here is the link to this problem- PALIN- The next Palindrome- SPOJAfter i explained it to him, he was able to solve it mostly except removing the leading zeros, which i did. Following is his solution of the problem -
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
Scanner in = new Scanner(System.in);
int t = Integer.parseInt(in.nextLine());
String[] numbersInString = new String[t];
for (int i = 0; i <t; i++) {
String str = in.nextLine();
numbersInString[i] = removeLeadingZeros(str);
}
for (int i = 0 ; i<t; i++) {
int K = Integer.parseInt(numbersInString[i]);
int answer = findTheNextPalindrome(K);
System.out.println(answer);
}
}catch(Exception e) {
return;
}
}
static boolean isPalindrome(int x) {
String str = Integer.toString(x);
int length = str.length();
StringBuffer strBuff = new StringBuffer();
for(int i = length - 1;i>=0;i--) {
char ch = str.charAt(i);
strBuff.append(ch);
}
String str1 = strBuff.toString();
if(str.equals(str1)) {
return true;
}
return false;
}
static int findTheNextPalindrome(int K) {
for(int i = K+1;i<9999999; i++) {
if(isPalindrome(i) == true) {
return i;
}
}
return -1;
}
static String removeLeadingZeros(String str) {
String retString = str;
if(str.charAt(0) != '0') {
return retString;
}
return removeLeadingZeros(str.substring(1));
}
}
It is giving correct answer in Eclipse on his computer, but it is failing in SPOJ. If someone helps this little boy in his first submission, it will definitely make him very happy. I couldn't find any problem with this solution... Thank you in advance...
This might be helpful
import java.io.IOException;
import java.util.Scanner;
public class ThenNextPallindrom2 {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = 0;
Scanner sc = new Scanner(System.in);
if(sc.hasNextInt()) {
t = sc.nextInt();
}
sc.nextLine();
int[] arr, arr2;
while(t > 0) {
t--;
String s = sc.nextLine();
arr = getStringToNumArray(s);
if(all9(arr)) {
arr2 = new int[arr.length + 1];
arr2[0] = 1;
for(int i=0;i<arr.length;i++) {
arr2[i+1] = 0;
}
arr2[arr2.length -1] = 1;
arr = arr2;
} else{
int mid = arr.length/ 2;
int left = mid-1;
int right = arr.length % 2 == 1 ? mid + 1 : mid;
boolean left_small = false;
while(left >= 0 && arr[left] == arr[right]) {
left--;
right++;
}
if(left < 0 || arr[left] < arr[right]) left_small = true;
if(!left_small) {
while(left >= 0) {
arr[right++] = arr[left--];
}
} else {
mid = arr.length/ 2;
left = mid-1;
int carry = 1;
if(arr.length % 2 == 0) {
right = mid;
} else {
arr[mid] += carry;
carry = arr[mid]/10;
arr[mid] %= 10;
right = mid + 1;
}
while(left >= 0) {
arr[left] += carry;
carry = arr[left] / 10;
arr[left] %= 10;
arr[right++] = arr[left--];
}
}
}
printArray(arr);
}
}
public static boolean all9(int[] arr) {
for(int i=0;i<arr.length;i++) {
if(arr[i] != 9)return false;
}
return true;
}
public static void printArray(int[] arr) {
for(int i=0;i<arr.length;i++) {
System.out.print(arr[i]);
}
System.out.println();
}
public static int[] getStringToNumArray(String s) {
int[] arr = new int[s.length()];
for(int i=0; i<s.length();i++) {
arr[i] = Integer.parseInt(String.valueOf(s.charAt(i)));
}
return arr;
}
}
I am making SOS game with "stack".No other data structures such as normal array, string etc.
Gameboard size 3*6 and I am using random for determining letters.
But I don't know is the game over or not.
How can I control is there any horizontal or vertical sos at the console
Here is my code(I am beginner)
public static void main(String[] args) {
Random rnd = new Random();
Stack S1 = new Stack(6);
Stack S2 = new Stack(6);
Stack S3 = new Stack(6);
Stack ts1 = new Stack(6);
Stack ts2 = new Stack(6);
Stack ts3 = new Stack(6);
int Round = 0;
int satir;
boolean winner = false;
// for S1
int i = 0;
while (i != 6) { // 6time
i++;
// System.out.println(i);
if (rnd.nextBoolean()) {
S1.push("S");
} else {
S1.push("O");
}
}
// for S2
i = 0;
while (i != 6) {
i++;
if (rnd.nextBoolean()) {
S2.push("S");
} else {
S2.push("O");
}
}
// for S3
i = 0;
while (i != 6) {
i++;
if (rnd.nextBoolean()) {
S3.push("S");
} else {
S3.push("O");
}
}
int a = 0, b = 0, c = 0;
int tempa = a;
int tempb = b;
int tempc = c;
while (!winner) {
System.out.println("User" + ((Round % 2) + 1) + ":"); // User1:
// User2:
satir = rnd.nextInt(3) + 1;
if (satir == 1 && a != 6) {
a++;
}
if (satir == 2 && b != 6) {
b++;
}
if (satir == 3 && c != 6) {
c++;
}
tempa = a;
tempb = b;
tempc = c;
System.out.print("S1 ");
while (a > 0) { // S1 in icini yazdir bosalt ts1e kaydet
System.out.print(S1.peek() + " ");
ts1.push(S1.pop());
a--;
}
a = tempa;
while (!ts1.isEmpty()) { // S1i tekrar doldur
S1.push(ts1.pop());
}
System.out.println();
System.out.print("S2 ");
while (b > 0) { // S2 in icini yazdir bosalt ts1e kaydet
System.out.print(S2.peek() + " ");
ts2.push(S2.pop());
b--;
}
b = tempb;
while (!ts2.isEmpty()) { // S2i tekrar doldur
S2.push(ts2.pop());
}
System.out.println();
System.out.print("S3 ");
while (c > 0) { // S3 in icini yazdir bosalt ts1e kaydet
System.out.print(S3.peek() + " ");
ts3.push(S3.pop());
c--;
}
c = tempc;
while (!ts3.isEmpty()) { // S3i tekrar doldur
S3.push(ts3.pop());
}
System.out.println();
// check if there is sos winner=true;
if (a > 2) { //check S1 Horizontal
for (int yatay1 = 0; yatay1 < a; yatay1++) {
if (S1.peek().equals("S")) {
ts1.push(S1.pop());
if (S1.peek().equals("O")) {
ts1.push(S1.pop());
if (S1.peek().equals("S")) {
System.out.println("SOS-wow");
winner = true;
} else {
while (!ts1.isEmpty()) {
S1.push(ts1.pop());
}
}
} else {
S1.push(ts1.pop());
}
}
}
}
Round++;
} // end of loop
}
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
Java won't stop reading from input.
I understand that maybe this while loop might have something to do with it:
while(input.hasMoreTokens());
{
array1[counter] = input.nextToken();
counter++;
}
But I don't see why the loop should be a problem because I am already calling .nextToken() which should advance the token.
Here's the full source code:
import java.io.*;
import java.util.*;
class HelloWorld
{
static String ReadLn (int maxLg) // utility function to read from stdin
{
byte lin[] = new byte [maxLg];
int lg = 0, car = -1;
String line = "";
try
{
while (lg < maxLg)
{
car = System.in.read();
if ((car < 0) || (car == '\n')) break;
lin [lg++] += car;
}
}
catch (IOException e)
{
return (null);
}
if ((car < 0) && (lg == 0)) return (null); // eof
return (new String (lin, 0, lg));
}
public static void main (String args[]) // entry point from OS
{
HelloWorld myWork = new HelloWorld(); // create a dinamic instance
myWork.Begin(); // the true entry point
}
void Begin()
{
String idata;
StringTokenizer input;
while ((idata = HelloWorld.ReadLn (255)) != null)
{
input = new StringTokenizer (idata);
String[] array1 = {};
int counter = 0;
while(input.hasMoreTokens());
{
array1[counter] = input.nextToken();
counter++;
}
int[] array2 = {};
for(int a = 0; a < array1.length; a++)
{
array2[a] = Integer.parseInt(array1[a]);
}
int[] array3 = {};
for(int b = 0; b < array2.length; b++)
{
if ( array2[b] != 42)
{
array3[b] = array2[b];
}
else
{
break;
}
}
String string = "";
for( int c = 0; c < array3.length; c++)
{
if( c < array3.length - 1)
{
string += array3[c] + "\n";
}
else
{
string += array3[c];
}
}
System.out.println(string);
}
}
}
You have a stray semicolon at the end of the while:
while(input.hasMoreTokens());
^ REMOVE THIS
I am struggling with a "find supersequence" algorithm.
The input is for set of strings
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
the result would be properly aligned set of strings (and next step should be merge)
String E = "ca ag cca cc ta cat c a";
String F = "c gag ccat ccgtaaa g tt g";
String G = " aga acc tgc taaatgc t a ga";
Thank you for any advice (I am sitting on this task for more than a day)
after merge the superstring would be
cagagaccatgccgtaaatgcattacga
The definition of supersequence in "this case" would be something like
The string R is contained in supersequence S if and only if all characters in a string R are present in supersequence S in the order in which they occur in the input sequence R.
The "solution" i tried (and again its the wrong way of doing it) is:
public class Solution4
{
static boolean[][] map = null;
static int size = 0;
public static void main(String[] args)
{
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
Stack data = new Stack();
data.push(A);
data.push(B);
data.push(C);
Stack clone1 = data.clone();
Stack clone2 = data.clone();
int length = 26;
size = max_size(data);
System.out.println(size+" "+length);
map = new boolean[26][size];
char[] result = new char[size];
HashSet<String> chunks = new HashSet<String>();
while(!clone1.isEmpty())
{
String a = clone1.pop();
char[] residue = make_residue(a);
System.out.println("---");
System.out.println("OLD : "+a);
System.out.println("RESIDUE : "+String.valueOf(residue));
String[] r = String.valueOf(residue).split(" ");
for(int i=0; i<r.length; i++)
{
if(r[i].equals(" ")) continue;
//chunks.add(spaces.substring(0,i)+r[i]);
chunks.add(r[i]);
}
}
for(String chunk : chunks)
{
System.out.println("CHUNK : "+chunk);
}
}
static char[] make_residue(String candidate)
{
char[] result = new char[size];
for(int i=0; i<candidate.length(); i++)
{
int pos = find_position_for(candidate.charAt(i),i);
for(int j=i; j<pos; j++) result[j]=' ';
if(pos==-1) result[candidate.length()-1] = candidate.charAt(i);
else result[pos] = candidate.charAt(i);
}
return result;
}
static int find_position_for(char character, int offset)
{
character-=((int)'a');
for(int i=offset; i<size; i++)
{
// System.out.println("checking "+String.valueOf((char)(character+((int)'a')))+" at "+i);
if(!map[character][i])
{
map[character][i]=true;
return i;
}
}
return -1;
}
static String move_right(String a, int from)
{
return a.substring(0, from)+" "+a.substring(from);
}
static boolean taken(int character, int position)
{ return map[character][position]; }
static void take(char character, int position)
{
//System.out.println("taking "+String.valueOf(character)+" at "+position+" (char_index-"+(character-((int)'a'))+")");
map[character-((int)'a')][position]=true;
}
static int max_size(Stack stack)
{
int max=0;
while(!stack.isEmpty())
{
String s = stack.pop();
if(s.length()>max) max=s.length();
}
return max;
}
}
Finding any common supersequence is not a difficult task:
In your example possible solution would be something like:
public class SuperSequenceTest {
public static void main(String[] args) {
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
int iA = 0;
int iB = 0;
int iC = 0;
char[] a = A.toCharArray();
char[] b = B.toCharArray();
char[] c = C.toCharArray();
StringBuilder sb = new StringBuilder();
while (iA < a.length || iB < b.length || iC < c.length) {
if (iA < a.length && iB < b.length && iC < c.length && (a[iA] == b[iB]) && (a[iA] == c[iC])) {
sb.append(a[iA]);
iA++;
iB++;
iC++;
}
else if (iA < a.length && iB < b.length && a[iA] == b[iB]) {
sb.append(a[iA]);
iA++;
iB++;
}
else if (iA < a.length && iC < c.length && a[iA] == c[iC]) {
sb.append(a[iA]);
iA++;
iC++;
}
else if (iB < b.length && iC < c.length && b[iB] == c[iC]) {
sb.append(b[iB]);
iB++;
iC++;
} else {
if (iC < c.length) {
sb.append(c[iC]);
iC++;
}
else if (iB < b.length) {
sb.append(b[iB]);
iB++;
} else if (iA < a.length) {
sb.append(a[iA]);
iA++;
}
}
}
System.out.println("SUPERSEQUENCE " + sb.toString());
}
}
However the real problem to solve is to find the solution for the known problem of Shortest Common Supersequence http://en.wikipedia.org/wiki/Shortest_common_supersequence,
which is not that easy.
There is a lot of researches which concern the topic.
See for instance:
http://www.csd.uwo.ca/~lila/pdfs/Towards%20a%20DNA%20solution%20to%20the%20Shortest%20Common%20Superstring%20Problem.pdf
http://www.ncbi.nlm.nih.gov/pubmed/14534185
You can try finding the shortest combination like this
static final char[] CHARS = "acgt".toCharArray();
public static void main(String[] ignored) {
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
String expected = "cagagaccatgccgtaaatgcattacga";
List<String> ABC = new Combination(A, B, C).findShortest();
System.out.println("expected: " + expected.length());
System.out.println("Merged: " + ABC.get(0).length() + " " + ABC);
}
static class Combination {
int shortest = Integer.MAX_VALUE;
List<String> shortestStr = new ArrayList<>();
char[][] chars;
int[] pos;
int count = 0;
Combination(String... strs) {
chars = new char[strs.length][];
pos = new int[strs.length];
for (int i = 0; i < strs.length; i++) {
chars[i] = strs[i].toCharArray();
}
}
public List<String> findShortest() {
findShortest0(new StringBuilder(), pos);
return shortestStr;
}
private void findShortest0(StringBuilder sb, int[] pos) {
if (allDone(pos)) {
if (sb.length() < shortest) {
shortestStr.clear();
shortest = sb.length();
}
if (sb.length() <= shortest)
shortestStr.add(sb.toString());
count++;
if (++count % 100 == 1)
System.out.println("Searched " + count + " shortest " + shortest);
return;
}
if (sb.length() + maxLeft(pos) > shortest)
return;
int[] pos2 = new int[pos.length];
int i = sb.length();
sb.append(' ');
for (char c : CHARS) {
if (!tryChar(pos, pos2, c)) continue;
sb.setCharAt(i, c);
findShortest0(sb, pos2);
}
sb.setLength(i);
}
private int maxLeft(int[] pos) {
int maxLeft = 0;
for (int i = 0; i < pos.length; i++) {
int left = chars[i].length - pos[i];
if (left > maxLeft)
maxLeft = left;
}
return maxLeft;
}
private boolean allDone(int[] pos) {
for (int i = 0; i < chars.length; i++)
if (pos[i] < chars[i].length)
return false;
return true;
}
private boolean tryChar(int[] pos, int[] pos2, char c) {
boolean matched = false;
for (int i = 0; i < chars.length; i++) {
pos2[i] = pos[i];
if (pos[i] >= chars[i].length) continue;
if (chars[i][pos[i]] == c) {
pos2[i]++;
matched = true;
}
}
return matched;
}
}
prints many solutions which are shorter than the one suggested.
expected: 28
Merged: 27 [acgaagccatccgctaaatgctatcga, acgaagccatccgctaaatgctatgca, acgaagccatccgctaacagtgctaga, acgaagccatccgctaacatgctatga, acgaagccatccgctaacatgcttaga, acgaagccatccgctaacatgtctaga, acgaagccatccgctacaagtgctaga, acgaagccatccgctacaatgctatga, acgaagccatccgctacaatgcttaga, acgaagccatccgctacaatgtctaga, acgaagccatcgcgtaaatgctatcga, acgaagccatcgcgtaaatgctatgca, acgaagccatcgcgtaacagtgctaga, acgaagccatcgcgtaacatgctatga, acgaagccatcgcgtaacatgcttaga, acgaagccatcgcgtaacatgtctaga, acgaagccatcgcgtacaagtgctaga, acgaagccatcgcgtacaatgctatga, acgaagccatcgcgtacaatgcttaga, acgaagccatcgcgtacaatgtctaga, acgaagccatgccgtaaatgctatcga, acgaagccatgccgtaaatgctatgca, acgaagccatgccgtaacagtgctaga, acgaagccatgccgtaacatgctatga, acgaagccatgccgtaacatgcttaga, acgaagccatgccgtaacatgtctaga, acgaagccatgccgtacaagtgctaga, acgaagccatgccgtacaatgctatga, acgaagccatgccgtacaatgcttaga, acgaagccatgccgtacaatgtctaga, cagaagccatccgctaaatgctatcga, cagaagccatccgctaaatgctatgca, cagaagccatccgctaacagtgctaga, cagaagccatccgctaacatgctatga, cagaagccatccgctaacatgcttaga, cagaagccatccgctaacatgtctaga, cagaagccatccgctacaagtgctaga, cagaagccatccgctacaatgctatga, cagaagccatccgctacaatgcttaga, cagaagccatccgctacaatgtctaga, cagaagccatcgcgtaaatgctatcga, cagaagccatcgcgtaaatgctatgca, cagaagccatcgcgtaacagtgctaga, cagaagccatcgcgtaacatgctatga, cagaagccatcgcgtaacatgcttaga, cagaagccatcgcgtaacatgtctaga, cagaagccatcgcgtacaagtgctaga, cagaagccatcgcgtacaatgctatga, cagaagccatcgcgtacaatgcttaga, cagaagccatcgcgtacaatgtctaga, cagaagccatgccgtaaatgctatcga, cagaagccatgccgtaaatgctatgca, cagaagccatgccgtaacagtgctaga, cagaagccatgccgtaacatgctatga, cagaagccatgccgtaacatgcttaga, cagaagccatgccgtaacatgtctaga, cagaagccatgccgtacaagtgctaga, cagaagccatgccgtacaatgctatga, cagaagccatgccgtacaatgcttaga, cagaagccatgccgtacaatgtctaga, cagagaccatccgctaaatgctatcga, cagagaccatccgctaaatgctatgca, cagagaccatccgctaacagtgctaga, cagagaccatccgctaacatgctatga, cagagaccatccgctaacatgcttaga, cagagaccatccgctaacatgtctaga, cagagaccatccgctacaagtgctaga, cagagaccatccgctacaatgctatga, cagagaccatccgctacaatgcttaga, cagagaccatccgctacaatgtctaga, cagagaccatcgcgtaaatgctatcga, cagagaccatcgcgtaaatgctatgca, cagagaccatcgcgtaacagtgctaga, cagagaccatcgcgtaacatgctatga, cagagaccatcgcgtaacatgcttaga, cagagaccatcgcgtaacatgtctaga, cagagaccatcgcgtacaagtgctaga, cagagaccatcgcgtacaatgctatga, cagagaccatcgcgtacaatgcttaga, cagagaccatcgcgtacaatgtctaga, cagagaccatgccgtaaatgctatcga, cagagaccatgccgtaaatgctatgca, cagagaccatgccgtaacagtgctaga, cagagaccatgccgtaacatgctatga, cagagaccatgccgtaacatgcttaga, cagagaccatgccgtaacatgtctaga, cagagaccatgccgtacaagtgctaga, cagagaccatgccgtacaatgctatga, cagagaccatgccgtacaatgcttaga, cagagaccatgccgtacaatgtctaga, cagagccatcctagctaaagtgctaga, cagagccatcctagctaaatgctatga, cagagccatcctagctaaatgcttaga, cagagccatcctagctaaatgtctaga, cagagccatcctgactaaagtgctaga, cagagccatcctgactaaatgctatga, cagagccatcctgactaaatgcttaga, cagagccatcctgactaaatgtctaga, cagagccatcctgctaaatgctatcga, cagagccatcctgctaaatgctatgca, cagagccatcctgctaacagtgctaga, cagagccatcctgctaacatgctatga, cagagccatcctgctaacatgcttaga, cagagccatcctgctaacatgtctaga, cagagccatcctgctacaagtgctaga, cagagccatcctgctacaatgctatga, cagagccatcctgctacaatgcttaga, cagagccatcctgctacaatgtctaga]