We Keep Coding

Hanging Letter Program

I was practicing problems in JAVA for the last few days and I got a problem like this:
I/p: I Am A Good Boy
O/p:
I A A G B
m o o
o y
d
This is my code.
System.out.print("Enter sentence: ");
String s = sc.nextLine();
s+=" ";
String s1="";
for(int i=0;i<s.length();i++)
{
char c = s.charAt(i);
if(c!=32)
{s1+=c;}
else
{
for(int j=0;j<s1.length();j++)
{System.out.println(s1.charAt(j));}
s1="";
}
}
The problem is I am not able to make this design.My output is coming as each character in each line.

First, you need to divide your string with space as a delimiter and store them in an array of strings, you can do this by writing your own code to divide a string into multiple strings, Or you can use an inbuilt function called split()
After you've 'split' your string into array of strings, just iterate through the array of strings as many times as your longest string appears, because that is the last line you want to print ( as understood from the output shared) i.e., d from the string Good, so iterate through the array of strings till you print the last most character in the largest/ longest string, and exit from there.
You need to handle any edge cases while iterating through the array of strings, like the strings that does not have any extra characters left to print, but needs to print spaces for the next string having characters to be in the order of the output.
Following is the piece of code that you may refer, but remember to try the above explained logic before reading further,
import java.io.*;
import java.util.*;
public class MyClass {
public static void main(String args[]) throws IOException{
//BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
Scanner sc = new Scanner(System.in);
String[] s = sc.nextLine().split(" ");
// Split is a String function that uses regular function to split a string,
// apparently you can strings like a space given above, the regular expression
// for space is \\s or \\s+ for multiple spaces
int max = 0;
for(int i=0;i<s.length;i++) max = Math.max(max,s[i].length()); // Finds the string having maximum length
int count = 0;
while(count<max){ // iterate till the longest string exhausts
for(int i=0;i<s.length;i++){
if(count<s[i].length()) System.out.print(s[i].charAt(count)+" "); // exists print the character
else System.out.print(" "); // Two spaces otherwise
}
System.out.println();count++;
}
}
}
Edit: I am sharing the output below for the string This is a test Input
T i a t I
h s e n
i s p
s t u
t

Working Sturcture of Arrays. binarySearch() method in java

I'm trying solve problem in Strings,finding matching characters in to String.
I solve it using Character Array and inner loop but i think it has more time complexity. so try to solve it in Arrays binary search but it gives inappropriate result.i want working structure of binary search method in java.
I set matched value in the String two to duplicate char '#',because don't want to match another char.
public static void main(String[] args) {
Scanner s= new Scanner(System.in);
String team1 = s.next();
String team2 = s.next();
char[] teamA = team1.toCharArray();
char[] teamB = team2.toCharArray();
Arrays.sort(teamB);
int count = 0;
for(int a=0;a< teamA.length;a++) {
int index = Arrays.binarySearch(teamB, teamA[a]);
if(index >= 0) {
count++;
teamB[index] = '#';
}
}
System.out.println(count);
}
if i give input of two strings
"aabc" and "zbaa" expected output is 3
but my program gives output 2.

The problem is that once you update the teamB array in the loop the array is no longer sorted. And in unsorted array binary search will give unexpected outputs.

How to compare character input by user to dictionary file in Java?

I need to read the user input and compare this to a dictionary.txt. The user may input any number of characters and the program must return all the words in the English language that can be made from these characters. The letters can be used in any order and may only be used once.
For example:
User Input: "odg"
Output: "dog" , "god" ... and any others
After quite a substantial amount of research, I have come up with the following partial solution:
Read user input
Convert to an array of characters
Loop through the document depending on array length
Using indexOf to compare each character in this array to each line, then printing the word/s which do not return -1
How do I compare a set of characters inputted by the user to those found in a text file (dictionary) ? The characters do not have to be in any order to match .(as seen in the example used above)
Bear with me here, I know this must be one of the most inefficient ways to do such a task! Any further ideas on how to implement my original idea would be appreciated, while I am also open to any new and more efficient methods to perform this operation.
Below is what I have come up with thus far:
public static void main(String[] args) throws FileNotFoundException {
BufferedReader reader1 = new BufferedReader(new FileReader(FILENAME));
Scanner sc = new Scanner(System.in);
String line;
ArrayList<String> match = new ArrayList<>();
System.out.println("Enter characters to see which english words match: ");
String userInput = sc.next();
char arr[] = userInput.toCharArray();
int i;
try {
while ((line = reader1.readLine()) != null) {
for (i=0; i < arr.length; i++)
{
if ((line.indexOf(userInput.charAt(i)) != -1) && (line.length() == arr.length)) {
match.add(line);
}
else {
// System.out.println("no matches");
}
}
}
System.out.println(match);
}
catch (IOException e) {
e.printStackTrace();
}
**Current results: **
Words in text file:
cab
dog
god
back
dogs
quick
User input: "odg"
Program output:
[god, god, god, dog, dog, dog]
The program should return all words in the dictionary that can be made out of the string entered by the user I am managing to return both instances in this case, however, each are displayed for three times (arr.length).

First of all, interesting question. I implemented my solution and Ole V.V's solution. Here are the codes based on your post. I test the only test case you provided, not sure whether this is what you want. Let me know if it is not working as you expected.
Solution One: counting O(nk)
public static void main(String[] args) throws IOException {
BufferedReader reader1 = new BufferedReader(new FileReader(FILENAME));
Scanner sc = new Scanner(System.in);
System.out.println("Enter characters to see which english words match: ");
String userInput = sc.next();
Map<Character, Integer> counter = count(userInput);
String line;
while ((line = reader1.readLine()) != null) {
Map<Character, Integer> lineCounter = count(line);
if(lineCounter.equals(counter)) {
System.out.println(line);
}
}
}
public static Map<Character, Integer> count(String input) {
Map<Character, Integer> result = new HashMap<Character, Integer>();
for (char c: input.toCharArray()) {
result.putIfAbsent(c, 0);
result.put(c, result.get(c) + 1);
}
return result;
}
Solution Two: sorting O(nk)
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new FileReader(FILENAME));
Scanner sc = new Scanner(System.in);
System.out.println("Enter characters to see which english words match: ");
String userInput = sc.next();
userInput = sort(userInput);
String line;
while ((line = reader.readLine()) != null) {
String sortedLine = sort(line);
if(sortedLine.equals(userInput)) {
System.out.println(new String(line));
}
}
}
// counting sort
public static String sort(String input) {
char c[] = input.toCharArray();
int length = c.length;
char output[] = new char[length];
int count[] = new int[256];
for (int i = 0; i < length; i++) {
count[c[i]] = count[c[i]] + 1;
}
for (int i = 1; i <= 255; i++) {
count[i] += count[i - 1];
}
for (int i = 0; i < length; i++) {
output[count[c[i]] - 1] = c[i];
count[c[i]] = count[c[i]] - 1;
}
return new String(output);
}

The standard solution to this kind of problem is: sort the characters of the user input. So odg will become dgo and back will become abck. For each word in the dictionary, do the same sorting. So cab will become abc and dog will be dgo — hey, that’s the same as the first user input, so now we know that this word should be output.
The strong point with this solution is you make sure every letter is used exactly once. It even takes duplicate letters into account: if the same letter comes twice in the user input, it will only find words that also contain that letter exactly twice.
If you like, you can prepare your word list in advance by building a map where the keys are the alphabetically sorted words and the values are lists of words that contain those same letters. So key dgo will map to a list of [dog, god]. Then you just have to sort the input and make a lookup.

I'll show you a solution that is easy to understand and implement but not the fastest available:
Possible solution: Array sorting
Treat input string and dictionary word as array of chars, sort them, then compare them:
public static boolean stringsMatchSort(String a, String b) {
// Different length? Definitely no match!
if (a.length() != b.length()) {
return false;
}
// Turn both Strings to char arrays
char[] charsA = a.toCharArray();
char[] charsB = b.toCharArray();
// Sort both arrays
Arrays.sort(charsA);
Arrays.sort(charsB);
// Compare them, if equal: match!
return Arrays.equals(charsA, charsB);
}
Note how I made the meat of your program / problem into a method. You can then easily use that method in a loop that iterates over all words of your dictionary. The method doesn't care where the words come from: a file, a collection, additional user input, the network, etc.
It also helps to simplify your program by dividing it into smaller parts, each with a smaller responsibility. This is commonly known as divide & conquer and is one of the most valuable strategies for both, new and old programmers alike, when it comes to tackling complicated problems.
Other solutions: Prime numbers, HashMaps, ...
There are other (including faster and more elegant) solutions available. Take a look at these related questions, which yours is pretty much a duplicate of:
"How to check if two words are anagrams"
"finding if two words are anagrams of each other"
Additional notes
Depending on your application, it might be a good idea to first read the dictionary into a suitable collection. This would be especially helpful if you perform multiple "queries" against the same dictionary. Or, if the dictionary is really huge, you could already strip out duplicates during the creation of the collection.

Java Word Count

I am just starting out in Java so I appreciate your patience. Anyways, I am writing a word count program as you can tell by the title, I am stuck at the numWords function below the for loop, I am not sure what I should set it equal to. If someone could set me in the right direction that would be awesome. Thank you. Here is all of my code thus far, let me know if I not specific enough in what I am asking, this is my first post. Thanks again.
import java.util.Scanner;
public class WCount {
public static void main (String[] args) {
Scanner stdin = new Scanner(System.in);
String [] wordArray = new String [10000];
int [] wordCount = new int [10000];
int numWords = 0;
while(stdin.hasNextLine()){
String s = stdin.nextLine();
String [] words = s.replaceAll("[^a-zA-Z ]", "").toLowerCase().split("\\s\
+");
for(int i = 0; i < words.length; i++){
numWords = 0;
}
}
}
}

If your code is intended to just count words, then you don't need to iterate through the words array at all. In other words, replace your for loop with just:
numWords += words.length;
Most likely a simpler approach would be to look for sequences of alpha characters:
Matcher wordMatch = Pattern.compile("\\w+").matcher();
while (wordMatch.find())
numWords++;
If you need to do something with the words (such as store them in a map to a count) then this approach will make that simpler:
Map<String,Integer> wordCount = new HashMap<>();
Matcher wordMatch = Pattern.compile("\\w+").matcher();
while (wordMatch.find()) {
String word = wordMatch.group();
int count = wordCount.getOrDefault(word, 0);
wordCount.put(word, count + 1);
}

Don't worry. We were all beginners once.
First of all, you don't need to do the loop because "length" attribute already has it. But, if you want to practice with loops is so easy as increasing the counter each time the iterator advances and that's it.
numWords++;

Hint: Read the input
String sentence = stdin.nextLine();
Split the string
String [] words = sentence.split(" ");
Number of words in a sentence
System.out.println("number of words in a sentence are " + words.length);
You mentioned in comments that you would also like to print the line in alphabetical order. For that Java got you covered:
Arrays.sort(words);

The best way to count the amount of words in a String String phrase is simply to get a String array from it using the String method split String[] words = phrase.split(" ") and giving it as argument the space itself, this will return a String array with each different words, then you can simple check its lengthwords.length and this will give you the exact number.

Print out prime Number less than a given number N

Print out the prime numbers less than a given number N. For bonus points your solution should run in N*log(N) time or better. You may assume that N is always a positive integer.
Input sample:
Your program should accept as its first argument a path to a filename. Each line in this file is one test case. Each test case will contain an integer n < 4,294,967,295.
E.g.
10
20
100
Output sample:
For each line of input, print out the prime numbers less than N, in ascending order, comma delimited. (There should not be any spaces between the comma and numbers) E.g.
2,3,5,7
2,3,5,7,11,13,17,19
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97
Here is my solution:
public class problem1 {
public static void main(String [] args) throws Exception
{
File f=new File("C://Users/Rahul/Documents/Projects/r.txt");
FileReader fr=new FileReader(f);
List<Integer> l=new ArrayList<>();
int p;
BufferedReader br = new BufferedReader(fr);
String s;
while( (s= br.readLine()) != null ) {
int a=Integer.parseInt(s);
for(int i=2;i<a;i++)
{
p=0;
for(int j=2;j<i;j++)
{
if(i%j==0)
p=1;
}
if(p==0)
l.add(i);
}
String st=l.toString();
st=st.replaceAll("\\[", "").replaceAll("\\]", "").replace(", ", ",");
System.out.print(st);
System.out.println("\t");
}
fr.close();
}
}
My input is :
10
50
And output is :
2,3,5,7
2,3,5,7,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
But when i submit this solution they are not accepting this solution.
But when i put content in document like this:
10 50
30
I am trying that java program ignore this 50. How to do it ?
Any better solution then this ?
Give me some idea!

To ignore the extra number in your file you can take only the first number of each line.
Your solution is probably not accepted because in your second line you have printed 2,3,5,7 twice (i.e. the primes of the previous line)
See the example below to fix both problems
while( (s= br.readLine()) != null ) {
String [] numbers = s.split(" "); // split the line
int a = Integer.parseInt(numbers[0]); // take only the first one
....
System.out.print(st);
System.out.println("\t");
l.clear(); // clear the list before trying to find primes for the new line
}

"Your program should accept as its first argument a path to a filename"
You have a hardcoded filename in your solution - use args[0] instead.
Othwerwise, your solutions looks OK, although there is some room for improvements regarding the efficiency.