Java String Rotation Interview Questions & Answers

This guide covers logic and code for checking and performing string rotations (left and right) in Java.

1. How to check if one string is a rotation of another?

The simplest logic to check if String B is a rotation of String A is to concatenate A with itself and check if B is a substring of the result.

The Logic:

1. Concatenate the original string with itself: concatenated = original + original.
2. Check if the target string is a substring: concatenated.contains(target).
3. Constraint: Both strings must be of the same length.

Example:

• Original: DECODE
• Rotated (Left 2): CODEDE
• Concatenated: DECODE + DECODE = DECODEDECODE
• Result: DECODEDECODE contains CODEDE → True.

2. Left Rotation Logic

Left rotation moves characters from the beginning of the string to the end.

Formula: Result = original.substring(r) + original.substring(0, r) (where r is the number of characters to rotate)

Example (Rotate Left by 2):

• String: DECODE (Length 6, r = 2)
• substring(2) → CODE
• substring(0, 2) → DE
• Result: CODEDE

3. Right Rotation Logic

Right rotation moves characters from the end of the string to the front. To simplify this, we calculate a "partition" point.

Logic:

1. Partition (P): length - r
2. Formula: Result = original.substring(P) + original.substring(0, P)

Example (Rotate Right by 2):

• String: DECODE (Length 6, r = 2)
• Partition (P): 6 - 2 = 4
• substring(4) → DE
• substring(0, 4) → DECO
• Result: DEDECO

Java Implementation

public class StringRotation {
    public static void main(String[] args) {
        String str = "DECODE";
        int r = 2;

        System.out.println("Left Rotation: " + leftRotate(str, r));
        System.out.println("Right Rotation: " + rightRotate(str, r));
    }

    public static String leftRotate(String str, int r) {
        return str.substring(r) + str.substring(0, r);
    }

    public static String rightRotate(String str, int r) {
        int partition = str.length() - r;
        return str.substring(partition) + str.substring(0, partition);
    }
}