You are given two string str1 and str2 of same length. In a single shift you can rotate one string (str2) by 1 element such that its 1st element becomes the last and second one becomes the first like “abcd” will change to “bcda” after one shift operation. You have to find the minimum shift operation required to get common prefix of maximum length from str1 and str2.
Examples:
Input : str1[] = "geeks",
str2 = "dgeek"
Output : Shift = 1,
Prefix = geek
Input : str1[] = "practicegeeks",
str2 = "coderpractice"
Output : Shift = 5
Prefix = practice
Naive Approach : Shift second string one by one and keep track the length of longest prefix for each shift, there are total of n shifts and for each shift finding the length of common prefix will take O(n) time. Hence, overall time complexity for this approach is O(n^2).
Better Approach : If we will add second string at the end of itself that is str2 = str2 + str2 then there is no need of finding prefix for each shift separately. Now, after adding str2 to itself we have to only find the longest prefix of str1 present in str2 and the starting position of that prefix in str2 will give us the actual number of shift required. For finding longest prefix we can use KMP pattern search algorithm.
So, in this way our time-complexity will reduces to O(n) only.
C++
// CPP program to find longest common prefix // after rotation of second string. #include <bits/stdc++.h> using namespace std; // function for KMP search void KMP(int m, int n, string str2, string str1) { int pos = 0, len = 0; // preprocessing of longest proper prefix int p[m + 1]; int k = 0; p[1] = 0; for (int i = 2; i <= n; i++) { while (k > 0 && str1[k] != str1[i - 1]) k = p[k]; if (str1[k] == str1[i - 1]) ++k; p[i] = k; } // find out the longest prefix and position for (int j = 0, i = 0; i < m; i++) { while (j > 0 && str1[j] != str2[i]) j = p[j]; if (str1[j] == str2[i]) j++; // for new position with longer prefix in str2 // update pos and len if (j > len) { len = j; pos = i - j + 1; } } // print result cout << "Shift = " << pos << endl; cout << "Prefix = " << str1.substr(0, len); } // driver function int main() { string str1 = "geeksforgeeks"; string str2 = "forgeeksgeeks"; int n = str1.size(); str2 = str2 + str2; KMP(2 * n, n, str2, str1); return 0; } |
Java
// Java program to find longest common prefix // after rotation of second string. class GFG { // function for KMP search static void KMP(int m, int n, String str2, String str1) { int pos = 0, len = 0; int []p = new int[m + 1]; int k = 0; //p[1] = 0; char []ch1 = str1.toCharArray(); char []ch2 = str2.toCharArray(); for (int i = 2; i <= n; i++) { while (k > 0 && ch1[k] != ch1[i - 1]) k = p[k]; if (ch1[k] == ch1[i - 1]) ++k; p[i] = k; } // find out the longest prefix and position for (int j = 0, i = 0; i < m; i++) { while (j > 0 && j < n && ch1[j] != ch2[i]) j = p[j]; if (j < n && ch1[j] == ch2[i]) j++; // for new position with longer prefix in str2 // update pos and len if (j > len) { len = j; pos = i - j + 1; } } // print result System.out.println("Shift = " + pos); System.out.println("Prefix = " + str1.substring(0,len)); } // Driver code public static void main(String[] args) { String str1 = "geeksforgeeks"; String str2 = "forgeeksgeeks"; int n = str1.length(); str2 = str2 + str2; KMP(2 * n, n, str2, str1); } } // This code is contributed by Ita_c. |
Python3
# Python3 program to find longest common prefix # after rotation of second string. # function for KMP search def KMP(m, n, str2, str1): pos = 0 Len = 0 # preprocessing of longest proper prefix p = [0 for i in range(m + 1)] k = 0 for i in range(2, n + 1): while (k > 0 and str1[k] != str1[i - 1]): k = p[k] if (str1[k] == str1[i - 1]): k += 1 p[i] = k # find out the longest prefix and position j = 0 for i in range(m): while (j > 0 and j < n and str1[j] != str2[i]): j = p[j] if (j < n and str1[j] == str2[i]): j += 1 # for new position with longer prefix # in str2 update pos and Len if (j > Len): Len = j pos = i - j + 1 # prresult print("Shift = ", pos) print("Prefix = ", str1[:Len]) # Driver Code str1 = "geeksforgeeks"str2 = "forgeeksgeeks"n = len(str1) str2 = str2 + str2 KMP(2 * n, n, str2, str1) # This code is contributed by Mohit kumar 29 |
C#
// C# program to find longest common prefix // after rotation of second string. using System; class GFG { // function for KMP search static void KMP(int m, int n, String str2, String str1) { int pos = 0, len = 0; int []p = new int[m + 1]; int k = 0; //p[1] = 0; char []ch1 = str1.ToCharArray(); char []ch2 = str2.ToCharArray(); for (int i = 2; i <= n; i++) { while (k > 0 && ch1[k] != ch1[i - 1]) k = p[k]; if (ch1[k] == ch1[i - 1]) ++k; p[i] = k; } // find out the longest prefix and position for (int j = 0, i = 0; i < m; i++) { while (j > 0 && j < n && ch1[j] != ch2[i]) j = p[j]; if (j < n && ch1[j] == ch2[i]) j++; // for new position with longer prefix in str2 // update pos and len if (j > len) { len = j; pos = i - j + 1; } } // print result Console.WriteLine("Shift = " + pos); Console.WriteLine("Prefix = " + str1.Substring(0,len)); } // Driver code public static void Main(String[] args) { String str1 = "geeksforgeeks"; String str2 = "forgeeksgeeks"; int n = str1.Length; str2 = str2 + str2; KMP(2 * n, n, str2, str1); } } /* This code contributed by PrinciRaj1992 */ |
Output:
Shift = 8 Prefix = geeksforgeeks
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
