Check if a string is rotation of another WITHOUT concatenating

Question

There are 2 strings , how can we check if one is a rotated version of another ?

For Example : hello --- lohel

One simple solution is by co

Answer 1

You can do it in O(n) time and O(1) space:

def is_rot(u, v):
    n, i, j = len(u), 0, 0
    if n != len(v):
        return False
    while i < n and j < n:
        k = 1
        while k <= n and u[(i + k) % n] == v[(j + k) % n]:
            k += 1
        if k > n:
            return True
        if u[(i + k) % n] > v[(j + k) % n]:
            i += k
        else:
            j += k
    return False

See my answer here for more details.

Answer 2

input1= "hello" input2="llohe" input3="lohel"(input3 is special case)

if length's of input 1 & input2 are not same return 0.Let i and j be two indexes pointing to input1 and input2 respectively and initialize count to input1.length. Have a flag called isRotated which is set to false

while(count != 0){

When the character's of input1 matches input2

1. increment i & j
2. decrement count

If the character's donot match

1. if isRotated = true(it means even after rotation there's mismatch) so break;
2. else Reset j to 0 as there's a mismatch. Eg:

Please find the code below and let me know if it fails for some other combination I may not have considered.

    public boolean isRotation(String input1, String input2) {
    boolean isRotated = false;
    int i = 0, j = 0, count = input1.length();

    if (input1.length() != input2.length())
        return false;

    while (count != 0) {

        if (i == input1.length() && !isRotated) {
            isRotated = true;
            i = 0;
        }

        if (input1.charAt(i) == input2.charAt(j)) {
            i++;
            j++;
            count--;
        }

        else {
            if (isRotated) {
                break;
            }
            if (i == input1.length() - 1 && !isRotated) {
                isRotated = true;
            }
            if (i < input1.length()) {
                j = 0;
                count = input1.length();

            }
            /* To handle the duplicates. This is the special case.
             * This occurs when input1 contains two duplicate elements placed side-by-side as "ll" in "hello" while 
             * they may not be side-by-side in input2 such as "lohel" but are still valid rotations. 
             Eg: "hello" "lohel"
            */
            if (input1.charAt(i) == input2.charAt(j)) {
                i--;
            }
            i++;
        }
    }
    if (count == 0)
        return true;
    return false;
}

public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println(new StringRotation().isRotation("harry potter",
            "terharry pot"));
    System.out.println(new StringRotation().isRotation("hello", "llohe"));
    System.out.println(new StringRotation().isRotation("hello", "lohell"));
    System.out.println(new StringRotation().isRotation("hello", "hello"));
    System.out.println(new StringRotation().isRotation("hello", "lohe"));
}

Answer 3

There's no need to concatenate at all.

First, check the lengths. If they're different then return false.

Second, use an index that increments from the first character to the last of the source. Check if the destination starts with all the letters from the index to the end, and ends with all the letters before the index. If at any time this is true, return true.

Otherwise, return false.

EDIT:

An implementation in Python:

def isrot(src, dest):
  # Make sure they have the same size
  if len(src) != len(dest):
    return False

  # Rotate through the letters in src
  for ix in range(len(src)):
    # Compare the end of src with the beginning of dest
    # and the beginning of src with the end of dest
    if dest.startswith(src[ix:]) and dest.endswith(src[:ix]):
      return True

  return False

print isrot('hello', 'lohel')
print isrot('hello', 'lohell')
print isrot('hello', 'hello')
print isrot('hello', 'lohe')

Answer 4

Simple solution in Java. No need of iteration or concatenation.

private static boolean isSubString(String first, String second){
        int firstIndex = second.indexOf(first.charAt(0));
        if(first.length() == second.length() && firstIndex > -1){

            if(first.equalsIgnoreCase(second))
                return true;

            int finalPos = second.length() - firstIndex ; 
            return second.charAt(0) == first.charAt(finalPos)
            && first.substring(finalPos).equals(second.subSequence(0, firstIndex));
        }
        return false;

    }

Test case:

String first = "bottle";
String second = "tlebot";

Logic:

Take the first string's first character, find the index in the second string. Subtract the length of the second with the index found, check if first character of the second at 0 is same as character at the difference of length of the second and index found and substrings between those 2 characters are the same.

Answer 5

Solving the problem in O(n)

void isSubstring(string& s1, string& s2)
{
    if(s1.length() != s2.length())
        cout<<"Not rotation string"<<endl;
    else
    {
        int firstI=0, secondI=0;
        int len = s1.length();

        while( firstI < len )
        {
            if(s1[firstI%len] == s2[0] && s1[(firstI+1) %len] == s2[1])
                break;

            firstI = (firstI+1)%len;
        }

        int len2 = s2.length(); 
        int i=0;
        bool isSubString = true; 
        while(i < len2)
        {
            if(s1[firstI%len] != s2[i])
            {
                isSubString = false;
                break;
            }
            i++;
        }

        if(isSubString)
            cout<<"Is Rotation String"<<endl;
        else
            cout<<"Is not a rotation string"<<endl;
    }
}