I would like to be able to generate all possible strings from a given length, and I frankly don't know how to code that. So for further explanation, I and a friend would like to demonstrate some basic hacking techniques, so bruteforcing comes up. Of course, he will be my victim, no illegal thing there.
However, the only thing he told me is that his PW will be 4-char-long, but I'm pretty sure his PW won't be in any dictionnary, that would be toi easy.
So I came up with the idea of generating EVERY 4-char-long-string possible, containing a-z characters (no caps).
Would someone have a lead to follow to code such an algorithm ? I don't really bother with performances, if it takes 1 night to generate all PW, that's no problem.
Don't forget, that's only on demonstration purposes.
You can do it just how you'd do it with numbers. Start with aaaa. Then increment the 'least significant' part, so aaab. Keep going until you get to aaaz. Then increment to aaba. Repeat until you get to zzzz.
So all you need to do is implement is
String getNext(String current)
To expand on this; It possibly isnt the quickest way of doing things, but it is the simplest to get right.
As the old adage goes - 'first make it right, then make it fast'. Getting a working implementation that passes all your tests (you do have tests, right?) is what you do first. You then rewrite it to make it fast, using your tests as reassurance you're not breaking the core functionality.
The absolutely simplest way is to use four nested loops:
char[] pw = new char[4];
for (pw[0] = 'a' ; pw[0] <= 'z' ; pw[0]++)
for (pw[1] = 'a' ; pw[1] <= 'z' ; pw[1]++)
for (pw[2] = 'a' ; pw[2] <= 'z' ; pw[2]++)
for (pw[3] = 'a' ; pw[3] <= 'z' ; pw[3]++)
System.out.println(new String(pw));
This does not scale well, because adding extra characters requires adding a level of nesting. Recursive approach is more flexible, but it is harder to understand:
void findPwd(char[] pw, int pos) {
if (pos < 0) {
System.out.println(new String(pwd));
return;
}
for (pw[pos] = 'a' ; pw[pos] <= 'z' ; pw[pos]++)
findPwd(pw, pos-1);
}
Call recursive method like this:
char[] pw = new char[4];
findPwd(pw, 3);
private static void printAllStringsOfLength(int len) {
char[] guess = new char[len];
Arrays.fill(guess, 'a');
do {
System.out.println("Current guess: " + new String(guess));
int incrementIndex = guess.length - 1;
while (incrementIndex >= 0) {
guess[incrementIndex]++;
if (guess[incrementIndex] > 'z') {
if (incrementIndex > 0) {
guess[incrementIndex] = 'a';
}
incrementIndex--;
}
else {
break;
}
}
} while (guess[0] <= 'z');
}
public class GenerateCombinations {
public static void main(String[] args) {
List<Character> characters = new ArrayList<Character>();
for (char c = 'a'; c <= 'z'; c++) {
characters.add(c);
}
List<String> allStrings = new ArrayList<String>();
for (Character c : characters) {
for (Character d : characters) {
for (Character e : characters) {
for (Character f : characters) {
String s = "" + c + d + e + f;
allStrings.add(s);
}
}
}
}
System.out.println(allStrings.size()); // 456 976 combinations
}
}
This is something you can do recursively.
Lets define every (n)-character password the set of all (n-1)-character passwords, prefixed with each of the letters a thru z. So there are 26 times as many (n)-character passwords as there are (n-1)-character passwords. Keep in mind that this is for passwords consisting of lower-case letters. Obviously, you can increase the range of each letter quite easily.
Now that you've defined the recursive relationship, you just need the terminating condition.
That would be the fact that there is only one (0)-character password, that being the empty string.
So here's the recursive function:
def printNCharacterPasswords (prefix, num):
if num == 0:
print prefix
return
foreach letter in 'a'..'z':
printNCharacterPasswords (prefix + letter, num - 1)
to be called with:
printNCharacterPasswords ("", 4)
And, since Python is such a wonderful pseudo-code language, you can see it in action with only the first five letters:
def printNCharacterPasswords (prefix, num):
if num == 0:
print prefix
return
for letter in ('a', 'b', 'c', 'd', 'e'):
printNCharacterPasswords (prefix + letter, num - 1)
printNCharacterPasswords ("", 2)
which outputs:
aa
ab
ac
ad
ae
ba
bb
bc
bd
be
ca
cb
cc
cd
ce
da
db
dc
dd
de
ea
eb
ec
ed
ee
A aroth points out, using a digit counter approach is faster. To make this even faster, you can use a combination of an inner loop for the last digit and a counter for the rest (so the number of digits can be variable)
public static void main(String... args) {
long start = System.nanoTime();
int letters = 26;
int count = 6;
final int combinations = (int) Math.pow(letters, count);
char[] chars = new char[count];
Arrays.fill(chars, 'a');
final int last = count - 1;
OUTER:
while (true) {
for (chars[last] = 'a'; chars[last] <= 'z'; chars[last]+=2) {
newComination(chars);
chars[last]++;
newComination(chars);
}
UPDATED:
{
for (int i = last - 1; i >= 0; i--) {
if (chars[i]++ >= 'z')
chars[i] = 'a';
else
break UPDATED;
}
// overflow;
break OUTER;
}
}
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to generate %,d combinations%n", time / 1e9, combinations);
}
private static void newComination(char[] chars) {
}
prints
Took 0.115 seconds to generate 308,915,776 combinations
Note: the loop is so simple, its highly likely that the JIT can eliminate key pieces of code (after in-lining newCombination) and the reason its so fast is its not really calculating every combination.
A simpler way to generate combinations.
long start = System.nanoTime();
int letters = 26;
int count = 6;
final int combinations = (int) Math.pow(letters, count);
StringBuilder sb = new StringBuilder(count);
for (int i = 0; i < combinations; i++) {
sb.setLength(0);
for (int j = 0, i2 = i; j < count; j++, i2 /= letters)
sb.insert(0, (char) ('a' + i2 % letters));
// System.out.println(sb);
}
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to generate %,d combinations%n", time / 1e9, combinations);
prints
aaaa
aaab
aaac
....
zzzx
zzzy
zzzz
Took 0.785 seconds to generate 456,976 combinations
It spends most of its time waiting for the screen to update. ;)
If you comment out the line which prints the combinations, and increase the count to 5 and 6
Took 0.671 seconds to generate 11,881,376 combinations
Took 15.653 seconds to generate 308,915,776 combinations
public class AnagramEngine {
private static int[] anagramIndex;
public AnagramEngine(String str) {
AnagramEngine.generate(str);
}
private static void generate(String str) {
java.util.Map<Integer, Character> anagram = new java.util.HashMap<Integer, Character>();
for(int i = 0; i < str.length(); i++) {
anagram.put((i+1), str.charAt(i));
}
anagramIndex = new int[size(str.length())];
StringBuffer rev = new StringBuffer(AnagramEngine.start(str)+"").reverse();
int end = Integer.parseInt(rev.toString());
for(int i = AnagramEngine.start(str), index = 0; i <= end; i++){
if(AnagramEngine.isOrder(i))
anagramIndex[index++] = i;
}
for(int i = 0; i < anagramIndex.length; i++) {
StringBuffer toGet = new StringBuffer(anagramIndex[i] + "");
for(int j = 0; j < str.length(); j++) {
System.out.print(anagram.get(Integer.parseInt(Character.toString(toGet.charAt(j)))));
}
System.out.print("\n");
}
System.out.print(size(str.length()) + " iterations");
}
private static boolean isOrder(int num) {
java.util.Vector<Integer> list = new java.util.Vector<Integer>();
String str = Integer.toString(num);
char[] digits = str.toCharArray();
for(char vecDigits : digits)
list.add(Integer.parseInt(Character.toString(vecDigits)));
int[] nums = new int[str.length()];
for(int i = 0; i < nums.length; i++)
nums[i] = i+1;
for(int i = 0; i < nums.length; i++) {
if(!list.contains(nums[i]))
return false;
}
return true;
}
private static int start(String str) {
StringBuffer num = new StringBuffer("");
for(int i = 1; i <= str.length(); i++)
num.append(Integer.toString(i));
return Integer.parseInt(num.toString());
}
private static int size(int num) {
int size;
if(num == 1) {
return 1;
}
else {
size = num * size(num - 1);
}
return size;
}
public static void main(final String[] args) {
final java.util.Scanner sc = new java.util.Scanner(System.in);
System.out.printf("\n%s\t", "Entered word:");
String word = sc.nextLine();
System.out.printf("\n");
new AnagramEngine(word);
}
}
Put all the characters you expect the password to contain into an array. Write a stub function to test if your algorithm finds the correct password. Start with passwords of length 1, work your way up to 4 and see if your fake password is found on each iteration.
you can use the following code for getting random string. It will return you a string of 32 chars. you can get string of desired length by using substring(). Like if you want a string with 10 chars then:
import java.security.SecureRandom;
import java.math.BigInteger;
SecureRandom srandom = new SecureRandom();
String rand = new BigInteger(176, srandom).toString(32);
rand.substring(0,7);
Related
I am currently a beginner in programming and I am trying to write a program in java to convert binary in hexadecimal numbers.
I know that the program will have to divide the number in groups of 4 and convert them to hexadecimal.
Ex: 11101111 (b2) --> E + F --- EF
However, since I used ints to do the conversion of the numbers, I'm stuck when I need to print a letter because it is a String.
Can someone point me to the right way? What am I doing wrong? I've also tried another version with an auxiliary array to store each group of 4 digits but I can't manage to insert a proper dimension to the array.
Unfortunately I am not allowed to use any function other than Scanner and Math, the method lenght and charAt and the basic stuff. I can't modify the public static line either.
EDIT: So after your inputs and so many tries, I managed to get this code. However it gives me an error if I insert too many numbers, eg: 0111011010101111. I've tried to change int to double but that didn't fix the problem.
import java.util.Scanner;
public class Bin2HexString {
public static void main(String[] args) {
Scanner keyb = new Scanner(System.in);
System.out.println("Valor?");
int vlr = keyb.nextInt();
String num = "";
int aux = vlr;
// Hexadecimal numbers
String arr[] = {"0","1","2","3","4","5","6","7","8","9","A", "B", "C", "D", "E", "F"};
String bits[] = {"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
String letters = "";
//Divide in groups of 4
int r;
for (; aux > 0; ) {
r = aux % 10000;
aux = aux / 10000;
num = "" + r;
for (;num.length() < 4;) { //add missing zeros
String zero = "0";
num = zero + num;
}
int charint = 0,bitint = 0;
for (int i = 0; i < arr.length;i++) {
String aux2 = bits[i];
String aux3 = arr[i];
for (int j = 0; j < num.length();j++) { // compare each group with arr[i]
char charvl = num.charAt(j);
char bitsvl = aux2.charAt(j);
charint = ((int) (charvl)-'0');
bitint = ((int) (bitsvl) - '0');
if (bitint != charint)
break;
}
if (bitint == charint)
letters = aux3 + "" + letters;
}
}
System.out.println(letters);
}
}
Having thought about this for a while to determine the most effective and useful way to do this is to write methods which convert a string from any base between 2 and 16 to an int and back to a string again.
This way you have useful methods for other things. And note that they methods can be easily changed and names to simply hard code the desired radix into the method to limit it to binary and hex methods.
The indexOf utility method was written to avoid using the builtin String method.
final static String hex = "0123456789ABCDEF";
static int stringToInt(String str, int radix) {
if (radix < 2 || radix > 16) {
System.out.println("Base must be between 2 and 16 inclusive");
return -1;
}
int v = 0;
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
int idx = indexOf(hex, c);
if (idx < 0 || idx > radix) {
System.out.println("Illegal character in string (" + c + ")");
}
v = v * radix + idx;
}
return v;
}
static String intToBase(int v, int radix) {
if (radix < 2 || radix > 16) {
System.out.println("Base must be between 2 and 16 inclusive");
return null;
}
String s = "";
while (v > 0) {
int idx = v % radix;
s = hex.charAt(idx) + s;
v /= radix;
}
return s;
}
static int indexOf(String str, char c) {
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == c) {
return i;
}
}
return -1;
}
And here is an example of their use.
// generate some test data
Random r = new Random(23);
String[] bitStrings =
r.ints(20, 20, 4000).mapToObj(Integer::toBinaryString).toArray(
String[]::new);
for (String bitstr : bitStrings) {
int v = baseToInt(bitstr, 2);
String hex = intToBase(v, 16);
System.out.printf("%12s = %s%n", bitstr, hex);
}
Which prints the following:
101110000011 = B83
111001111100 = E7C
10001110111 = 477
100110001111 = 98F
111001010 = 1CA
111001001111 = E4F
111000011010 = E1A
100001010010 = 852
11011001101 = 6CD
111010010111 = E97
Just some quick notes:
First this is wrong:
//Divide in groups of 4
for (; aux > 0; ) {
r = aux % 10000;
aux = aux / 10000;
Not at all what you want to do. Try it by hand and see what happens. Take a simple number that you know the answer to, and try it. You won't get the right answer. A good test is 17, which is 11 hex.
Try this instead: convert directly to the base you want. Hex is base 16 (its radix is 16), so you use 16 instead.
//Divide in groups of 4
for (; aux > 0; ) {
r = aux % 16;
aux = aux / 16;
Try those numbers with the test case, which is 17, and see what you get. That will get you much closer.
I'm assuming by "without methods" in the title, you are attempting to write your own integer parsing method instead of using Scanner.nextInt(int radix). In that case, my first advice would be work with a string instead of an integer - you'll be able to handle larger numbers and you can simply make an array of substrings (length 4) to convert to letters.
So, if you use the string approach - first scan in a string, not an int. Then I'd recommend a hash table with the 4-bit strings as keys and the hexadecimal equivalents as values. That should make calculation quite fast.
e.g.
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class HashMapBin2Hex
{
public static void main(String[] args)
{
//Read the string in
Scanner sc = new Scanner(System.in);
System.out.println("Binary number?");
String bin = sc.nextLine();
//Pad the bitstring with leading zeros to make a multiple of four
String zeros = "";
int i;
if (bin.length() % 4 != 0)
{
for (i = 0; i < 4 - (bin.length() % 4); i++)
{
zeros += "0";
}
}
bin = zeros + bin;
//Split the padded string into 4-bit chunks
String[] chunks = new String[bin.length() / 4];
for (i = 0; (i * 4) < bin.length() - 1; i++)
{
chunks[i] = bin.substring(i * 4, (i * 4) + 4);
}
//Convert the chunks to hexadecimal
String hex = "";
Map<String, String> bin2hex = new HashMap<>();
bin2hex.put("0000", "0");
bin2hex.put("0001", "1");
bin2hex.put("0010", "2");
bin2hex.put("0011", "3");
bin2hex.put("0100", "4");
bin2hex.put("0101", "5");
bin2hex.put("0110", "6");
bin2hex.put("0111", "7");
bin2hex.put("1000", "8");
bin2hex.put("1001", "9");
bin2hex.put("1010", "A");
bin2hex.put("1011", "B");
bin2hex.put("1100", "C");
bin2hex.put("1101", "D");
bin2hex.put("1110", "E");
bin2hex.put("1111", "F");
for (String s : chunks)
{
hex += bin2hex.get(s);
}
System.out.println("Hexadecimal: " + hex);
sc.close();
}
}
Further iterations could have some error checking to prevent catastrophic failure in the case of characters other than 0 or 1.
And of course, if you're fine with the other way (builtins), the following is far easier and more robust (ie will throw an exception if the string contains anything other than 0s and 1s):
import java.util.Scanner;
public class BuiltinBin2Hex
{
public static void main(String[] args)
{
//Read the binary number in
Scanner sc = new Scanner(System.in);
System.out.println("Binary number?");
int bin = sc.nextInt(2);
//And print as hexadecimal
System.out.println("Hexadecimal: " + Integer.toString(bin, 16));
sc.close();
}
}
import java.util.Calendar;
public class MainClass
{
public static void main(String args[])
{
String s = new String("ABCD");
long swapStart = Calendar.getInstance().getTimeInMillis();
for(int i=0; i<s.length()/2;i++)
{
char left = s.charAt(i);
char right = s.charAt(s.length()-(i+1));
s=s.substring(0, i)+right+s.substring(i+1, s.length()-(i+1))+left+s.substring(s.length()-i, s.length());
}
long swapStop = Calendar.getInstance().getTimeInMillis();
long bufStart = Calendar.getInstance().getTimeInMillis();
String str = new String("ABCD");
StringBuffer strBuf = new StringBuffer(str);
str = strBuf.reverse().toString();
long bufStop = Calendar.getInstance().getTimeInMillis();
System.out.println(swapStop-swapStart);
System.out.println(bufStop-bufStart);
}
}
***** in the new String("ABCD") of the string if i provide a really big string say couple of hundreds of alpha numerics
*****in the console output is :
61
0
*****the stringbuffer always calculated in 0 milli seconds and my char swapping algo takes as per the string size
Q. how come my swap algo can't do it in 0 milli seconds and why stringbuffer always does it in 0 milli seconds ?
I checked the Java Source Code and StringBuffer.reverse() is implemented as follows :
public AbstractStringBuilder reverse() {
boolean hasSurrogate = false;
int n = count - 1;
for (int j = (n-1) >> 1; j >= 0; --j) {
char temp = value[j];
char temp2 = value[n - j];
if (!hasSurrogate) {
hasSurrogate = (temp >= Character.MIN_SURROGATE && temp <= Character.MAX_SURROGATE)
|| (temp2 >= Character.MIN_SURROGATE && temp2 <= Character.MAX_SURROGATE);
}
value[j] = temp2;
value[n - j] = temp;
}
if (hasSurrogate) {
// Reverse back all valid surrogate pairs
for (int i = 0; i < count - 1; i++) {
char c2 = value[i];
if (Character.isLowSurrogate(c2)) {
char c1 = value[i + 1];
if (Character.isHighSurrogate(c1)) {
value[i++] = c1;
value[i] = c2;
}
}
}
}
return this;
}
Q. Please explain the surrogate thing.
Q1: There is a performance difference because your swapping code creates a lot of String objects that get operated on, while the other routine works directly with a char array and requires no additional object creation. Let's examine your line of code:
s=s.substring(0, i)+right+s.substring(i+1, s.length()-(i+1))+left+s.substring(s.length()-i, s.length());
It does this kind of work:
s = [new String 1] + char + [new String 2] + char + [new String 3]
That's 4 new String objects (the three shown, plus the one resulting from the addition of String objects. Also, each of the 3 new strings shown has a call to substring which takes time to process. In addition you do all of this work in a for-loop so it gets repeated for each character!
String manipulation is convenient but expensive. Array manipulation is direct, requires no additional object or blocks of memory, so is much faster.
Q2: Surrogates are a special case of unicode character created to handle longer unicode characters. See this article for more details. The order of the hi/low parts of the surrogate are important so it would be wrong for the swap code to reverse those two characters, so if they are found, their order is put back the way it should be.
I want to exhaustively test a String matching algorithm, named myAlgo(Char[] a, Char[] b)
The exhaustive test includes a no. of different char letters, alplhabet " l ", in an "n" long array. The test then computes all combinations, while comparing it with all combinations of another array with similar properties (Like truth tables),e.g.
I have not been able to either compute something that would generate every combination of the array of size n and alphabet l, niether have I been able to make code that is able to combine the computation into iterative testcases (test all the combinations of the two arrays compared), though with code that would be able to generate the combinations, making a nested for-loop should do the required testing.
My goal is to break my algorithm by making it compute something it should not compute.
Test(char[] l, int n)
l = [a;b] //a case could be
n = 2 //a case could be
myAlgo([a;a],[a;a]); //loops over my algorithm in the following way
myAlgo([a;b],[a;a]);
myAlgo([b;a],[a;a]);
myAlgo([b;b],[a;a]);
myAlgo([a;a],[a;b]);
myAlgo([a;b],[a;b]);
myAlgo([b;a],[a;b]);
myAlgo([b;b],[a;b]);
myAlgo([a;a],[b;a]);
myAlgo([a;b],[b;a]);
...
myAlgo([b;b],[b;b]);
My own solution (only works for a finite set of "l") and also starts printing wierd outputs on later iterations.
public class Test {
//aux function to format chars
public static String concatChar(char [] c){
String s = "";
for(char cc : c){
s += cc;
}
return s;
}
public static void main(String[] args) {
String ss1 = "AA"; //TestCases, n = 2
String ss2 = "AA";
char[] test1 = ss1.toCharArray();
char[] test2 = ss2.toCharArray();
Fordi fordi = new Fordi(); //my algorithm
TestGenerator tGen = new TestGenerator(); //my testGenerator
for(int i=0; i<Math.pow(4.0, 2.0);i++){ //to test all different cases
for(int j=0; j<Math.pow(4.0, 2.0);j++){
int k = fordi.calculate(test1, test2); //my algorithm
String mys1 = concatChar(test1); //to print result
String mys2 = concatChar(test2); //to print result
System.out.println(mys1 + " - " + mys2);
System.out.println(k);
test2 = tGen.countArray(test2); //"flip" one number
}
test2 = ss1.toCharArray();
test1 = tGen.countArray(test1); //"flip"
}
}
}
My arrayflipper code:
public char[] countArray(char[] a){
int i=0;
while(i<a.length){
switch (a[i]){
case 'A':
a[i]='B';
clearBottom(a,i);
return a;
case 'B':
a[i]='C';
clearBottom(a,i);
return a;
case 'C':
a[i]='D';
clearBottom(a,i);
return a;
case 'D':
i++;
break;
default:
System.out.println("Something went terribly wrong!");
}
}
return a;
}
public char[] clearBottom(char [] a, int i){
while(i >0){
i--;
a[i] = 'A';
}
return a;
}
As I understand it, your goal is to create all n-character long strings (stored individually as elements in an array) consisting of letters in the L letter alphabet?
One way to accomplish this is to order your letters (A=0, B=1, C=2, etc). Then you can, from a starting string of AAA...AAA (n-characters long) just keep adding 1. Essentially you implement an addition algorithm. Adding 1 would turn an A=0 into a B=1. For example, n=3 and L=3:
start: AAA (0,0,0).
Adding 1 becomes AAB (0,0,1)
Adding 1 again become AAC (0, 0, 2)
Adding 1 again (since we are out of letters, now we carry a bit over) ABA (0, 1, 0).
You can boil the process down to looking for the right-most number that is not maxed out and add 1 to it (then all digits to the right of that digit go back to zero). So in the string ABCCC, the B digit is the right-most not maxed out digit, it goes up by 1 and becomes a C, then all the maxed out digits to the right go back to 0 (A) leaving ACAAA as the next string.
Your algorithm just repeatedly adds 1 until all the elements in the string are maxed out.
Instead of using a switch statement, I recommend putting every character you want to test (A, B, C, D) into an array, and then using the XOR operation to calculate the index of each character from the iteration number in a manner similar to the following:
char[] l = new char[]{'A','B','C','D'};
int n = 2;
char[] test1 = new char[n];
char[] test2 = new char[n];
int max = (int)Math.pow(l.length, n);
for (int i = 0; i < max; i++) {
for (int k = 0; k < n; k++) {
test2[k] = l[(i % (int)Math.pow(l.length, k + 1)) / (int)Math.pow(l.length, k)];
}
for (int j = 0; j < max; j++) {
for (int k = 0; k < n; k++) {
test1[k] = l[(j % (int)Math.pow(l.length, k + 1)) / (int)Math.pow(l.length, k)];
}
int k = fordi.calculate(test1, test2);
System.out.println(new String(test1) + "-" + new String(test2));
System.out.println(k);
}
}
You can add more characters to l as well as increase n and it should still work. Of course, this can be further optimized, but you should get the idea. Hope this answer helps!
I have some random string with unknown content, what is known is that the content is alphanumeric and in lower case.
I am looking for a simple method to upper case a random number of the alpha characters in that string. The higher the randomness the better.
I can think of a few ways to do this, but none of them seem very optimal.
alright first solution:
public String randomizeCase(String myString){
Random rand = new Random();
StringBuilder build = new StringBuilder();
for(char c: myString.toCharArray()){
String s = new String(c);
if(Character.isLetter(c) && rand.nextBoolean()){
s = s.toUpperCase();
}
build.append(s);
}
return build.toString();
}
I dont like this solution because:
50% chance that every char is uppercased does not equal 50% chance that 50% of the chars are uppercased
There is a chance that nothing is upped cased
char to string conversion is ugly
The solution depends on the probabilistic model you choose. If for example you decide on binomial distribution, then you can traverse the chars, and switch every char to uppercase with a fixed probability p. The expected number of uppercase letters will be p * str.length():
public static String randomUpper(String str, double p) {
StringBuilder sb = new StringBuilder(str.length());
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (Character.isLetter(c) && Math.random() < p)
c = Character.toUpperCase(c);
sb.append(c);
}
return sb.toString();
}
If on the other hand you want to decide on the exact number of upercase letters for a given string, then the problem becomes a random sample problem (i.e. choose M positions to switch out of N positions in the string). This can be much faster than the first approach, when M is much smaller than N (though with Java's immutable strings the difference becomes minor because you have to copy the whole string anyway).
-- edit --
Now that you clarified the requirements, consider the following:
public static String randomUpper2(String str, double p) {
int letters = 0;
for (int i = 0; i < str.length(); i++) {
if (Character.isLetter(str.charAt(i)))
letters++;
}
int toChoose = (int) (p * letters);
StringBuilder sb = new StringBuilder(str.length());
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (Character.isLetter(c)) {
if (Math.random() < (toChoose/(double)letters)) {
c = Character.toUpperCase(c);
toChoose--;
}
letters--;
}
sb.append(c);
}
return sb.toString();
}
This code performs a random sample "on the fly", considering only alpha chars, as required. Use p=0.5 to switch exactly half of the letters.
Here is the code snippet for random sample problem (thanks Eyal for naming it). Not sure if that is what you are looking for.
Be aware, that this solution would run into an infinete loop if not enough lowercase letters are in the string. So you would need to tackle that as well, but I guess it is a starting point. ;-)
String myString = "9aie3ra3nr23rr5r21t";
System.out.println(upperCaseRandom(myString, 10));
public static String upperCaseRandom(String input, int n) {
StringBuilder output = new StringBuilder(input);
Random r = new Random();
for (int i = 0; i < n; i++) {
// Pick a place
int position = r.nextInt(input.length());
// Check if lowercase alpha
if (Character.isLowerCase(output.charAt(position))) {
output.setCharAt(position, Character.toUpperCase(output.charAt(position)));
} else {
i--;
}
}
return output.toString();
}
Edit:
Here is an improved version. It does change exactly n lowercase letters into uppercase letters (if there are enough, otherwise it changes all of them). The programm does not run into infinite loops, but still running time is a problem though.
public static String upperCaseRandom(String input, int n) {
final int length = input.length();
final StringBuilder output = new StringBuilder(input);
final boolean[] alreadyChecked = new boolean[length];
final Random r = new Random();
for (int i = 0, checks = 0; i < n && checks < length; i++) {
// Pick a place
int position = r.nextInt(length);
// Check if lowercase alpha
if (!alreadyChecked[position]) {
if (Character.isLowerCase(output.charAt(position))) {
output.setCharAt(position, Character.toUpperCase(output.charAt(position)));
} else {
i--;
}
checks++;
alreadyChecked[position] = true;
} else {
i--;
}
}
return output.toString();
}
I tried with
String lowerCasedRandomString = "4210281f-76ac-96b5-ed54-5458abf788d0";
String upperCasedRandomString = "4210281F-76AC-96B5-ED54-5458ABF788D0";
System.out.println(lowerCasedRandomString.toUpperCase());
System.out.println(upperCasedRandomString.toLowerCase());
I got the output
4210281F-76AC-96B5-ED54-5458ABF788D0
4210281f-76ac-96b5-ed54-5458abf788d0
What is the most efficient way to reverse a string in Java? Should I use some sort of xor operator? The easy way would be to put all the chars in a stack and put them back into a string again but I doubt that's a very efficient way to do it.
And please do not tell me to use some built in function in Java. I am interested in learning how to do it not to use an efficient function but not knowing why it's efficient or how it's built up.
You say you want to know the most efficient way and you don't want to know some standard built-in way of doing this. Then I say to you: RTSL (read the source, luke):
Check out the source code for AbstractStringBuilder#reverse, which gets called by StringBuilder#reverse. I bet it does some stuff that you would not have considered for a robust reverse operation.
The following does not deal with UTF-16 surrogate pairs.
public static String reverse(String orig)
{
char[] s = orig.toCharArray();
int n = s.length;
int halfLength = n / 2;
for (int i=0; i<halfLength; i++)
{
char temp = s[i];
s[i] = s[n-1-i];
s[n-1-i] = temp;
}
return new String(s);
}
You said you don't want to do it the easy way, but for those Googling you should use StringBuilder.reverse:
String reversed = new StringBuilder(s).reverse().toString();
If you need to implement it yourself, then iterate over the characters in reverse order and append them to a StringBuilder. You have to be careful if there are (or can be) surrogate pairs, as these should not be reversed. The method shown above does this for you automatically, which is why you should use it if possible.
An old post & question, however still did not see answers pertaining to recursion. Recursive method reverse the given string s, without relaying on inbuilt jdk functions
public static String reverse(String s) {
if (s.length() <= 1) {
return s;
}
return reverse(s.substring(1)) + s.charAt(0);
}
`
The fastest way would be to use the reverse() method on the StringBuilder or StringBuffer classes :)
If you want to implement it yourself, you can get the character array, allocate a second character array and move the chars, in pseudo code this would be like:
String reverse(String str) {
char[] c = str.getCharArray
char[] r = new char[c.length];
int end = c.length - 1
for (int n = 0; n <= end; n++) {
r[n] = c[end - n];
}
return new String(r);
}
You could also run half the array length and swap the chars, the checks involved slow things down probably.
I'm not really sure by what you mean when you say you need an efficient algorithm.
The ways of reversing a string that I can think of are (they are all already mentioned in other answers):
Use a stack (your idea).
Create a new reversed String by adding characters one by one in reverse order from the original String to a blank String/StringBuilder/char[].
Exchange all characters in the first half of the String with its corresponding position in the last half (i.e. the ith character gets swapped with the (length-i-1)th character).
The thing is that all of them have the same runtime complexity: O(N). Thus it cannot really be argued that any one is any significantly better than the others for very large values of N (i.e. very large strings).
The third method does have one thing going for it, the other two require O(N) extra space (for the stack or the new String), while it can perform swaps in place. But Strings are immutable in Java so you need to perform swaps on a newly created StringBuilder/char[] anyway and thus end up needing O(N) extra space.
public class ReverseInPlace {
static char[] str=null;
public static void main(String s[]) {
if(s.length==0)
System.exit(-1);
str=s[0].toCharArray();
int begin=0;
int end=str.length-1;
System.out.print("Original string=");
for(int i=0; i<str.length; i++){
System.out.print(str[i]);
}
while(begin<end){
str[begin]= (char) (str[begin]^str[end]);
str[end]= (char) (str[begin]^str[end]);
str[begin]= (char) (str[end]^str[begin]);
begin++;
end--;
}
System.out.print("\n" + "Reversed string=");
for(int i=0; i<str.length; i++){
System.out.print(str[i]);
}
}
}
I think that if you REALLY don't have performance problem you should just go with the most readable solution which is:
StringUtils.reverse("Hello World");
private static String reverse(String str) {
int i = 0;
int j = str.length()-1;
char []c = str.toCharArray();
while(i <= j){
char t = str.charAt(i);
c[i] = str.charAt(j);
c[j]=t;
i++;
j--;
}
return new String(c);
}
If you do not want to use any built in function, you need to go back with the string to its component parts: an array of chars.
Now the question becomes what is the most efficient way to reverse an array? The answer to this question in practice also depends upon memory usage (for very large strings), but in theory efficiency in these cases is measured in array accesses.
The easiest way is to create a new array and fill it with the values you encounter while reverse iterating over the original array, and returning the new array. (Although with a temporary variable you could also do this without an additional array, as in Simon Nickersons answer).
In this way you access each element exactly once for an array with n elements. Thus giving an efficiency of O(n).
I would simply do it this way without a use of any single util function. Just the String class is sufficient.
public class MyStringUtil {
public static void main(String[] args) {
String reversedString = reverse("StringToReverse");
System.out.println("Reversed String : " + reversedString);
}
/**
* Reverses the given string and returns reversed string
*
* #param s Input String
* #returns reversed string
*/
private static String reverse(String s) {
char[] charArray = s.toCharArray(); // Returns the String's internal character array copy
int j = charArray.length - 1;
for (int i = 0; charArray.length > 0 && i < j; i++, j--) {
char ch = charArray[i];
charArray[i] = charArray[j];
charArray[j] = ch;
}
return charArray.toString();
}
}
Check it. Cheers!!
Using String:
String abc = "abcd";
int a= abc.length();
String reverse="";
for (int i=a-1;i>=0 ;i--)
{
reverse= reverse + abc.charAt(i);
}
System.out.println("Reverse of String abcd using invert array is :"+reverse);
Using StringBuilder:
String abc = "abcd";
int a= abc.length();
StringBuilder sb1 = new StringBuilder();
for (int i=a-1;i>=0 ;i--)
{
sb1= sb1.append(abc.charAt(i));
}
System.out.println("Reverse of String abcd using StringBuilder is :"+sb1);
One variant can be, swapping the elements.
int n = length - 1;
char []strArray = str.toCharArray();
for (int j = 0; j < n; j++) {
char temp = strArray[j];
char temp2 = strArray[n];
strArray[j] = temp2;
strArray[n] = temp;
n--;
}
public static void main(String[] args){
String string ="abcdefghijklmnopqrstuvwxyz";
StringBuilder sb = new StringBuilder(string);
sb.reverse();
System.out.println(sb);
}
public static String Reverse(String word){
String temp = "";
char[] arr = word.toCharArray();
for(int i = arr.length-1;i>=0;i--){
temp = temp+arr[i];
}
return temp;
}
char* rev(char* str)
{
int end= strlen(str)-1;
int start = 0;
while( start<end )
{
str[start] ^= str[end];
str[end] ^= str[start];
str[start]^= str[end];
++start;
--end;
}
return str;
}
=========================
Wondering how it works?
First operation:
x1 = x1 XOR x2
x1: 1 0 0
x2: 1 1 1
New x1: 0 1 1
Second operation
x2 = x2 XOR x1
x1: 0 1 1
x2: 1 1 1
New x2: 1 0 0
//Notice that X2 has become X1 now
Third operation:
x1 = x1 XOR x2
x1: 0 1 1
x2: 1 0 0
New x1: 1 1 1
//Notice that X1 became X2
public static string getReverse(string str)
{
char[] ch = str.ToCharArray();
string reverse = "";
for (int i = str.Length - 1; i > -1; i--)
{
reverse += ch[i];
}
return reverse;
}
//using in-built method reverse of Array
public static string getReverseUsingBulidingFunction(string str)
{
char[] s = str.ToCharArray();
Array.Reverse(s);
return new string(s);
}
public static void Main(string[] args)
{
string str = "123";
Console.WriteLine("The reverse string of '{0}' is: {1}",str,getReverse(str));
Console.WriteLine("The reverse string of '{0}' is: {1}", str, getReverseUsingBulidingFunction(str));
Console.ReadLine();
}
Using multiple threads to swap the elements:
final char[] strArray = str.toCharArray();
IntStream.range(0, str.length() / 2).parallel().forEach(e -> {
final char tmp = strArray[e];
strArray[e] = strArray[str.length() - e - 1];
strArray[str.length() - e - 1] = tmp;
});
return new String(strArray);
Of course this is the most efficient way:
String reversed = new StringBuilder(str).reverse().toString();
But if you don't like using that then I recommend this instead:
public String reverseString(String str)
{
String output = "";
int len = str.length();
for(int k = 1; k <= str.length(); k++, len--)
{
output += str.substring(len-1,len);
}
return output;
}
static String ReverseString(String input) {
var len = input.Length - 1;
int i = 0;
char[] revString = new char[len+1];
while (len >= 0) {
revString[i] = input[len];
len--;
i++;
}
return new string(revString);
}
why can't we stick with the simplest loop and revere with character read and keep adding to the char array, I have come across with a whiteboard interview, where interviewer set restrictions on not to use StringBuilder and inbuilt functions.
This is the optimal way to reverse a string with O(log n) complexity.
public char[] optimisedArrayReverse(char[] chars) {
char[] reversedChars = chars;
int length = chars.length;
int center = (length / 2) - 1;
int reversedIndex = chars.length - 1;
if (center < 0) {
return chars;
}
for (int index = 0; index <= center; index++) {
//Swap values
char temp = reversedChars[index];
reversedChars[index] = chars[reversedIndex];
reversedChars[reversedIndex] = temp;
reversedIndex --;
}
return reversedChars;
}
public static String reverseString(String str)
{
StringBuilder sb = new StringBuilder();
for (int i = str.length() - 1; i >= 0; i--)
{
sb.append(str[i]);
}
return sb.toString();
}