Check if a string can be obtained by rotating another string d places
Given two strings str1 and str2 and an integer d, the task is to check whether str2 can be obtained by rotating str1 by d places (either to the left or to the right).
Examples:
Input: str1 = “abcdefg”, str2 = “cdefgab”, d = 2
Output: Yes
Rotate str1 2 places to the left.Input: str1 = “abcdefg”, str2 = “cdfdawb”, d = 6
Output: No
Approach: An approach to solve the same problem has been discussed here. In this article, reversal algorithm is used to rotate the string to the left and to the right in O(n). If any one of the rotations of str1 is equal to str2 then print Yes else print No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to reverse an array from left // index to right index (both inclusive) void ReverseArray(string& arr, int left, int right) { char temp; while (left < right) { temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } // Function that returns true if str1 can be // made equal to str2 by rotating either // d places to the left or to the right bool RotateAndCheck(string& str1, string& str2, int d) { if (str1.length() != str2.length()) return false; // Left Rotation string will contain // the string rotated Anti-Clockwise // Right Rotation string will contain // the string rotated Clockwise string left_rot_str1, right_rot_str1; bool left_flag = true, right_flag = true; int str1_size = str1.size(); // Copying the str1 string to left rotation string // and right rotation string for (int i = 0; i < str1_size; i++) { left_rot_str1.push_back(str1[i]); right_rot_str1.push_back(str1[i]); } // Rotating the string d positions to the left ReverseArray(left_rot_str1, 0, d - 1); ReverseArray(left_rot_str1, d, str1_size - 1); ReverseArray(left_rot_str1, 0, str1_size - 1); // Rotating the string d positions to the right ReverseArray(right_rot_str1, 0, str1_size - d - 1); ReverseArray(right_rot_str1, str1_size - d, str1_size - 1); ReverseArray(right_rot_str1, 0, str1_size - 1); // Compairing the rotated strings for (int i = 0; i < str1_size; i++) { // If cannot be made equal with left rotation if (left_rot_str1[i] != str2[i]) { left_flag = false; } // If cannot be made equal with right rotation if (right_rot_str1[i] != str2[i]) { right_flag = false; } } // If both or any one of the rotations // of str1 were equal to str2 if (left_flag || right_flag) return true; return false; } // Driver code int main() { string str1 = "abcdefg"; string str2 = "cdefgab"; // d is the rotating factor int d = 2; // In case length of str1 < d d = d % str1.size(); if (RotateAndCheck(str1, str2, d)) cout << "Yes"; else cout << "No"; return 0; } |
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Python3
# Python3 implementation of the approach # Function to reverse an array from left # index to right index (both inclusive) def ReverseArray(arr, left, right) : while (left < right) : temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left += 1; right -= 1; # Function that returns true if str1 can be # made equal to str2 by rotating either # d places to the left or to the right def RotateAndCheck(str1, str2, d) : if (len(str1) != len(str2)) : return False; # Left Rotation string will contain # the string rotated Anti-Clockwise # Right Rotation string will contain # the string rotated Clockwise left_rot_str1 = []; right_rot_str1 = []; left_flag = True; right_flag = True; str1_size = len(str1); # Copying the str1 string to left rotation string # and right rotation string for i in range(str1_size) : left_rot_str1.append(str1[i]); right_rot_str1.append(str1[i]); # Rotating the string d positions to the left ReverseArray(left_rot_str1, 0, d - 1); ReverseArray(left_rot_str1, d, str1_size - 1); ReverseArray(left_rot_str1, 0, str1_size - 1); # Rotating the string d positions to the right ReverseArray(right_rot_str1, 0, str1_size - d - 1); ReverseArray(right_rot_str1, str1_size - d, str1_size - 1); ReverseArray(right_rot_str1, 0, str1_size - 1); # Compairing the rotated strings for i in range(str1_size) : # If cannot be made equal with left rotation if (left_rot_str1[i] != str2[i]) : left_flag = False; # If cannot be made equal with right rotation if (right_rot_str1[i] != str2[i]) : right_flag = False; # If both or any one of the rotations # of str1 were equal to str2 if (left_flag or right_flag) : return True; return False; # Driver code if __name__ == "__main__" : str1 = list("abcdefg"); str2 = list("cdefgab"); # d is the rotating factor d = 2; # In case length of str1 < d d = d % len(str1); if (RotateAndCheck(str1, str2, d)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
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Output:
Yes
Time Complexity: O(n)
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