C++ 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-

1. Left (Or anticlockwise) rotate the given string by d elements (where d <= n)
2. 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 : 

C++

// C program for Left Rotation and Right
// Rotation of a String
#include<bits/stdc++.h>
using namespace std;

// In-place rotates s towards left by d
void leftrotate(string &s, int d)
{
    reverse(s.begin(), s.begin()+d);
    reverse(s.begin()+d, s.end());
    reverse(s.begin(), s.end());
}

// In-place rotates s towards right by d
void rightrotate(string &s, int d)
{
   leftrotate(s, s.length()-d);
}

// Driver code
int main()
{
    string str1 = "GeeksforGeeks";
    leftrotate(str1, 2);
    cout << str1 << endl;

    string str2 = "GeeksforGeeks";
    rightrotate(str2, 2);
    cout << str2 << endl;
    return 0;
}

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!

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