Java Program for Left Rotation and Right Rotation of a String
Given a string of size n, write functions to perform the following operations on a string-
- Left (Or anticlockwise) rotate the given string by d elements (where d <= n)
- Right (Or clockwise) rotate the given string by d elements (where d <= n).
Examples:
Input : s = "GeeksforGeeks" d = 2 Output : Left Rotation : "eksforGeeksGe" Right Rotation : "ksGeeksforGee" Input : s = "qwertyu" d = 2 Output : Left rotation : "ertyuqw" Right rotation : "yuqwert"
A Simple Solution is to use a temporary string to do rotations. For left rotation, first, copy last n-d characters, then copy first d characters in order to the temporary string. For right rotation, first, copy last d characters, then copy n-d characters.
Can we do both rotations in-place and O(n) time?
The idea is based on a reversal algorithm for rotation.
// Left rotate string s by d (Assuming d <= n) leftRotate(s, d) reverse(s, 0, d-1); // Reverse substring s[0..d-1] reverse(s, d, n-1); // Reverse substring s[d..n-1] reverse(s, 0, n-1); // Reverse whole string. // Right rotate string s by d (Assuming d <= n) rightRotate(s, d) // We can also call above reverse steps // with d = n-d. leftRotate(s, n-d)
Below is the implementation of the above steps :
Java
// Java program for Left Rotation and Right // Rotation of a String import java.util.*; import java.io.*; class GFG { // function that rotates s towards left by d static String leftrotate(String str, int d) { String ans = str.substring(d) + str.substring( 0 , d); return ans; } // function that rotates s towards right by d static String rightrotate(String str, int d) { return leftrotate(str, str.length() - d); } // Driver code public static void main(String args[]) { String str1 = "GeeksforGeeks" ; System.out.println(leftrotate(str1, 2 )); String str2 = "GeeksforGeeks" ; System.out.println(rightrotate(str2, 2 )); } } // This code is contributed by rachana soma |
Output:
Left rotation: eksforGeeksGe Right rotation: ksGeeksforGee
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Left Rotation and Right Rotation of a String for more details!