Develop Reference

Replacing String in Java to get all variations

I'm trying to get a printout of all variations of a certain String. For example, we have this input: AB0C0. The 0 in the 3rd and 5th spots should be treated as variables. The variable characters are 1, 2, and 3 to be placed in the spot of 0. This means there would be all possible variations of this input:
AB1C1
AB2C1
AB3C1
AB1C2
AB1C3
AB2C2
AB2C3
AB3C2
AB3C3
This is just an example. A 5-character long string is a place for 1 to 5 variables. The issue I'm facing is, that it should generate all variations no matter how many variables are in the input in no matter in which place they are.
Scanner scanner = new Scanner (System.in);
System.out.println("Enter the key consisting of 5 characters:");
String input = scanner.next();
String strOutput1 = input.replaceFirst("0","1");
String strOutput1A = input.replace("0","1");
String strOutput2 = input.replaceFirst("0","2");
String strOutput3 = input.replaceFirst("0","3");
String strOutput4 = input.replaceFirst("0","4");
String strOutput5 = input.replaceFirst("0","5");
System.out.println(strOutput1.toUpperCase());
System.out.println(strOutput1A.toUpperCase());
System.out.println(strOutput2.toUpperCase());
System.out.println(strOutput3.toUpperCase());
System.out.println(strOutput4.toUpperCase());
System.out.println(strOutput5.toUpperCase());

What about this:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws Exception {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the key consisting of 5 characters:");
String input = scanner.next();
//find positions of '0' in input
List<Integer> varPositions = findVarPositions(input);
//create permutations
List<String> permutations = new ArrayList<>();
permutations.add(input);//AB0C0
for (int position : varPositions) {
permutations = permutateAtPosition(permutations, position);
}
//print permutations
for (String permutation : permutations) {
System.out.println(permutation.toUpperCase());
}
}
private static List<Integer> findVarPositions(String input) {
List<Integer> varPositions = new ArrayList<>();
int lastVarPosition = -1;
while ((lastVarPosition = input.indexOf('0', lastVarPosition + 1)) != -1) {
varPositions.add(lastVarPosition);
}
return varPositions;
}
private static List<String> permutateAtPosition(List<String> partialyPermutated, int position) {
List<String> result = new ArrayList<>();
char[] replacements = {'1', '2', '3', '4', '5'};
for (String item : partialyPermutated) {
for (int i = 0; i < replacements.length; i++) {
String output = replaceCharAt(item, position, replacements[i]);
result.add(output);
}
}
return result;
}
private static String replaceCharAt(String input, int position, char replacement) {
//converting to char array, because there's no method like
//String.replaceAtPosition(position, char)
char[] charArray = input.toCharArray();
charArray[position] = replacement;
return new String(charArray);
}
}
It's not fixed to a number of variables.
The idea is to extract positions of '0' and subsequently call the method permutateAtPosition, which takes a partially permutated list and permutates it by one more level.
For "a0b0c0" and values 1-2 it would be ['a0b0c0'], then ['a1b0c0','a2b0c0'], then ['a1b1c0','a1b2c0','a2b1c0','a2b2c0'], and finally ['a1b1c1','a1b1c2','a1b2c1','a1b2c2','a2b1c1','a2b1c2','a2b2c1''a2b2c2'].
This solution keeps everything in memory, so in the general case (unlimited input string) it would be wiser to go with depth-first instead.

I've got another solution for you.
First step, getting the amount of variables:
int variableCount = 0;
for (int i = 0; i < 5; i++) {
if (input.charAt(i) == '0') {
variableCount++;
}
}
Then calculating the amount of results we are expecting:
int countMax = (int)Math.pow(4,variableCount);
Lastly, count up in base 4. Pad the number with 0's and replace the original input 0's:
for (int i = 0; i < countMax; i++) {
String paddedNumbers = format("%" + variableCount + "s",Integer.toString(i, 4)).replace(" ", "0");
int replacedCount = 0;
char[] outputChars = input.toCharArray();
for (int j = 0; j < 5; j++) {
if (input.charAt(j) == '0') {
outputChars[j] = paddedNumbers.charAt(replacedCount);
replacedCount++;
}
}
System.out.println(outputChars);
}

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;
}
}

Displaying a string with hidden characters

I have a test for a class that displays a word randomly chosen from an array.
I'm trying to display the word with several chars hidden
I have taken the string, then converted it to an array of chars, but I'm confused as to where to go from here.
import java.util.Scanner;
public class wordTest {
public static void main (String args[])
{
Scanner scanner = new Scanner(System.in);
String readString = scanner.nextLine();
char[] stringArray;
String [] gamewords = { "dog", "cat", "coffee", "tag", "godzilla", "gamera", "lightning", "flash", "spoon", "steak", "moonshine", "whiskey", "tango", "foxtrot", "ganymede"
, "saturn", "enterprise", "reliant", "defiant", "doom", "galapagos", "jidai", "sengoku"};
arrayWords wl = new arrayWords();
// Words w = new Words();
Word n = new Word();
int a = 0;
int b = gamewords.length;
RandNum rand = new RandNum(a,b);
n.setWord(gamewords[rand.nextRandomIntegerInRange()]);
stringArray = n.getWord().toCharArray();
int blank1 = 1;
int blank2 = 4;
RandNum blanks = new RandNum(blank1,blank2);
n.setWord(gamewords[rand.nextRandomIntegerInRange()]);
do{
int i = 0;
//scanner.nextLine();
for( i = 0; i < stringArray.length; i++){
for( i = 0 ; i < blanks.nextRandomIntegerInRange() ; i++ ){
stringArray[i] = '*';
}
System.out.println(stringArray[i]);
}
}while(scanner.nextLine().equals(""));
}
}

Since you haven't given clear definition on what you want to do, here I assume for every string, you are randomly masking 2 characters, in pseudo code, it looks like:
if inputString.length < 2 {
mask all character
} else {
loop until 2 character masked {
r = random from 0 to inputString.length-1
if (inputString[r] is not masked) {
set inputString[r] to mask character
}
}
}
some hints:
to make "inputString" modifiable, make use of a StringBuilder
Way to check if certain position is masked, you can either simply check if the character in the string == mask character, or you can use a Set to keep all masked position
In order to find out number of position masked, you can keep a counter, or simply use the size of the Set in 2 if you choose to use a Set.

Okay, so I think I've found the solution:
for (int i = 0; i < stringArray.length ; i++) {
stringArray[blanks.nextRandomIntegerInRange()] = '_';
System.out.print(stringArray[i] + " " );
}

Getting TLE in CodeChef and need for improvement of code

I am trying to solve a practice question of CodeChef . In this problem we are given N numbers Ai...An and we first have to sort(ascending order) the numbers and then add the alternate numbers starting from the last and show the output for each test cases , the test cases has 2 parts :
1>Constraints :
1 ≤ Ai ≤ 109
1 ≤ N ≤ 1000
2>Constraints:
1 ≤ Ai ≤ 109
1 ≤ N ≤ 105
You can see the full problem here.
The first part of my problem was successfully submitted but second part showed NZEC because I was using long to add those numbers(which was beyond that range). So I decided to use Strings to add up my numbers here is the method :
public static String myStringWayToAdd(String first , String second){
String temp = "";
if(first.length() < second.length()){
temp = first;
first = second;
second = temp;
}
temp = "";
int carry = 0;
for(int i=1;i<=first.length();++i){
if(i <= second.length()){
carry += Integer.parseInt(first.charAt(first.length()-i)+"") + Integer.parseInt(second.charAt(second.length()-i)+"");
}
else{
carry += Integer.parseInt(first.charAt(first.length()-i)+"");
}
temp += carry%10;
carry = carry/10;
}
if(carry != 0)
temp += carry;
StringBuilder myResult = new StringBuilder(temp);
return(myResult.reverse().toString());
}
But now it shows TLE(Time Limit Expire) , So then I thought to use BigInteger(which I am not pretty much Aware of but I saw some tutorials) :
BigInteger big = new BigInteger("0");
big = big.add(BigInteger.valueOf(mySort.get(j))); //for addition and mySort is my ArrayList
But this gave me NZEC I don't know whywell now I want to use double variable but there is a problem with that too, because with double large numbers will be in form of exponential value like :
1.243536E15 which will not be accepted by the machine, so is there any good way to solve this problem and not getting any Time Limit Expiry?.
Any help will really be appreciated. Thank you in Advance.
Edit 1 :
I changed baxck the variable to long and run and this time strangely I got TLE here is my code :
import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
import java.math.BigInteger;
import java.lang.Number;
class CFEA{
public static void main(String[] s){
Scanner scan = new Scanner(System.in);
int testCases = scan.nextInt();
for(int i = 0 ; i<testCases;++i){
long sum = 0;
//BigInteger big = new BigInteger("0");
ArrayList<Integer> mySort = new ArrayList<Integer>();
int n = scan.nextInt();
for(int j = 1 ; j <= n ; ++j){
mySort.add(scan.nextInt());
}
Collections.sort(mySort);
for(int j = mySort.size()-1 ; j >= 0 ; j=j-2){
sum += mySort.get(j);
}
System.out.println(sum);
}
}
}
And here is Link to my submission.Is there Anything I can optimize in my code?

The sum of all number is at most 10^9 * 10^5 = 10^14. It is small enough to fit into long. There is no need to use BigInteger.
java.util.Scanner has performance issues. You can implement a custom scanner(using BufferedReader) to speed up your code.
Here is my implementation of a scanner:
import java.io.*;
import java.util.StringTokenizer;
public class FastScanner {
private BufferedReader reader;
private StringTokenizer tokenizer;
public FastScanner(InputStream inputStream) {
reader = new BufferedReader(new InputStreamReader(inputStream));
}
public String next() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
String line = reader.readLine();
if (line == null)
throw new IOException();
tokenizer = new StringTokenizer(line);
}
return tokenizer.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public void close() {
try {
reader.close();
} catch (IOException e) {
//ignore
}
}
}

I did some changes in my Program and it was All Accepted a much relief after submitting for about 20 times , here is my new code :
import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
class CFEA{
public static void main(String[] s){
Scanner scan = new Scanner(System.in);
byte testCases = Byte.parseByte(scan.nextLine()); //used byte for test cases instead of int
for(int i = 0 ; i<testCases;++i){
long sum = 0;
//BigInteger big = new BigInteger("0");
ArrayList<Integer> mySort = new ArrayList<Integer>();
int n = Integer.parseInt(scan.nextLine());
String input = scan.nextLine();
String[] my = input.split(" ");
for(String myString : my){
mySort.add(Integer.parseInt(myString));
}
Collections.sort(mySort);
for(int j = mySort.size()-1 ; j >= 0 ; j=j-2){
sum += mySort.get(j);
}
System.out.println(sum);
}
}
}
I think the main villain was that I was scanning for Integers N number of times as in this :
for(int j = 1 ; j <= n ; ++j){
mySort.add(scan.nextInt());
}
When N was something Like 100000 then this really slows it down.So i used 1 String for complete line and then Split it into Integers using split method as in :
String input = scan.nextLine(); //only 1 Scanner
String[] my = input.split(" ");
for(String myString : my){
mySort.add(Integer.parseInt(myString));
}
Although My code got submitted I still think there is Further scope for optimization , so please do answer if you have something better

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();
}
}
}