Since last night I have been working very hard trying to write a solution to the problem "0-1 Sequences" (https://open.kattis.com/problems/sequences) on Kattis. At this point, I am at my wits end. I keep getting the wrong answer on the third test case, and I've tried everything I can possibly think of to fix my code. I'm fairly certain that the formula I am using to compute the result is correct, and I haven't been able to find an example where it gives the wrong answer. I suspect the issue is something to do with how I am calculating the modulus, but I've written and rewritten the relevant parts of the code and I am still failing each time. Below is my code (sorry in advance, I've been working like a madman with not much regard for writing "nice" code).
import java.util.*;
import java.util.ArrayList;
public class oneZeroSequences {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
//leading zero's have no effect on the number of inversions,
//and we therefore get rid of them all right off the bat.
while(s.charAt(0)=='0'){
s=s.substring(1);
if(s.length()==0){
System.out.println(0);
System.exit(0);
}
}
System.out.println(findSum(s));
System.exit(0);
}
public static long findSum(String s) {
int m=1000000007;
long res = 0;
//if passed an empty string, the sum of inversions is obviously 0
if(s.length()==0){
return 0;
}
while(s.charAt(0)=='0'){
s=s.substring(1);
if(s.length()==0){
return 0;
}
}
//if the first character of the string is a question mark, we handle the case
//where that question mark is 1 first, and then remove the question mark since
//leading 0's have no effect on inversions.
while(s.charAt(0)=='?'){
res+=findSum("1"+s.substring(1))%m;
s=s.substring(1);
if(s.length()==0){
return res%m;
}
}
ArrayList<Integer> qArray = new ArrayList<Integer>();
int start=0;
int found;
//here we construct an array of the indices of question marks in the string
while(true){
found=s.indexOf('?', start);
if(found==-1){
break;
}
qArray.add(found);
start=found+1;
}
ArrayList<Integer> zeroArray = new ArrayList<Integer>();
start=0;
found=0;
//here we construct an array of the indices of 0's in the string
while(true){
found=s.indexOf('0', start);
if(found==-1){
break;
}
zeroArray.add(found);
start=found+1;
}
long k = qArray.size();
long a = zeroArray.size();
long zeroSum = sumArrayList(zeroArray);
long qSum = sumArrayList(qArray);
Double dres = 0.0;
//the formula for sum of inversions is as follows:
//2^k*(zeroSum) + 2^(k-1)*(qSum) - (a^2-a)*2^(k-1) - (2a-1)*2^(k-2) - (k^2+k)*2^(k-3)
//the following lines of code apply this formula, taking the modulus at each step.
if(k>=3){
res += (expBySquaringMod(2, k)*(zeroSum%m))%m;
//res=(res%m+m)%m;
res +=((expBySquaringMod(2, k-1))*(qSum%m))%m;
//res=(res%m+m)%m;
res -= ((((((a%m)*(a%m))%m)-a%m)%m)*expBySquaringMod(2,k-1))%m;
//res=(res%m+m)%m;
res -= (((((2*a-1)%m)*(k%m))%m)*expBySquaringMod(2,k-2))%m;
//res=(res%m+m)%m;
res-= (((((k%m)*(k%m))%m+k%m)%m)*expBySquaringMod(2, k-3))%m;
//res=(res%m+m)%m;
return res%m;
}
//the expBySquaringMod function will give incorrect
//results for negative exponents, hence the following code.
dres += ((Math.pow(2,k))*(zeroSum%m))%m;
dres=(dres%m+m)%m;
dres +=((Math.pow(2,k-1))*(qSum%m))%m;
dres=(dres%m+m)%m;
dres -= (((((a%m)*(a%m))%m-a%m)%m)*(Math.pow(2,k-1)))%m;
dres=(dres%m+m)%m;
dres -=((((2*a-1)%m*(k%m)%m)*(Math.pow(2,k-2))))%m;
dres=(dres%m+m)%m;
dres-=((((k%m)*(k%m)%m+k%m))%m*(Math.pow(2,k-3)))%m;
dres=(dres%m+m)%m;
res+=dres.intValue()%m;
return res%m;
}
public static int sumArrayList(ArrayList<Integer> list) {
int res = 0;
for(int d : list) {
res+=d;
}
return res;
}
//this function was written to deal with the massive powers of two required for the solution
public static long expBySquaringMod(int base, long exp) {
int m=1000000007;
long res=1;
if(exp%2==0) {
for(int i=0; i<exp/2;i++) {
res=(res*base*base)%m;
}
} else{
res=base;
for(int i=0; i<exp/2;i++) {
res=(res*base*base)%m;
}
}
return res;
}
}
Any help will be greatly appreciated. This problem has been causing me a lot of distress.
Related
I am trying to write a program that will receive a function as a String and solve it. For ex. "5*5+2/2-8+5*5-2" should return 41
I wrote the code for multiplication and divisions and it works perfectly:
public class Solver
{
public static void operationS(String m)
{
ArrayList<String> z = new ArrayList<String>();
char e= ' ';
String x= " ";
for (int i =0; i<m.length();i++)
{
e= m.charAt(i);
x= Character.toString(e);
z.add(x);
}
for (int i =0; i<z.size();i++)
{
System.out.print(z.get(i));
}
other(z);
}
public static void other(ArrayList<String> j)
{
int n1=0;
int n2=0;
int f=0;
String n= " ";
for (int m=0; m<j.size();m++)
{
if ((j.get(m)).equals("*"))
{
n1 = Integer.parseInt(j.get(m-1));
n2 = Integer.parseInt(j.get(m+1));
f= n1*n2;
n = Integer.toString(f);
j.set(m,n);
j.remove(m+1);
j.remove(m-1);
m=0;
}
for (int e=0; e<j.size();e++)
{
if ((j.get(e)).equals("/"))
{
n1 = Integer.parseInt(j.get(e-1));
n2 = Integer.parseInt(j.get(e+1));
f= n1/n2;
n = Integer.toString(f);
j.set(e,n);
j.remove(e+1);
j.remove(e-1);
e=0;
}
}
}
System.out.println();
for (int i1 =0; i1<j.size();i1++)
{
System.out.print(j.get(i1)+",");
}
However, for adding and subtracting, since there isnt an order for adding and subtracting, just whichever comes first, I wrote the following:
int x1=0;
int x2=0;
int x3=0;
String z = " ";
for (int g=0; g<j.size();g++)
{
if ((j.get(g)).equals("+"))
{
x1= Integer.parseInt(j.get(g-1));
x2= Integer.parseInt(j.get(g+1));
x3= x1+x2;
z = Integer.toString(x3);
j.set(g,z);
j.remove(g+1);
j.remove(g-1);
g=0;
}
g=0;
if ((j.get(g)).equals("-"))
{
x1= Integer.parseInt(j.get(g-1));
x2= Integer.parseInt(j.get(g+1));
x3= x1-x2;
z = Integer.toString(x3);
j.set(g,z);
j.remove(g+1);
j.remove(g-1);
g=0;
}
g=0;
}
System.out.println();
for (int i1 =0; i1<j.size();i1++)
{
System.out.print(j.get(i1)+",");
}
After this, it prints:
25,+,1,-,8,+,25,–,2,
. What am I doing wrong? Multiplication and dividing seem to be working perfectly
You have 2 problems:
1) g=0; statements after if and else blocks will make you go into an infinite loop.
2) From the output you gave, the first minus (-) is Unicode character HYPHEN-MINUS (U+002D), while the second minus (–) is Unicode character EN DASH (U+2013), so (j.get(g)).equals("-") fails for the second minus as they are not equal.
Going for an answer that doesn't help with your exact specific problem, but that hopefully helps you much further than that.
On a first glance, there are various problems with your code:
Your are using super-short variable names all over the place. That saves you maybe 1 minute of typing overall; and costs you 5, 10, x minutes every time you read your code; or show it to other people. So: dont do that. Use names that say what the thing behind that name is about.
You are using a lot of low-level code. You use a "couting-for" loop to iterate a list (called j, that is really really horrible!) for example. Meaning: you make your code much more complicated to read than it ought to be.
In that way, it looks like nobody told you so far, but the idea of code is: it should be easy to read and understand. Probably you dont get grades for that, but believe me: in the long run, learning to write readable code is a super-important skill. If that got you curious, see if you can get a hand on "Clean code" by Robert Martin. And study that book. Then study it again. And again.
But the real problem is your approach to solve this problem. As I assume: this is some part of study assignment. And the next step will be that you don't have simple expressions such as "1+2*3"; but that you are asked to deal with something like "sqrt(2) + 3" and so on. Then you will be asked to add variables, etc. And then your whole approach breaks apart. Because your simple string operations won't do it any more.
In that sense: you should look into this question, and carefully study the 2nd answer by Boann to understand how to create a parser that dissects your input string into expressions that are then evaluated. Your code does both things "together"; thus making it super-hard to enhance the provided functionality.
You can use the built-in Javascript engine
public static void main(String[] args) throws Exception{
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
String code = "5*5+2/2-8+5*5-2";
System.out.println(engine.eval(code));
}
Primarily Don't Repeat Yourself (the DRY principle). And use abstractions (full names, extracting methods when sensible). Static methods are a bit cumbersome, when using several methods. Here it is handy to use separate methods.
Maybe you want something like:
Solver solver = new Solver();
List<String> expr = solver.expression("5*5+2/2-8+5*5-2");
String result = solver.solve(expr);
A more abstract Solver class would do:
class Solver {
List<String> expression(String expr) {
String[] args = expr.split("\\b");
List<String> result = new ArrayList<>();
Collections.addAll(result, args);
return result;
}
String solve(List<String> args) {
solveBinaryOps(args, "[*/]");
solveBinaryOps(args, "[-+]");
return args.stream().collect(Collectors.joining(""));
}
The above solveBinaryOps receives a regular expression pattern or alternatively simply in some form the operators you want to tackle.
It takes care of operator precedence.
private void solveBinaryOps(List<String> args, String opPattern) {
for (int i = 1; i + 1 < args.length; ++i) {
if (args.get(i).matches(opPattern)) {
String value = evalBinaryOp(args.get(i - 1), args.get(i), args.get(i + 1));
args.set(i, value);
args.remove(i + 1);
args.remove(i - 1);
--i; // Continue from here.
}
}
}
private String evalBinaryOp(String lhs, String op, String rhs) {
int x = Integer.parseInt(lhs);
int y = Integer.parseInt(rhs);
int z = 0;
switch (op) {
case "*":
z = x * y;
break;
case "/":
z = x / y;
break;
case "+":
z = x + y;
break;
case "-":
z = x - y;
break;
}
return Integer.toString(z);
}
}
The above can be improved at several points. But it is readable, and rewritable.
public class Solver {
public static void main(String args[]) {
operation("5+2*5-6/2+1+5*12/3");
}
public static void operation(String m) {
ArrayList<Object> expressions = new ArrayList<Object>();
String e;
String x = "";
for (int i = 0; i < m.length(); i++) {
e = m.substring(i, i + 1);
if (!(e.equals("*") || e.equals("/") || e.equals("+") || e
.equals("-"))) {
x += e;
continue;
} else {
if (!x.equals("") && x.matches("[0-9]+")) {
int oper = Integer.parseInt(x);
expressions.add(oper);
expressions.add(m.charAt(i));
x = "";
}
}
}
if (!x.equals("") && x.matches("[0-9]+")) {
int oper = Integer.parseInt(x);
expressions.add(oper);
x = "";
}
for (int i = 0; i < expressions.size(); i++) {
System.out.println(expressions.get(i));
}
evaluateExpression(expressions);
}
public static void evaluateExpression(ArrayList<Object> exp) {
//Considering priorities we calculate * and / first and put them in a list mulDivList
ArrayList<Object> mulDivList=new ArrayList<Object>();
for (int i = 0; i < exp.size(); i++) {
if (exp.get(i) instanceof Character) {
if ((exp.get(i)).equals('*')) {
int tempRes = (int) exp.get(i - 1) * (int) exp.get(i + 1);
exp.set(i - 1, null);
exp.set(i, null);
exp.set(i + 1, tempRes);
}
else if ((exp.get(i)).equals('/')) {
int tempRes = (int) exp.get(i - 1) / (int) exp.get(i + 1);
exp.set(i - 1, null);
exp.set(i, null);
exp.set(i + 1, tempRes);
}
}
}
//Create new list with only + and - operations
for(int i=0;i<exp.size();i++)
{
if(exp.get(i)!=null)
mulDivList.add(exp.get(i));
}
//Calculate + and - .
for(int i=0;i<mulDivList.size();i++)
{
if ((mulDivList.get(i)).equals('+')) {
int tempRes = (int) mulDivList.get(i - 1) + (int) mulDivList.get(i + 1);
mulDivList.set(i - 1, null);
mulDivList.set(i, null);
mulDivList.set(i + 1, tempRes);
}
else if ((mulDivList.get(i)).equals('-')) {
int tempRes = (int) mulDivList.get(i - 1) - (int) mulDivList.get(i + 1);
mulDivList.set(i - 1, null);
mulDivList.set(i, null);
mulDivList.set(i + 1, tempRes);
}
}
System.out.println("Result is : " + mulDivList.get(mulDivList.size() - 1));
}
}
Having a String representation of a number(no decimals), what's the best way to convert it to either one of java.lang.Integer or java.lang.Long or java.math.BigInteger? The only condition is that the converted type should be of minimal datatype required to hold the number.
I've this current implementation that works fine, but I would like to know if there's a better code without exception handling.
package com.stackoverflow.programmer;
import java.math.BigInteger;
public class Test {
public static void main(String[] args) {
String number = "-12121111111111111";
Number numberObject = null;
try {
numberObject = Integer.valueOf(number);
} catch (NumberFormatException nfe) {
System.out.println("Number will not fit into Integer type. Trying Long...");
try {
numberObject = Long.valueOf(number);
} catch (NumberFormatException nfeb) {
System.out.println("Number will not fit into Long type. Trying BigInteger...");
numberObject = new BigInteger(number);
}
}
System.out.println(numberObject.getClass() + " : "
+ numberObject.toString());
}
}
From what you said, here is what I would have done:
import java.math.BigInteger;
import java.util.Arrays;
import java.util.List;
public class TestSO09_39463168_StringToMinimalNumber {
public static void main(String[] args) {
List<String> strNumbers = Arrays.asList("0", //int
"123", //int
"-456", //int
"2147483700", // Long
"-2147483700", // Long
"9223372036854775900", //BigInt
"-9223372036854775900" //BigInt
);
for(String strNumber : strNumbers){
Number number = stringToMinimalNumber(strNumber);
System.out.println("The string '"+strNumber+"' is a "+number.getClass());
}
}
public static Number stringToMinimalNumber(String s){
BigInteger tempNumber = new BigInteger(s);
if(tempNumber.compareTo(BigInteger.valueOf(Long.MAX_VALUE)) > 0 || tempNumber.compareTo(BigInteger.valueOf(Long.MIN_VALUE)) < 0){
return tempNumber;
} else if(tempNumber.compareTo(BigInteger.valueOf(Integer.MAX_VALUE)) > 0 || tempNumber.compareTo(BigInteger.valueOf(Integer.MIN_VALUE)) < 0){
return tempNumber.longValue(); //Autobox to Long
} else {
return tempNumber.intValue(); //Autobox to Integer
}
}
}
You must use a temporary BigInteger, or else you'll end up with lazarov's solution, which is correct, but you can't really do something like that for reason mentionned in the comments.
Anyway, every BigInteger (the ones that are not returned) will be garbage collected. As for autoboxing, I don't think it's that of a bad thing. You could also make "BigInteger.valueOf(Long.MAX_VALUE))" as a constant. Maybe the compiler or the JVM will do this on its own.
I'm not really sure of how efficient it is, and using only BigInteger might be a good idea (as Spotted did), because I serioulsy doubt it would really improve the rest of your code to use the right size, and it might even be error prone if you try to use these Numbers with each other ... But again, it all depend on what you need. (and yes, using Exception as flow control is a really bad idea, but you can add a try catch on the BigInteger tempNumber = new BigInteger(s); to throw your own exception if s is not a number at all)
For recreational purpose, I have made the solution without using a BigInteger, and only with String parsing (this is still not what I recommand to do, but it was fun :)
public static final String INT_MAX_VALUE = "2147483647";
public static final String LONG_MAX_VALUE = "9223372036854775807";
public static Number stringToMinimalNumberWithoutBigInteger(String numberStr){
//Removing the minus sign to test the value
String s = (numberStr.startsWith("-") ? numberStr.substring(1,numberStr.length()) : numberStr);
if(compareStringNumber(s, LONG_MAX_VALUE) > 0){
return new BigInteger(numberStr);
} else if(compareStringNumber(s, INT_MAX_VALUE) > 0){
return new Long(numberStr);
} else {
return new Integer(numberStr);
}
}
//return postive if a > b, negative if a < b, 0 if equals;
private static int compareStringNumber(String a, String b){
if(a.length() != b.length()){
return a.length() - b.length();
}
for(int i = 0; i < a.length(); i++){
if( a.codePointAt(i) != b.codePointAt(i) ){ //Or charAt()
return a.codePointAt(i) - b.codePointAt(i);
}
}
return 0;
}
Please don't use exceptions for handling flow control, this is a serious anti-pattern (also here).
As you mentionned in the comments, the real thing you've been asked is to convert a List<String> into a List<Number>.
Also, if I understand correctly, you know that:
You should encounter only numbers without decimals
The biggest value you can encounter is possibly unbound
Based on that, the following method will do the job in a more clever way:
private static List<Number> toNumbers(List<String> strings) {
return strings.stream()
.map(BigInteger::new)
.collect(Collectors.toList());
}
Eidt: if you're not very familiar with the stream concept, here's the equivalent code without streams:
private static List<Number> toNumbers(List<String> strings) {
List<Number> numbers = new ArrayList<>();
for (String s : strings) {
numbers.add(new BigInteger(s));
}
return numbers;
}
Well if you want to do it "by hand" try something like this:
We define the max values as strings :
String intMax = "2147483647";
String longMax = "9223372036854775807";
and our number:
String ourNumber = "1234567890"
Now our logic will be simple :
We will check lenghts of strings firstly
If our numbers length < int max length : IT IS INT
If our numbers length == int max length : Check is it INT or LONG
If our numbers length > int max length :
3.1 If our numbers length < long max length : IT IS LONG
3.2 If our numbers length == long max length : Check is it LONG or BIG INTEGER
3.3 If our numbers length > long max length : IT IS BIG INTEGER
The code should look something like this (I have not tried to compile it may have syntax or other errors) :
if(ourNumber.lenght() < intMax.length ){
System.out.println("It is an Integer");
} else if(ourNumber.lenght() == intMax.length){
// it can be int if the number is between 2000000000 and 2147483647
char[] ourNumberToCharArray = ourNumber.toCharArray();
char[] intMaxToCharArray = intMax.toCharArray();
int diff = 0;
for(int i = 0; i < ourNumberToCharArray.length; i++) {
diff = Character.getNumericValue(intMaxToCharArray[i]) - Character.getNumericValue(ourNumberToCharArray[i]);
if(diff > 0) {
System.out.println("It is a Long");
break;
} else if(diff < 0) {
System.out.println("It is an Integer");
break;
}
}
if(diff == 0){
System.out.println("It is an Integer");
}
} else {
if(ourNumber.lenght() < longMax.length()) {
System.out.println("It is a Long");
} else if(ourNumber.lenght() == longMax.length()){
char[] ourNumberToCharArray = ourNumber.toCharArray();
char[] longMaxToCharArray = longMax.toCharArray();
int diff = 0;
for(int i = 0; i < ourNumberToCharArray.length; i++) {
diff = Character.getNumericValue(longMaxToCharArray[i]) - Character.getNumericValue(ourNumberToCharArray[i]);
if(diff > 0) {
System.out.println("It is a BigInteger");
break;
} else if(diff < 0) {
System.out.println("It is a Long");
break;
}
}
if(diff == 0){
System.out.println("It is a Long");
}
} else {
System.out.println("It is a BigInteger");
}
}
Then logic that checks if the numbers match or not is the same in both cases you can but it in a function for example.
I am aware there are multiple threads like my assignment below, but I just can't figure it out. I can't exactly figure out the mistake. Help would be appreciated.
I am trying to do this program:
Everything works fine unless I input the same chains or similar (for example ACTG and ACTG or ACTG and ACTGCCCC), when it tells me
string index out of range
This is that part of my code:
int tries=0;
int pos=-1;
int k;
for (int i=0; i<longDNA.length(); i++) {
tries=0;
k=i;
for (int j=0; j<shortDNA.length(); j++) {
char s=shortDNA.charAt(j);
char l=longDNA.charAt(k);
if (canConnect(s,l)) {
tries+=1;
k+=1;
}
}
if (tries==shortDNA.length()-1) {
pos=i-1;
break;
}
}
Let's call the two DNA strings longer and shorter. In order for shorter to attach somewhere on longer, a sequence of bases complementary to shorter must be found somewhere in longer, e.g. if there is ACGT in shorter, then you need to find TGCA somewhere in longer.
So, if you take shorter and flip all of its bases to their complements:
char[] cs = shorter.toCharArray();
for (int i = 0; i < cs.length; ++i) {
// getComplement changes A->T, C->G, G->C, T->A,
// and throws an exception in all other cases
cs[i] = getComplement(cs[i]);
}
String shorterComplement = new String(cs);
For the examples given in your question, the complement of TTGCC is AACGG, and the complement of TGC is ACG.
Then all you have to do is to find shorterComplement within longer. You can do this trivially using indexOf:
return longer.indexOf(shorterComplement);
Of course, if the point of the exercise is to learn how to do string matching, you can look at well-known algorithms for doing the equivalent of indexOf. For instance, Wikipedia has a category for String matching algorithms.
I tried to replicate your full code as fast as I could, I'm not sure if I fixed the problem but you don't get any errors.
Please try it and see if it works.
I hope you get this in time and good luck!
import java.util.Arrays;
public class DNA {
public static void main(String[] args) {
System.out.println(findFirstMatchingPosition("ACTG", "ACTG"));
}
public static int findFirstMatchingPosition(String shortDNA, String longDNA) {
int positionInLong = 0;
int positionInShort;
while (positionInLong < longDNA.length()) {
positionInShort = 0;
while(positionInShort < shortDNA.length()) {
String s = shortDNA.substring(positionInShort, positionInShort + 1);
if(positionInShort + positionInLong + 1 > longDNA.length()) {
break;
}
String l = longDNA.substring(positionInShort + positionInLong, positionInShort + positionInLong + 1);
if(canConnect(s, l)) {
positionInShort++;
if(positionInShort == shortDNA.length()) {
return positionInLong;
}
} else {
break;
}
}
positionInLong++;
if(positionInLong == longDNA.length()) {
return -1;
}
}
return -1;
}
private static String[] connections = {
"AT",
"TA",
"GC",
"CG"
};
private static boolean canConnect(String s, String l) {
if(Arrays.asList(connections).contains((s+l).toUpperCase())) {
return true;
} else {
return false;
}
}
}
I finally changed something with the k as Faraz had mentioned above to make sure the charAt does not get used when k overrides the length of the string and the program worked marvelously!
The code was changed to the following:
int tries=0;
int pos=-1;
int k;
for (int i=0; i<longDNA.length(); i++) {
tries=0;
k=i;
for (int j=0; j<shortDNA.length(); j++) {
if (k<longDNA.length()) {
char s=shortDNA.charAt(j);
char l=longDNA.charAt(k);
if ((s=='A' && l=='T') || (s=='T' && l=='A') || (s=='G' && l=='C') || (s=='C' && l=='G')) {
tries+=1;
k+=1;
}
}
}
if (tries==shortDNA.length()) {
pos=i;
break;
}
}
I am not sure how aesthetically pleasing or correct this excerpt is but - it completely solved my problem, and just 2 minutes before the deadline! :)
A huge thanks to all of you for spending some time to help me!!
import java.util.*;
class A{
static int count=0;
static String s;
public static void main(String z[]){
int n;
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
System.out.println(noOfBouncy(n));
}
public static int noOfBouncy(int k){
int limit=(int)Math.pow(10,k);
s=new String("1");
int num=Integer.parseInt(s);
while(num<limit){
if(isIncreasing(s) || isDecreasing(s) ){
}
else{
count++;
}
num++;
s=new String(Integer.toString(Integer.parseInt(s)+1));
}
count=limit-count;
return count;
}
}
public static boolean isIncreasing(String s){
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)>s.charAt(i+1)){
return false;
}
}
return true;
}
public static boolean isDecreasing(String s){
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)<s.charAt(i+1)){
return false;
}
}
return true;
}
I have given the definitions to the two functions used isIncreasing() & isDecresing()
The program runs well for the value of n<7 but does not respond for values above it, Why ?
I accept the programming style is very immature,please ignore.
I've tried to execute it with n=7 and it finishes in 810ms, returning 30817.
However, I recommend to you to optimize the performance of your program by saving unnecessary object instantiation: It will be better if you maintain the counter in num, and convert it to string just once, at the beginning of the loop:
int num=1;
while (num < limit)
{
s=Integer.toString(num);
if (isIncreasing(s) || isDecreasing(s))
{
}
else
{
count++;
}
num++;
}
Like this it takes just 450ms to finish.
The program was not actually stuck but it is taking way too much time to complete its execution when value of 'n' is larger.
So now the question is, I need to optimize the code to take minimum time #Little have an optimization bit that's not enough.
Any hint would be appreciable.
To increase the performance you should avoid the conversation to String and do the check with numbers.
As it doesn't matter for the result if you start the comparison from left to right or from right to left one computational solution could be.
as pseudo code
1) compare the value of the right most digit with the digit on it's left
2) is it lower --> we found a decreasing pair
3) else check if it is bigger --> we found an increasing pair
4) else --> not a bouncy pair
5) if we found already one decreasing and one increasing pair it's bouncy number
6) divide the number by ten if it's bigger then ten repeat with step 1)
The method to check if it's a bouncy number could look like this
static boolean isBouncyNumber(int number) {
boolean increasingNumber = false;
boolean decreasingNumber = false;
int previousUnitPosition = number % 10;
int remainder = number / 10;
while (remainder > 0) {
// step 1
int currentUnitPosition = remainder % 10;
if (currentUnitPosition > previousUnitPosition) {
// step 2
decreasingNumber = true;
} else if (currentUnitPosition < previousUnitPosition) {
// step 3
increasingNumber = true;
}
// step 5
if (decreasingNumber && increasingNumber) {
return true;
}
// step 6
previousUnitPosition = currentUnitPosition;
remainder = remainder / 10;
}
return decreasingNumber && increasingNumber;
}
Over the past couple of weeks I've read through the book Error Control Coding: Fundamentals and Applications in order to learn about BCH (Bose, Chaudhuri, Hocquenghem) Codes for an junior programming role at a telecoms company.
This book mostly covers the mathematics and theory behind the subject, but I'm struggling to implement some of the concepts; primarily getting the next n codewords.I have a GUI (implemented through NetBeans, so I won't post the code as the file is huge) that passes a code in order to get the next n numbers:
Generating these numbers is where I am having problems. If I could go through all of these within just the encoding method instead of looping through using the GUI my life would be ten times easier.
This has been driving me crazy for days now as it is easy enough to generate 0000000000 from the input, but I am lost as to where to go from there with my code. What do I then do to generate the next working number?
Any help with generating the above code would be appreciated.
(big edit...) Playing with the code a bit more this seems to work:
import java.util.ArrayList;
import java.util.List;
public class Main
{
public static void main(final String[] argv)
{
final int startValue;
final int iterations;
final List<String> list;
startValue = Integer.parseInt(argv[0]);
iterations = Integer.parseInt(argv[1]);
list = encodeAll(startValue, iterations);
System.out.println(list);
}
private static List<String> encodeAll(final int startValue, final int iterations)
{
final List<String> allEncodings;
allEncodings = new ArrayList<String>();
for(int i = 0; i < iterations; i++)
{
try
{
final int value;
final String str;
final String encoding;
value = i + startValue;
str = String.format("%06d", value);
encoding = encoding(str);
allEncodings.add(encoding);
}
catch(final BadNumberException ex)
{
// do nothing
}
}
return allEncodings;
}
public static String encoding(String str)
throws BadNumberException
{
final int[] digit;
final StringBuilder s;
digit = new int[10];
for(int i = 0; i < 6; i++)
{
digit[i] = Integer.parseInt(String.valueOf(str.charAt(i)));
}
digit[6] = ((4*digit[0])+(10*digit[1])+(9*digit[2])+(2*digit[3])+(digit[4])+(7*digit[5])) % 11;
digit[7] = ((7*digit[0])+(8*digit[1])+(7*digit[2])+(digit[3])+(9*digit[4])+(6*digit[5])) % 11;
digit[8] = ((9*digit[0])+(digit[1])+(7*digit[2])+(8*digit[3])+(7*digit[4])+(7*digit[5])) % 11;
digit[9] = ((digit[0])+(2*digit[1])+(9*digit[2])+(10*digit[3])+(4*digit[4])+(digit[5])) % 11;
// Insert Parity Checking method (Vandermonde Matrix)
s = new StringBuilder();
for(int i = 0; i < 9; i++)
{
s.append(Integer.toString(digit[i]));
}
if(digit[6] == 10 || digit[7] == 10 || digit[8] == 10 || digit[9] == 10)
{
throw new BadNumberException(str);
}
return (s.toString());
}
}
class BadNumberException
extends Exception
{
public BadNumberException(final String str)
{
super(str + " cannot be encoded");
}
}
I prefer throwing the exception rather than returning a special string. In this case I ignore the exception which normally I would say is bad practice, but for this case I think it is what you want.
Hard to tell, if I got your problem, but after reading your question several times, maybe that's what you're looking for:
public List<String> encodeAll() {
List<String> allEncodings = new ArrayList<String>();
for (int i = 0; i < 1000000 ; i++) {
String encoding = encoding(Integer.toString(i));
allEncodings.add(encoding);
}
return allEncodings;
}
There's one flaw in the solution, the toOctalString results are not 0-padded. If that's what you want, I suggest using String.format("<something>", i) in the encoding call.
Update
To use it in your current call, replace a call to encoding(String str) with call to this method. You'll receive an ordered List with all encodings.
I aasumed, you were only interested in octal values - my mistake, now I think you just forgot the encoding for value 000009 in you example and thus removed the irretating octal stuff.