Write a program to compute if one string is a rotation of another.

Question

Yash · Accepted Answer

Simplest way is to concatenate the rotated array after itself and then see if the original array is a sub-string of that!
bool checkStrings (string s1, string s2){
     s1+=s1;
     if(s1.find(s2)

Josh · Answer

Answer by Yash is incorrect, the input:

11121, 111211

Is incorrect, but would return True.  Correct answer is:

def check_strings(str1, str2):
    return len(str1) == len(str2) and 0 <= (str1*2).find(str2) < len(str1)

alex · Answer

Does not look correct to me. Consider "123" and "321".

Liron · Answer

yes, you are right, the code above is only for checking if it's a regular permutation.
here is the correct code, although it O(n^2) complexity, and I could nit think of a more efficient solution, any ideas?

public static void main(String[] args) {
		String str1 = "1111111111111111111112";
		String str2 = "1111111111111111111121";
		
		
		char[] char1 = str1.toCharArray();
		
		char[] char2 = str2.toCharArray();
		
		int i=0;
		int j=0;
		for (int start = 0; start < char2.length; ++start){
			while (true && i

Liron · Answer

String str1 = "1111111111111111111112";
		String str2 = "1111111111111111111121";
		
		if (str1.length() != str2.length()){
			return false;
		}
		
		char[] char1 = str1.toCharArray();
		List list1 = new ArrayList();
		for (int i=0; i list2 = new ArrayList();
		for (int i=0; i

Cedric · Answer

public class TwoTimesACharm {
	
	public static void main(String[] args)
	{
		String[] test1 = {"1234", "4123", "3412", "2341", "1212", "12345678"};
		String original = "1234";
		
		TwoTimesACharm aCharm = new TwoTimesACharm();
		
		for(String s:test1)
		{
			System.out.println("It is " + aCharm.isRotation(original, s) + " that " + s + " is a rotation of " + original);
		}
		
		System.out.println("
 Second method test. the first method implementation doesnt show any skills");
		for(String s:test1)
		{
			System.out.println("It is " + aCharm.isRotation2(original, s) + " that " + s + " is a rotation of " + original);
		}
	}
	
	//so the reason that we repeat the original string twice is because the rotation would be in the string.
			//for example string "1234" possible rotations are 
			//shift right 4123, 3412, 2341
			//shift left 2341, 3412, 4123 
			// repeat original string "12341234" we will see those rotations are present in there. 

	private boolean isRotation(String originalString, String rotateString)
	{
		String repeatTwice = originalString+originalString;
		
		return (repeatTwice.indexOf(rotateString) > -1 ? true:false);
	}
	
	private boolean isRotation2(String originalString, String rotateString)
	{
		//before going any further make sure they are the same length.
		if(originalString.length() != rotateString.length())
			return false;
		
		//make sure that the inputs aren't the same.
		if(originalString.equals(rotateString))
			return true;
		
		//I would use StringBuilder for efficiency but for simplicity I used a string
		String repeatTwice = originalString+originalString;
		
		int i = 0;
		// we do the length of the string because the start of i in the repeatTwice string shouldn't be
		//pass the end of the original string or else we will get array out of bounds.
		while(i <= originalString.length())
		{
			//so basically we are iterating through the repeat string until we find a match.
			if(rotateString.equals(repeatTwice.substring(i, i+originalString.length())))
				return true;
			else
				i++;
		}
		return false;
	}


}

JJ · Answer

Why are we assuming that the string is a string of numbers?

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