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Find minimum shift for longest common prefix

  • Difficulty Level : Hard
  • Last Updated : 29 Apr, 2021

You are given two strings str1 and str2 of the same length. In a single shift, you can rotate one string (str2) by 1 element such that its 1st element becomes the last and the second one becomes the first like “abcd” will change to “bcda” after the one-shift operation. You have to find the minimum shift operation required to get the 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 of the length of the longest prefix for each shift, there is a total of n shifts and for each shift finding the length of the common prefix will take O(n) time. Hence, the overall time complexity for this approach is O(n^2). 
Better Approach: If we will add a second string at the end of itself that is str2 = str2 + str2 then there is no need of finding a 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 shifts required. For finding the longest prefix we can use KMP pattern search algorithm. 
So, in this way, our time-complexity will reduce 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 */

Javascript




<script>
 
 
// Javascript program to find longest common prefix
// after rotation of second string.
 
// function for KMP search
function KMP(m, n, str2, str1)
{
    var pos = 0, len = 0;
 
    // preprocessing of longest proper prefix
    var p = Array(m+1).fill(0);
    var k = 0;
    p[1] = 0;
 
    for (var 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 (var 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
    document.write( "Shift = " + pos + "<br>");
    document.write( "Prefix = " + str1.substr(0, len));
}
 
// driver function
var str1 = "geeksforgeeks";
var str2 = "forgeeksgeeks";
var n = str1.length;
str2 = str2 + str2;
KMP(2 * n, n, str2, str1);
 
 
</script>

Output: 
 

Shift = 8
Prefix = geeksforgeeks

 

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