Develop Reference

Removing a substring from a string, repeatedly

Problem:
Remove the substring t from a string s, repeatedly and print the number of steps involved to do the same.
Explanation/Working:
For Example: t = ab, s = aabb. In the first step, we check if t is
contained within s. Here, t is contained in the middle i.e. a(ab)b.
So, we will remove it and the resultant will be ab and increment the
count value by 1. We again check if t is contained within s. Now, t is
equal to s i.e. (ab). So, we remove that from s and increment the
count. So, since t is no more contained in s, we stop and print the
count value, which is 2 in this case.
So, here's what I have tried:
Code 1:
static int maxMoves(String s, String t) {
int count = 0,i;
while(true)
{
if(s.contains(t))
{
i = s.indexOf(t);
s = s.substring(0,i) + s.substring(i + t.length());
}
else break;
++count;
}
return count;
}
I am just able to pass 9/14 test cases on Hackerrank, due to some reason (I am getting "Wrong Answer" for rest of the cases). After a while, I found out that there is something called replace() method in Java. So, I tried using that by replacing the if condition and came up with a second version of code.
Code 2:
static int maxMoves(String s, String t) {
int count = 0,i;
while(true)
{
if(s.contains(t))
s.replace(t,""); //Marked Statement
else break;
++count;
}
return count;
}
But for some reason (I don't know why), the "Marked Statement" in the above code gets executed infinitely (this I noticed when I replaced the "Marked Statement" with System.out.println(s.replace(t,""));). I don't the reason for the same.
Since, I am passing only 9/14 test cases, there must be some logical error that is leading to a "Wrong Answer". How do I overcome that if I use Code 1? And if I use Code 2, how do I avoid infinite execution of the "Marked Statement"? Or is there anyone who would like to suggest me a Code 3?
Thank you in advance :)

Try saving the new (returned) string instead of ignoring it.
s = s.replace(t,"");
replace returns a new string; you seemed to think that it alters the given string in-place.

Try adding some simple parameter checks of the strings. The strings shouldn't be equal to null and they should have a length greater than 0 to allow for counts greater than 0.
static int maxMoves(String s, String t) {
int count = 0,i;
if(s == null || s.length() == 0 || t == null || t.length() == 0)
return 0;
while(true)
{
if(s.contains(t) && !s.equals(""))
s = s.replace(t,""); //Marked Statement
else break;
++count;
}
return count;
}

You might be missing on the edge cases in the code 1.
In code 2, you are not storing the new string formed after the replace function.
The replace function replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence.
Try this out:
public static int findCount(String s, String t){
if( null == s || "" == s || null == t || "" == t)
return 0;
int count =0;
while(true){
if(s.contains(t)){
count++;
int i = s.indexOf(t);
s = s.substring(0, i)+s.substring(i+t.length(), s.length());
// s = s.replace(t,"");
}
else
break;
}
return count;
}

String r1="ramraviraravivimravi";
String r2="ravi";
int count=0,i;
while(r1.contains(r2))
{
count++;
i=r1.indexOf(r2);
StringBuilder s1=new StringBuilder(r1);
s1.delete(i,i+r2.length());
System.out.println(s1.toString());
r1=s1.toString();
}
System.out.println(count);

First of all no logical difference in both the codes.
All the mentioned answers are to rectify the error of code 2 but none told how to pass all (14/14) cases.
Here I am mentioning a test case where your code will fail.
s = "abcabcabab";
t = "abcab"
Your answer 1
Expected answer 2
According to your code:
In 1st step, removig t from index 0 of s,
s will reduce to "cabab", so the count will be 1 only.
But actual answer should be 2
I first step, remove t from index 3 of s,
s will reduced to "abcab", count = 1.
In 2nd step removing t from index 0,
s will reduced to "", count = 2.
So answer would be 2.
If anyone know how to handle such cases, please let me know.

How do i find how many times a substring is used in a string?

public Object countOccurrences(String string)
{
int e = 0;
int i = 0;
while ( i < sentence.length())
{
if(sentence.contains(string))
{
e++;
i++;
return e;
}
else
{
break;
}
}
return e;
}
What is wrong with my code that won't allow me to pass the test? is there anyway I can use the substring method inside a loop to do this?
#Test
public void testCountOccurrences()
{
assertEquals(1, sc1.countOccurrences("ence"));
assertEquals(2, sc1.countOccurrences("en"));
assertEquals(1, sc2.countOccurrences("that"));
assertEquals(0, sc2.countOccurrences("This"));
}

What is wrong with my code that won't allow me to pass the test?
Before doing anything else, you should consider how you could have worked out what was wrong for your code to start with. Did you debug through it? At what point did it behave differently to how you expected it to? If you haven't learned how to use a debugger yet, now is a great time to start.
As for the next step, look at this code:
if(sentence.contains(string))
{
e++;
i++;
return e;
}
That condition doesn't depend on i or e, just on sentence and string. So as long as sentence is of length at least 1, you'll either return 1 or 0. Your code can never return more than 1.
That's what's wrong with your code at the moment - as for how to fix it, I'd start looking at String.indexOf(String, int). In other words, you want to find the first occurrence, then find the next occurrence, then the next occurrence, until you can't find any more. (Use the return value to work out where to start looking on the next iteration, as well as checking that there was a match.)
A couple of situations to be careful of:
How many times does "abbbc" contain "bb"?
How many times does "abbbc" contain ""?
I'd also urge a couple of other changes:
Your method has a return type of Object - why? Surely it's always going to return an integer, so a return type of int would be more appropriate
This is a great candidate for parameterized testing. Look into how you can effectively separate your single test into multiple test cases which can pass or fail independently, but without the source overhead of lots of test methods... (Hint: each test case should have the sentence, the text you're looking for, and the expected number of matches.)

public Object countOccurrences(String string) {
int e = 0;
int i = 0;
while (i <= (sentence.length() - string.length() + 1)) {
if (sentence.substr(i, string.length() - 1).equals(string)) {
e++;
}
i++;
}
return e;
}
I didn't got to try it myself but it should work.

Java - Complex Recursive Backtracking

For Java practice I started working on a method countBinary that accepts an integer n as a parameter that prints all binary numbers that have n digits in ascending order, printing each value on a separate line. Assuming n is non-negative and greater than 0, some example outputs would look like this.
I am getting pretty much nowhere with this. I am able to write a program that finds all possible letter combinations of a String and similar things, but I have been unable to make almost any progress with this specific problem using binary and integers.
Apparently the best way to go about this issue is by defining a helper method that accepts different parameters than the original method and by building up a set of characters as a String for eventual printing.
Important Note: I am NOT supposed to use for loops at all for this exercise.
Edit - Important Note: I need to have trailing 0's so that all outputs are the same length.
So far this is what I have:
public void countBinary(int n)
{
String s = "01";
countBinary(s, "", n);
}
private static void countBinary(String s, String chosen, int length)
{
if (s.length() == 0)
{
System.out.println(chosen);
}
else
{
char c = s.charAt(0);
s = s.substring(1);
chosen += c;
countBinary(s, chosen, length);
if (chosen.length() == length)
{
chosen = chosen.substring(0, chosen.length() - 1);
}
countBinary(s, chosen, length);
s = c + s;
}
}
When I run my code my output looks like this.
Can anyone explain to me why my method is not running the way I expect it to, and if possible show me a solution to my issue so that I might get the correct output? Thank you!

There are more efficient ways to do it, but this will give you a start:
public class BinaryPrinter {
static void printAllBinary(String s, int n) {
if (n == 0) System.out.println(s);
else {
printAllBinary(s + '0', n - 1);
printAllBinary(s + '1', n - 1);
}
}
public static void main(String [] args) {
printAllBinary("", 4);
}
}
I'll let you work out the more efficient way.

Common Substring of two strings

This particular interview-question stumped me:
Given two Strings S1 and S2. Find the longest Substring which is a Prefix of S1 and suffix of S2.
Through Google, I came across the following solution, but didnt quite understand what it was doing.
public String findLongestSubstring(String s1, String s2) {
List<Integer> occurs = new ArrayList<>();
for (int i = 0; i < s1.length(); i++) {
if (s1.charAt(i) == s2.charAt(s2.length()-1)) {
occurs.add(i);
}
}
Collections.reverse(occurs);
for(int index : occurs) {
boolean equals = true;
for(int i = index; i >= 0; i--) {
if (s1.charAt(index-i) != s2.charAt(s2.length() - i - 1)) {
equals = false;
break;
}
}
if(equals) {
return s1.substring(0,index+1);
}
}
return null;
}
My questions:
How does this solution work?
And how do you get to discovering this solution?
Is there a more intuitive / easier solution?

Part 2 of your question
Here is a shorter variant:
public String findLongestPrefixSuffix(String s1, String s2) {
for( int i = Math.min(s1.length(), s2.length()); ; i--) {
if(s2.endsWith(s1.substring(0, i))) {
return s1.substring(0, i);
}
}
}
I am using Math.min to find the length of the shortest String, as I don't need to and cannot compare more than that.
someString.substring(x,y) returns you the String you get when reading someString beginning from character x and stopping at character y. I go backwards from the biggest possible substring (s1 or s2) to the smallest possible substring, the empty string. This way the first time my condition is true it will be biggest possible substring the fulfills it.
If you prefer you can go the other way round, but you have to introduce a variable saving the length of the longest found substring fulfilling the condition so far:
public static String findLongestPrefixSuffix(String s1, String s2) {
if (s1.equals(s2)) { // this part is optional and will
return s1; // speed things up if s1 is equal to s2
} //
int max = 0;
for (int i = 0; i < Math.min(s1.length(), s2.length()); i++) {
if (s2.endsWith(s1.substring(0, i))) {
max = i;
}
}
return s1.substring(0, max);
}
For the record: You could start with i = 1 in the latter example for a tiny bit of extra performance. On top of this you can use i to specify how long the suffix has at least to be you want to get. ;) If you writ Math.min(s1.length(), s2.length()) - x you can use x to specify how long the found substring may be at most. Both of these things are possible with the first solution, too, but the min length is a bit more involving. ;)
Part 1 of your question
In the part above the Collections.reverse the author of the code searches for all positions in s1 where the last letter of s2 is and saves this position.
What follows is essentially what my algorithm does, the difference is, that he doesn't check every substring but only those that end with the last letter of s2.
This is some sort of optimization to speed things up. If speed is not that important my naive implementation should suffice. ;)

Where did you find that solution? Was it written by a credible, well-respected coder? If you're not sure of that, then it might not be worth reading it. One could write really complex and inefficient code to accomplish something really simple, and it will not be worth understanding the algorithm.
Rather than trying to understand somebody else's solution, it might be easier to come up with it on your own. I think you understand the problem much better that way, and the logic becomes your own. Over time and practice the thought process will start to come more naturally. Practice makes perfect.
Anyway, I put a more simple implementation in Python here (spoiler alert!). I suggest you first figure out the solution on your own, and compare it to mine later.

Apache commons lang3, StringUtils.getCommonPrefix()
Java is really bad in providing useful stuff via stdlib. On the plus side there's almost always some reasonable tool from Apache.

I converted the #TheMorph's answer to javascript. Hope this helps js developer
if (typeof String.prototype.endsWith !== 'function') {
String.prototype.endsWith = function(suffix) {
return this.indexOf(suffix, this.length - suffix.length) !== -1;
};
}
function findLongestPrefixSuffix(s2, s1) {
for( var i = Math.min(s1.length, s2.length); ; i--) {
if(s2.endsWith(s1.substring(0, i))) {
return s1.substring(0, i);
}
}
}
console.log(findLongestPrefixSuffix('abc', 'bcd')); // result: 'bc'

Recursive method to determine if a string is a hex number - Java

This is a homework question that I am having a bit of trouble with.
Write a recursive method that determines if a String is a hex number.
Write javadocs for your method.
A String is a hex number if each and every character is either
0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9
or a or A or b or B or c or C or d or D or e or E or f or f.
At the moment all I can see to test this is if the character at 0 of the string is one of these values he gave me then that part of it is a hex.
Any hints or suggestions to help me out?
This is what I have so far: `
public boolean isHex(String string){
if (string.charAt(0)==//what goes here?){
//call isHex again on s.substring(1)
}else return false;
}
`

If you're looking for a good hex digit method:
boolean isHexDigit(char c) {
return Character.isDigit(c) || (Character.toUpperCase(c) >= 'A' && Character.toUpperCase(c) <= 'F');
}
Hints or suggestions, as requested:
All recursive methods call themselves with a different input (well, hopefully a different input!)
All recursive methods have a stop condition.
Your method signature should look something like this
boolean isHexString(String s) {
// base case here - an if condition
// logic and recursion - a return value
}
Also, don't forget that hex strings can start with "0x". This might be (more) difficult to do, so I would get the regular function working first. If you tackle it later, try to keep in mind that 0xABCD0x123 shouldn't pass. :-)
About substring: Straight from the String class source:
public String substring(int beginIndex, int endIndex) {
if (beginIndex < 0) {
throw new StringIndexOutOfBoundsException(beginIndex);
}
if (endIndex > count) {
throw new StringIndexOutOfBoundsException(endIndex);
}
if (beginIndex > endIndex) {
throw new StringIndexOutOfBoundsException(endIndex - beginIndex);
}
return ((beginIndex == 0) && (endIndex == count)) ? this :
new String(offset + beginIndex, endIndex - beginIndex, value);
}
offset is a member variable of type int
value is a member variable of type char[]
and the constructor it calls is
String(int offset, int count, char value[]) {
this.value = value;
this.offset = offset;
this.count = count;
}
It's clearly an O(1) method, calling an O(1) constructor. It can do this because String is immutable. You can't change the value of a String, only create a new one. (Let's leave out things like reflection and sun.misc.unsafe since they are extraneous solutions!) Since it can't be changed, you also don't have to worry about some other Thread messing with it, so it's perfectly fine to pass around like the village bicycle.

Since this is homework, I only give some hints instead of code:
Write a method that always tests the first character of a String if it fulfills the requirements. If not, return false, if yes, call the method again with the same String, but the first character missing. If it is only 1 character left and it is also a hex character then return true.
Pseudocode:
public boolean isHex(String testString) {
If String has 0 characters -> return true;
Else
If first character is a hex character -> call isHex with the remaining characters
Else if the first character is not a hex character -> return false;
}

When solving problems recursively, you generally want to solve a small part (the 'base case'), and then recurse on the rest.
You've figured out the base case - checking if a single character is hex or not.
Now you need to 'recurse on the rest'.
Here's some pseudocode (Python-ish) for reversing a string - hopefully you will see how similar methods can be applied to your problem (indeed, all recursive problems)
def ReverseString(str):
# base case (simple)
if len(str) <= 1:
return str
# recurse on the rest...
return last_char(str) + ReverseString(all_but_last_char(str))

Sounds like you should recursively iterate the characters in string and return the boolean AND of whether or not the current character is in [0-9A-Fa-f] with the recursive call...

You have already received lots of useful answers. In case you want to train your recursive skills (and Java skills in general) a bit more I can recommend you to visit Coding Bat. You will find a lot of exercises together with automated tests.