public class newString {
public static void main (String args[]){
String title = "Book";
String title1;
title1 = title;
for(int i = 0; i < title.length(); i++){
for (int x = 0; x<title1.length(); x++){
if (title.charAt(i+x) == title1.charAt(x)){
System.out.print(title.charAt(0,1));
}
}
}
}
}
I really don't understand what I'm doing wrong here. What I need to do is define a string called "title", with "Book" in it, which I did, and create a second string called "title1". I need to create code to store the contents of title, into title1, but only every other character. For example: title1 should have "Bo" in it. What am I doing wrong?
Here's the looping solution with fewer operations. Instead of checking if i is even, just increment by 2.
String title1 = "Some title";
String title2 = "";
for (int i = 0; i < title1.length(); i += 2)
{
title2 += title1.charAt(i);
}
You algorithm is wrong, it seems what you need to do is to extract out every nth character from source string, for example:
String source = "Book";
End result should be "Bo"
The algorithm should be:
Iterate through each character in the original string, use a stride as you need, in this case, a stride of 2 should do (so rather than increment by one, increment by the required stride)
Take the character at that index and add it to your second string
The end result should be a string which holds every nth character.
I don't really understand what you are attempting, but I can tell you what you are doing. Your loop structure does the following:
when i = 0, it compares all characters in both strings (0 + n = n, so the inner loop goes from x - title1.length()).
when i = 1, compare all characters except the first one (for size x, 1 + n = x - 1 comparisons).
when i =2, compare x / 2 characters (for size x, 2 + n = x / 2)
when i = 3, compare x / 3 characters
... and so on
System.out.print(title.charAt(0,1)) Shouldn't even compile. charAt(int) is the correct call. And if title length is greater than 0, this will always print a String with a single character -- the first one in title. And it will always be the same unless you reassign title to a different String.
Also this code will always throw an IndexOutOfBoundsException at title.charAt(i+x) when i = title.length() - 1 and x > 0.
Related
Below Java code produces the valid output but it takes more time to execute. Code works fine in eclipse, but it do not work in an online compiler like hackerrank or hackerearth since it takes more time for execution.Someone help me to find the solution for my time complexity problem.
I have tried to find the solution of the problem but i wasn't able to fix the performance by reducing the time..
Scanner scan = new Scanner(System. in );
String s = "aab";
String s1 = "";
String s2 = "";
int n1 = 0;
int length = 0;
long n = 882787;
long count = 0;
while (s1.length() < n) {
s1 = s1 + s;
}
if (s1.length() > n) {
count = s1.length() - n;
n1 = (int) count;
}
for (int i = 0; i < s1.length() - n1; i++) {
if (s1.charAt(i) == 'a') {
length += 1;
}
}
System.out.println(length);
Explanation of the above program:
I have a string s,in lowercase English letters that .I have repeat the string for n times and I store it in the new string.
I have to find the number of occurrences of 'a' in my new string
How do i actually reduce the time complexity for the above program
Thanks in advance
I would use a regular expression to create a String based on the initial input consisting of only letter 'a'(s). Take the length of that String and multiply it by n. That is one line that looks like
System.out.println(s.replaceAll("[^a]+", "").length() * n);
You are going to add s to the string n/s.length() times, call this N:
int N = n / s.length();
Each time you add s to the string you are going to append the number of As in s:
int a = 0;
for (int i = 0; i < s.length(); ++i) {
a += s.charAt(i) == 'a' ? 1 : 0;
}
// Or int a = s.replaceAll("[^a]", "").length();
So multiple these together:
int length = a * N;
String is immutable. Modification of a string is in fact create a new String object and put both old and new String into Java String constant poom
If you don't want to change your algorithm, I'd suggest to use StringBuilder to improve the speed of the execution. Note that StringBuilder is not thread safe
String s="aab";
int n1 = 0;
StringBuilder sb1 = new StringBuilder();
int length=0;
long n=882787;
long count=0;
while(sb1.length() < n) {
sb1.append(s);
}
if(sb1.length()>n) {
count =sb1.length()-n;
n1=(int)count;
}
for(int i=0;i<sb1.length()- n1;i++) {
if(sb1.charAt(i)=='a') {
length+=1;
}
}
System.out.println(length);
From here
When to use which one :
If a string is going to remain constant throughout the program, then
use String class object because a String object is immutable. If a
string can change (example: lots of logic and operations in the
construction of the string) and will only be accessed from a single
thread, using a StringBuilder is good enough. If a string can change,
and will be accessed from multiple threads, use a StringBuffer because
StringBuffer is synchronous so you have thread-safety.
I see multiple possible optimizations:
a) One pattern that is not that good is creating lots of Strings through repeated string concatenation. Each "s1 = s1 + s;" creates a new instance of String which will be obsolet the next time the command runs (It increases the load, because the String instances will be additional work for the Garbage Collector).
b) Generally: If you find, that your algorithm takes too long, then you should think about a complete new way to solve the issue. So a different solution could be:
- You know the length you want to have (n) and the length of the small string (s1) that you use to create the big string. So you can calculate: How often will the small string be inside the target string? How many characters are left?
==> You can simply check the small string for the character you are looking for. That multiplied by the number how often the small string will be inside the big string is the first result that you get.
==> Now you need to check the substring of the small string that are missing.
Example: n=10, s1="aab", Looking for "a":
So first we check how often the s1 will fit into a new string of n Characters n/length(s1) => 3
So we check how often the "a" is inside "aab" -> 2
First result is 3*2 = 6
But we checked for 3*3 = 9 characters so far, but we want 10 characters. So we need to check n % length(s1) = 1 character of s1 and in this substring ("a"), wie have 1 a, so we have to add 1.
So the result is 7 which we got without building a big string (which is not required at all!)
Just check how many times the char occurs in the original and multiple it by n. Here's a simple way to do so without even using regex:
// take these as function input or w/e
String s = "aab";
String find = "a";
long n = 882787;
int count = s.length() - s.replaceAll(find, "").length();
System.out.println(count * n);
This question already has answers here:
Memory efficient power set algorithm
(5 answers)
Closed 8 years ago.
I'm trying to find every possible anagram of a string in Java - By this I mean that if I have a 4 character long word I want all the possible 3 character long words derived from it, all the 2 character long and all the 1 character long. The most straightforward way I tought of is to use two nested for loops and iterare over the string. This is my code as of now:
private ArrayList<String> subsets(String word){
ArrayList<String> s = new ArrayList<String>();
int length = word.length();
for (int c=0; c<length; c++){
for (int i=0; i<length-c; i++){
String sub = word.substring(c, c+i+1);
System.out.println(sub);
//if (!s.contains(sub) && sub!=null)
s.add(sub);
}
}
//java.util.Collections.sort(s, new MyComparator());
//System.out.println(s.toString());
return s;
}
My problem is that it works for 3 letter words, fun yelds this result (Don't mind the ordering, the word is processed so that I have a string with the letters in alphabetical order):
f
fn
fnu
n
nu
u
But when I try 4 letter words, it leaves something out, as in catq gives me:
a
ac
acq
acqt
c
cq
cqt
q
qt
t
i.e., I don't see the 3 character long word act - which is the one I'm looking for when testing this method. I can't understand what the problem is, and it's most likely a logical error I'm making when creating the substrings. If anyone can help me out, please don't give me the code for it but rather the reasoning behind your solution. This is a piece of coursework and I need to come up with the code on my own.
EDIT: to clear something out, for me acq, qca, caq, aqc, cqa, qac, etc. are the same thing - To make it even clearer, what happens is that the string gets sorted in alphabetical order, so all those permutations should come up as one unique result, acq. So, I don't need all the permutations of a string, but rather, given a 4 character long string, all the 3 character long ones that I can derive from it - that means taking out one character at a time and returning that string as a result, doing that for every character in the original string.
I hope I have made my problem a bit clearer
It's working fine, you just misspelled "caqt" as "acqt" in your tests/input.
(The issue is probably that you're sorting your input. If you want substrings, you have to leave the input unsorted.)
After your edits: see Generating all permutations of a given string Then just sort the individual letters, and put them in a set.
Ok, as you've already devised your own solution, I'll give you my take on it. Firstly, consider how big your result list is going to be. You're essentially taking each letter in turn, and either including it or not. 2 possibilities for each letter, gives you 2^n total results, where n is the number of letters. This of course includes the case where you don't use any letter, and end up with an empty string.
Next, if you enumerate every possibility with a 0 for 'include this letter' and a 1 for don't include it, taking your 'fnu' example you end up with:
000 - ''
001 - 'u'
010 - 'n'
011 - 'nu'
100 - 'f'
101 - 'fu' (no offense intended)
110 - 'fn'
111 - 'fnu'.
Clearly, these are just binary numbers, and you can derive a function that given any number from 0-7 and the three letter input, will calculate the corresponding subset.
It's fairly easy to do in java.. don't have a java compiler to hand, but this should be approximately correct:
public string getSubSet(string input, int index) {
// Should check that index >=0 and < 2^input.length here.
// Should also check that input.length <= 31.
string returnValue = "";
for (int i = 0; i < input.length; i++) {
if (i & (1 << i) != 0) // 1 << i is the equivalent of 2^i
returnValue += input[i];
}
return returnValue;
}
Then, if you need to you can just do a loop that calls this function, like this:
for (i = 1; i < (1 << input.length); i++)
getSubSet(input, i); // this doesn't do anything, but you can add it to a list, or output it as desired.
Note I started from 1 instead of 0- this is because the result at index 0 will be the empty string. Incidentally, this actually does the least significant bit first, so your output list would be 'f', 'n', 'fn', 'u', 'fu', 'nu', 'fnu', but the order didn't seem important.
This is the method I came up with, seems like it's working
private void subsets(String word, ArrayList<String> subset){
if(word.length() == 1){
subset.add(word);
return;
}
else {
String firstChar = word.substring(0,1);
word = word.substring(1);
subsets(word, subset);
int size = subset.size();
for (int i = 0; i < size; i++){
String temp = firstChar + subset.get(i);
subset.add(temp);
}
subset.add(firstChar);
return;
}
}
What I do is check if the word is bigger than one character, otherwise I'll add the character alone to the ArrayList and start the recursive process. If it is bigger, I save the first character and make a recursive call with the rest of the String. What happens is that the whole string gets sliced in characters saved in the recursive stack, until I hit the point where my word has become of length 1, only one character remaining.
When that happens, as I said at the start, the character gets added to the List, now the recursion starts and it looks at the size of the array, in the first iteration is 1, and then with a for loop adds the character saved in the stack for the previous call concatenated with every element in the ArrayList. Then it adds the character on its own and unwinds the recursion again.
I.E., with the word funthis happens:
f saved
List empty
recursive call(un)
-
u saved
List empty
recursive call(n)
-
n.length == 1
List = [n]
return
-
list.size=1
temp = u + list[0]
List = [n, un]
add the character saved in the stack on its own
List = [n, un, u]
return
-
list.size=3
temp = f + list[0]
List = [n, un, u, fn]
temp = f + list[1]
List = [n, un, u, fn, fun]
temp = f + list[2]
List = [n, un, u, fn, fun, fu]
add the character saved in the stack on its own
List = [n, un, u, fn, fun, fu, f]
return
I have been as clear as possible, I hope this clarifies what was my initial problem and how to solve it.
This is working code:
public static void main(String[] args) {
String input = "abcde";
Set<String> returnList = permutations(input);
System.out.println(returnList);
}
private static Set<String> permutations(String input) {
if (input.length() == 1) {
Set<String> a = new TreeSet<>();
a.add(input);
return a;
}
Set<String> returnSet = new TreeSet<>();
for (int i = 0; i < input.length(); i++) {
String prefix = input.substring(i, i + 1);
Set<String> permutations = permutations(input.substring(i + 1));
returnSet.add(prefix);
returnSet.addAll(permutations);
Iterator<String> it = permutations.iterator();
while (it.hasNext()) {
returnSet.add(prefix + it.next());
}
}
return returnSet;
}
for(int j = 1;j<fileArray.size();j++) {
if(str.contains(fileArray.get(end+j))) {
}
}
(assume end is some number such as 30).
The goal of this part is when having a window length of 30 and a fileArray size > 30, check if theres anything after index 30 that matches whatever is inside the window.
ex: "i like to eat piesss aaaabbbbpiesssbbbb"
starting from the beginning of the string add the first 17 characters to a arraylist called window. then i check the rest of the string starting from right after window to see if there's anything that matches. space doesnt match so you add it to the output. keep checking then you see "piesss" matches. Then i replace the second "piesss" with wherever the first "piesss" occurs.
So right now im using fileArray.get(end+j) to check if there's anything that matches within my string(str) except this doesn't really work. Is there a way I could fix this code segment?
The replacement part of your question is still unclear. As is any reasoning to use an ArrayList. I've written some code that does a 5 character window search for a match after splitting the string you provided. Note how with the 30 and 17 values you gave nothing is ever matched (see commented out code). However with tweaked values some matches can be found.
public static void main(String[] args) {
// 1 2 3
//012345678901234567890123456789012345678 <- shows the index
String test = "i like to eat piesss aaaabbbbpiesssbbbb";
// int first = 17;
// int end = 30;
int first = 20;
int end = 37;
String firstHalf = test.substring(0, first);
String secondHalf = test.substring(first, end);
int matchSize = 5;
for (int i = 0; i + matchSize < secondHalf.length() ; i++)
{
String window = secondHalf.substring(i, i + matchSize);
if ( firstHalf.contains(window) )
{
System.out.println(window);
}
}
System.out.println("Done searching.");
}
Displays:
piess
iesss
Done searching.
If this isn't what you meant PLEASE edit your question to make your needs clear.
So I'm still shaky on how basic java works, and here is a method I wrote but don't fully understand how it works anyone care to explain?
It's supposed to take a value of s in and return it in its reverse order.
Edit: Mainly the for loop is what is confusing me.
So say I input "12345" I would want my output to be "54321"
Public string reverse(String s){
String r = "";
for(int i=0; i<s.length(); i++){
r = s.charAt(i) + r;
}
return r;
}
We do a for loop to the last index of String a , add tha carater of index i to the String s , add here is a concatenation :
Example
String z="hello";
String x="world";
==> x+z="world hello" #different to z+x ="hello world"
for your case :
String s="";
String a="1234";
s=a.charAt(0)+s ==> s= "1" + "" = "1" ( + : concatenation )
s=a.charAt(1)+s ==> s='2'+"1" = "21" ( + : concatenation )
s=a.charAt(2)+s ==> s='3'+"21" = "321" ( + : concatenation )
s=a.charAt(3)+s ==> s='3'+"321" = "4321" ( + : concatenation )
etc..
public String reverse(String s){
String r = ""; //this is the ouput , initialized to " "
for(int i=0; i<s.length(); i++){
r = s.charAt(i) + r; //add to String r , the caracter of index i
}
return r;
}
What this code does is the following
Create a new variable r="";
then looping for the string in input lenght it adds at the beginning of r the current character of the loop.
i=0) r="1"
i=1) r="21"
i=2) r="321"
i=3) r="4321"
i=4) r="54321"
When you enter the loop you are having empty string in r.
Now r=""
In 1st iteration, you are taking first character (i=0) and appending r to it.
r = "1" + "";
Now r=1
In 2nd iteration, you are taking second character (i=1) and appending r to it
r = "2" + "1";
Now r=21
You can trace execution on a paper like this, then you will easily understand what is happening.
What the method is doing is taking the each character from the string s and putting it at the front of the new string r. Renaming the variables may help illustrate this.
public String reverse(String s){
String alreadyReversed = "";
for(int i=0; i<s.length(); i++){
//perform the following until count i is as long as string s
char thisCharacterInTheString = s.charAt(i); // for i==0 returns first
// character in passed String
alreadyReversed = thisCharacterInTheString + alreadyReversed;
}
return alreadyReversed;
}
So in the first iteration of the for loop alreadyReversed equals 1 + itself (an empty string).
In the second iteration alreadyReversed equals 2 + itself (1).
Then 3 + itself (21).
Then 4 + 321.
Then 5 + 4321.
GO back to your problem statement (take an input string and produce an output string in reverse order). Then consider how you would do this (not how to write Java code to do this).
You would probably come up with two alternatives:
Starting at the back of the input string, get one character at a time and form a new string (thus reversing its order).
Starting at the front of the string, get a character. Then for each next character, put it in front of all the characters you have created so far.
Your pseudo code results might be like the following
Option 1
let l = the length of the input string
set the output string to ""
while l > 0
add the "lth" character of the input string to the output string
subtract 1 from l
Option 2 left as an exercise for the questioner.
Then you would consider how to write Java to handle your algorithm. You will find that there are several ways to get the "lth" character of a string. First, in Java a string of length l has characters in position 0 through l-1. You can use string.charAt(loc) or string.substring(loc,loc+1) to get the character at position loc
I have a basic String variable that contains the letter x a total of three times.
I have attempted to find x within the String using charAt, and then print the char and the next two characters next to it.
I have hit a snag within my code and would appreciate any help.
Here is my code.
public class StringX{
public static void main(String[] args){
String ss = "xarxatxm";
char first = ss.charAt(0);
char last == ss.charAt(3);
if(first == "x"){
String findx = ss.substring(0, 2);
}
if(last == "x"){
String findX = ss.substring(3, 5);
}
System.out.print(findx + findX);
}
}
Also, is there a way to implement the for loop to cycle through the String looking for x also?
I just need some advice to see where my code is going wrong.
You cannot find characters using charAt - it's for getting a character once you know where it is.
Is there a way to implement the for loop to cycle through the String looking for x also?
You need to use indexOf for finding positions of characters. Pass the initial position which is the position of the last x that you found so far to get the subsequent position.
For example, the code below
String s = "xarxatxm";
int pos = -1;
while (true) {
pos = s.indexOf('x', pos+1);
if (pos < 0) break;
System.out.println(pos);
}
prints 0 3 6 for the three positions of 'x' in the string.