From an online notes, I read the following java code snippet for reversing a string, which is claimed to have quadratic time complexity. It seems to me that the “for” loop for i just iterates the whole length of s. How does it cause a quadratic time complexity?
public static String reverse(String s)
{
String rev = new String();
for (int i = (s.length()-1); i>=0; i--) {
rev = rev.append(s.charAt(i));
}
return rev.toString();
}
public static String reverse(String s)
{
String rev = " ";
for (int i=s.length()-1; i>=0; i--)
rev.append(s.charAt(i); // <--------- This is O(n)
Return rev.toString();
}
I copy pasted your code. I'm not sure where you get this but actually String doesn't have append method. Maybe rev is a StringBuilder or another Appendable.
Possibly because the append call does not execute in constant time. If it's linear with the length of the string, that would explain it.
append has to find the end of the string, which is Ο(n). So, you have an Ο(n) loop executed Ο(n) times.
I don't think String has an append method. So, this code won't compile.
But, coming to the problem of quadratic complexity, let us assume that you are actually appending the string with a character using '+' operator or the String.concat() method.
The String objects are immutable. So, whenever you append to a string, a new string of bigger length is created, old string contents are copied to it and then the final character is appended, and the previous string is destroyed. So, this process takes more and more time as the string grows.
The appending loop takes O(n) time but for every loop you take O(n) time to copy the string character by character. This leads to quadratic complexity.
It would be better to use StringBuilder or StringBuffer. However, I guess the time complexity you mentioned would be with older java compilers. But, new advanced compilers would actually optimize the '+' operation with StringBuilder.
Related
I wrote a function that does this, and it works; however, I'm a little bit uncertain about the time complexity of my function. I feel as though it's pretty inefficient, but I'm not sure if perhaps it's just the nature of the problem. I'm imagining there is a better way to do this opposed to how I did it. My initial thought was that this would be O(n^2) time complexity, but I think it may actually be worst, because of the split function I used. What's a better way to do this? Also, am I correct in thinking that this is actually worse than O(n^2)?
public static String wordReverse(String string){
//Split the string into an array such that each word is an element in the array
String[] arr = string.split(" ");
String result = "";
//Iterate throught the elements in the array
for(String value : arr){
String word = "";
//Reverse the letters of the element, and append them to a temp string
for(int i = value.length(); i > 0; i--){
word += value.charAt(i-1);
}
//Build the result string
result += word += " ";
}
//Return result string
return(result);
}
You're not worse that O(n^2). Actually, you're O(n).
split() is O(n), since it passed the array once.
You're reverse is O(n) too, since it checks all characters.
And concatenation is O(n) at worst.
The complexity of your approach is O(n) - linear time. The time increases linearly with the number of words and characters.
say I use the following methods to search for a palindrome. I know the first one is O(n) because it goes through entire string. Does the .reverse() in the StringBuffer also do O(n)? Im not worried about finding a better way to problem Im trying to understand if the reverse method physically reverses the string or is it much more efficient than that?? Thanks
public static boolean isAPalindrome(String s1){
String tmp = "";
int length = s1.length();
for(int i = 0; i < s1.length(); i++){
tmp += s1.charAt(s1.length()-i-1);
}
if (s1.equals(tmp)) return true;
return false;
}
public static boolean isAPalindrome(String s1){
StringBuffer a = new StringBuffer(s1);
return s1.equals(a.reverse().toString());
}
You could implement a reverse that is effectively o(1), but not with the typical string classes: You could do it by implementing a string class with a direction member, that can take the values "forward" or "backward". But that would only make sense if the reverting dominates your performance over all other computations.
Reversing a string can't be faster than O(n) (under normal representation of a String).
You have to "operate" on every character of the string, so it can't theoretically be faster than that.
StringBuffer extends AbstractStringBuilder which implements reverse. You may look at the souce code yourself here.
Note: this is not saying that StringBuilder/Buffers reverse will run as fast as an O(n) reverse method that you will write. It will probably run much faster. But asymptotically it will still be O(n).
I have:
String str = "Hello, how, are, you";
I want to create a helper method that removes the commas from any string. Which of the following is more accurate?
private static String removeComma(String str){
if(str.contains(",")){
str = str.replaceAll(",","");
}
return str;
}
OR
private static String removeComma(String str){
str = str.replaceAll(",","");
return str;
}
Seems like I don't need the IF statement but there might be a case where I do.
If there is a better way let me know.
Both are functionally equivalent but the former is more verbose and will probably be slower because it runs an extra operation.
Also note that you don't need replaceAll (which accepts a regular expression): replace will do.
So I would go for:
private static String removeComma(String str){
return str.replace(",", "");
}
The IF statement is unnecessary, unless you're handling "large" strings (we're talking megabytes or more).
If you're using the IF statement, your code will first search for the first occurance of a comma, and then execute the replacement. This could be costly if the comma is near the end of the string and your string is large, since it will have to be traversed twice.
Without the IF statement, commas will be replaced if they exist. If the answer is negative, your string will be untouched.
Bottom rule: use the version without the IF statement.
Both are correct, but the second one is cleaner since the IF statement of the first alternative is not needed.
It's a matter of what is the probability to have strings with comma in your universe of strings.
If you have a high probability, call the method replaceAll without checking first.
BUT If you are not using extremely huge strings, I guess you will see no difference in perfomance at all.
Just another solution with time complexity O(n), space complexity O(n):
public static String removeComma(String str){
int length = str.length();
StringBuffer sb = new StringBuffer();
for (int i = 0; i < length; i++) {
char c = str.charAt(i);
if (c != ',') {
sb.append(c);
}
}
return sb.toString();
}
I'm learning Java and am wondering what's the best way to modify strings here (both for performance and to learn the preferred method in Java). Assume you're looping through a string and checking each character/performing some action on that index in the string.
Do I use the StringBuilder class, or convert the string into a char array, make my modifications, and then convert the char array back to a string?
Example for StringBuilder:
StringBuilder newString = new StringBuilder(oldString);
for (int i = 0; i < oldString.length() ; i++) {
newString.setCharAt(i, 'X');
}
Example for char array conversion:
char[] newStringArray = oldString.toCharArray();
for (int i = 0; i < oldString.length() ; i++) {
myNameChars[i] = 'X';
}
myString = String.valueOf(newStringArray);
What are the pros/cons to each different way?
I take it that StringBuilder is going to be more efficient since the converting to a char array makes copies of the array each time you update an index.
I say do whatever is most readable/maintainable until you you know that String "modification" is slowing you down. To me, this is the most readable:
Sting s = "foo";
s += "bar";
s += "baz";
If that's too slow, I'd use a StringBuilder. You may want to compare this to StringBuffer. If performance matters and synchronization does not, StringBuilder should be faster. If sychronization is needed, then you should use StringBuffer.
Also it's important to know that these strings are not being modified. In java, Strings are immutable.
This is all context specific. If you optimize this code and it doesn't make a noticeable difference (and this is usually the case), then you just thought longer than you had to and you probably made your code more difficult to understand. Optimize when you need to, not because you can. And before you do that, make sure the code you're optimizing is the cause of your performance issue.
What are the pros/cons to each different way. I take it that StringBuilder is going to be more efficient since the convering to a char array makes copies of the array each time you update an index.
As written, the code in your second example will create just two arrays: one when you call toCharArray(), and another when you call String.valueOf() (String stores data in a char[] array). The element manipulations you are performing should not trigger any object allocations. There are no copies being made of the array when you read or write an element.
If you are going to be doing any sort of String manipulation, the recommended practice is to use a StringBuilder. If you are writing very performance-sensitive code, and your transformation does not alter the length of the string, then it might be worthwhile to manipulate the array directly. But since you are learning Java as a new language, I am going to guess that you are not working in high frequency trading or any other environment where latency is critical. Therefore, you are probably better off using a StringBuilder.
If you are performing any transformations that might yield a string of a different length than the original, you should almost certainly use a StringBuilder; it will resize its internal buffer as necessary.
On a related note, if you are doing simple string concatenation (e.g, s = "a" + someObject + "c"), the compiler will actually transform those operations into a chain of StringBuilder.append() calls, so you are free to use whichever you find more aesthetically pleasing. I personally prefer the + operator. However, if you are building up a string across multiple statements, you should create a single StringBuilder.
For example:
public String toString() {
return "{field1 =" + this.field1 +
", field2 =" + this.field2 +
...
", field50 =" + this.field50 + "}";
}
Here, we have a single, long expression involving many concatenations. You don't need to worry about hand-optimizing this, because the compiler will use a single StringBuilder and just call append() on it repeatedly.
String s = ...;
if (someCondition) {
s += someValue;
}
s += additionalValue;
return s;
Here, you'll end up with two StringBuilders being created under the covers, but unless this is an extremely hot code path in a latency-critical application, it's really not worth fretting about. Given similar code, but with many more separate concatenations, it might be worth optimizing. Same goes if you know the strings might be very large. But don't just guess--measure! Demonstrate that there's a performance problem before you try to fix it. (Note: this is just a general rule for "micro optimizations"; there's rarely a downside to explicitly using a StringBuilder. But don't assume it will make a measurable difference: if you're concerned about it, you should actually measure.)
String s = "";
for (final Object item : items) {
s += item + "\n";
}
Here, we're performing a separate concatenation operation on each loop iteration, which means a new StringBuilder will be allocated on each pass. In this case, it's probably worth using a single StringBuilder since you may not know how large the collection will be. I would consider this an exception to the "prove there's a performance problem before optimizing rule": if the operation has the potential to explode in complexity based on input, err on the side of caution.
Which option will perform the best is not an easy question.
I did a benchmark using Caliper:
RUNTIME (NS)
array 88
builder 126
builderTillEnd 76
concat 3435
Benchmarked methods:
public static String array(String input)
{
char[] result = input.toCharArray(); // COPYING
for (int i = 0; i < input.length(); i++)
{
result[i] = 'X';
}
return String.valueOf(result); // COPYING
}
public static String builder(String input)
{
StringBuilder result = new StringBuilder(input); // COPYING
for (int i = 0; i < input.length(); i++)
{
result.setCharAt(i, 'X');
}
return result.toString(); // COPYING
}
public static StringBuilder builderTillEnd(String input)
{
StringBuilder result = new StringBuilder(input); // COPYING
for (int i = 0; i < input.length(); i++)
{
result.setCharAt(i, 'X');
}
return result;
}
public static String concat(String input)
{
String result = "";
for (int i = 0; i < input.length(); i++)
{
result += 'X'; // terrible COPYING, COPYING, COPYING... same as:
// result = new StringBuilder(result).append('X').toString();
}
return result;
}
Remarks
If we want to modify a String, we have to do at least 1 copy of that input String, because Strings in Java are immutable.
java.lang.StringBuilder extends java.lang.AbstractStringBuilder. StringBuilder.setCharAt() is inherited from AbstractStringBuilder and looks like this:
public void setCharAt(int index, char ch) {
if ((index < 0) || (index >= count))
throw new StringIndexOutOfBoundsException(index);
value[index] = ch;
}
AbstractStringBuilder internally uses the simplest char array: char value[]. So, result[i] = 'X' is very similar to result.setCharAt(i, 'X'), however the second will call a polymorphic method (which probably gets inlined by JVM) and check bounds in if, so it will be a bit slower.
Conclusions
If you can operate on StringBuilder until the end (you don't need String back) - do it. It's the preferred way and also the fastest. Simply the best.
If you want String in the end and this is the bottleneck of your program, then you might consider using char array. In benchmark char array was ~25% faster than StringBuilder. Be sure to properly measure execution time of your program before and after optimization, because there is no guarantee about this 25%.
Never concatenate Strings in the loop with + or +=, unless you really know what you do. Usally it's better to use explicit StringBuilder and append().
I'd prefer to use StringBuilder class where original string is modified.
For String manipulation, I like StringUtil class. You'll need to get Apache commons dependency to use it
which of the following is an efficient way to reverse words in a string ?
public String Reverse(StringTokenizer st){
String[] words = new String[st.countTokens()];
int i = 0;
while(st.hasMoreTokens()){
words[i] = st.nextToken();i++}
for(int j = words.length-1;j--)
output = words[j]+" ";}
OR
public String Reverse(StringTokenizer st, String output){
if(!st.hasMoreTokens()) return output;
output = st.nextToken()+" "+output;
return Reverse(st, output);}
public String ReverseMain(StringTokenizer st){
return Reverse(st, "");}
while the first way seems more readable and straight forward, there are two loops in it. In the 2nd method, I've tried doing it in tail-recursive way. But I am not sure whether java does optimize tail-recursive code.
you could do this in just one loop
public String Reverse(StringTokenizer st){
int length = st.countTokens();
String[] words = new String[length];
int i = length - 1;
while(i >= 0){
words[i] = st.nextToken();i--}
}
But I am not sure whether java does optimize tail-recursive code.
It doesn't. Or at least the Sun/Oracle Java implementations don't, up to and including Java 7.
References:
"Tail calls in the VM" by John Rose # Oracle.
Bug 4726340 - RFE: Tail Call Optimization
I don't know whether this makes one solution faster than the other. (Test it yourself ... taking care to avoid the standard micro-benchmarking traps.)
However, the fact that Java doesn't implement tail-call optimization means that the 2nd solution is liable to run out of stack space if you give it a string with a large (enough) number of words.
Finally, if you are looking for a more space efficient way to implement this, there is clever way that uses just a StringBuilder.
Create a StringBuilder from your input String
Reverse the characters in the StringBuilder using reverse().
Step through the StringBuilder, identifying the start and end offset of each word. For each start/end offset pair, reverse the characters between the offsets. (You have to do this using a loop.)
Turn the StringBuilder back into a String.
You can test results by timing both of them on a large amount of results
eg. You reverse 100000000 strings and see how many seconds it takes. You could also compare start and end system timestamps to get the exact difference between the two functions.
StringTokenizer is not deprecated but if you read the current JavaDoc...
StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead.
String[] strArray = str.split(" ");
StringBuilder sb = new StringBuilder();
for (int i = strArray.length() - 1; i >= 0; i--)
sb.append(strArray[i]).append(" ");
String reversedWords = sb.substring(0, sb.length -1) // strip trailing space