Count number of ways to convert string S to T by performing K cyclic shifts

Given two strings S and T and a number K, the task is to count the number of ways to convert string S to string T by performing K cyclic shifts.

The cyclic shift is defined as the string S can be split into two non-empty parts X + Y and in one operation we can transform S to Y + X from X + Y.

Note: Since count can be very large print the answer to modulo 109 + 7.

Examples:

Input: S = “ab”, T = “ab”, K = 2
Output: 1
Explanation:
The only way to do this is to convert [ab to ba] in the first move and then [ba to ab] in the second move.



Input: S = “ababab”, T = “ababab”, K = 1
Output: 2
Explanation:
One possible way to convert S to T in one move is [ab | abab] -> [ababab], the second way is [abab | ab] -> [ababab]. So there are total two ways.

Approach: This problem can be solved using Dynamic Programming. Let us call a cyclic shift ‘good’ if at the end we are at string T and the vice versa for ‘bad’. Below are the steps:

  1. Precompute the number of good(denoted by a) and bad(denoted by b) cyclic shifts.
  2. Initialize two dp arrays such that dp1[i] denote the number of ways to get to a good shift in i moves and dp2[i] denotes the number of ways to get to a bad shift in i moves.
  3. For transition, we are only concerned about previous state i.e., (i – 1)th state and the answer to this question is dp1[K].
  4. So the number of ways to reach a good state in i moves is equal to the number of ways of reaching a good shift in i-1 moves multiplied by (a-1) (as last shift is also good)
  5. So the number of ways of reaching a bad shift in i-1 moves multiplied by (a)(as next move can be any of the good shifts).

Below is the recurrence relation for the good and bad shifts:

So for good shifts we have:
dp1[i]= dp1[i-1]*(a-1) + dp2[i-1]*a

Similarly, for bad shifts we have:
dp2[i]=dp1[i-1]*b + dp2[i-1]*(b-1)

Below is the implementation of above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define mod 10000000007
  
// Function to count number of ways to
// convert string S to string T by
// performing K cyclic shifts
long long countWays(string s, string t,
                    int k)
{
    // Calculate length of string
    int n = s.size();
  
    // 'a' is no of good cyclic shifts
    // 'b' is no of bad cyclic shifts
    int a = 0, b = 0;
  
    // Iterate in the string
    for (int i = 0; i < n; i++) {
  
        string p = s.substr(i, n - i)
                   + s.substr(0, i);
  
        // Precompute the number of good
        // and bad cyclic shifts
        if (p == t)
            a++;
        else
            b++;
    }
  
    // Initialize two dp arrays
    // dp1[i] to store the no of ways to
    // get to a good shift in i moves
  
    // dp2[i] to store the no of ways to
    // get to a bad shift in i moves
    vector<long long> dp1(k + 1), dp2(k + 1);
  
    if (s == t) {
        dp1[0] = 1;
        dp2[0] = 0;
    }
    else {
        dp1[0] = 0;
        dp2[0] = 1;
    }
  
    // Calculate good and bad shifts
    for (int i = 1; i <= k; i++) {
  
        dp1[i]
            = ((dp1[i - 1] * (a - 1)) % mod
               + (dp2[i - 1] * a) % mod)
              % mod;
  
        dp2[i]
            = ((dp1[i - 1] * (b)) % mod
               + (dp2[i - 1] * (b - 1)) % mod)
              % mod;
    }
  
    // Return the required number of ways
    return dp1[k];
}
  
// Driver Code
int main()
{
    // Given Strings
    string S = "ab", T = "ab";
  
    // Given K shifts required
    int K = 2;
  
    // Function Call
    cout << countWays(S, T, K);
    return 0;
}

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Java

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// Java program for above approach
class GFG{ 
      
static long mod = 10000000007L;
  
// Function to count number of ways to
// convert string S to string T by
// performing K cyclic shifts
static long countWays(String s, String t,
                      int k)
{
      
    // Calculate length of string
    int n = s.length();
  
    // 'a' is no of good cyclic shifts
    // 'b' is no of bad cyclic shifts
    int a = 0, b = 0;
  
    // Iterate in the string
    for(int i = 0; i < n; i++)
    {
       String p = s.substring(i, n - i) + 
                  s.substring(0, i);
         
       // Precompute the number of good
       // and bad cyclic shifts
       if (p == t)
           a++;
       else
           b++;
    }
  
    // Initialize two dp arrays
    // dp1[i] to store the no of ways to
    // get to a good shift in i moves
  
    // dp2[i] to store the no of ways to
    // get to a bad shift in i moves
    long dp1[] = new long[k + 1];
    long dp2[] = new long[k + 1];
  
    if (s == t)
    {
        dp1[0] = 1;
        dp2[0] = 0;
    }
    else
    {
        dp1[0] = 0;
        dp2[0] = 1;
    }
  
    // Calculate good and bad shifts
    for(int i = 1; i <= k; i++)
    {
       dp1[i] = ((dp1[i - 1] * (a - 1)) % mod +
                 (dp2[i - 1] * a) % mod) % mod;
       dp2[i] = ((dp1[i - 1] * (b)) % mod +
                 (dp2[i - 1] * (b - 1)) % mod) % mod;
    }
  
    // Return the required number of ways
    return dp1[k];
}
  
// Driver code 
public static void main(String[] args) 
      
    // Given Strings
    String S = "ab", T = "ab";
  
    // Given K shifts required
    int K = 2;
  
    // Function Call
    System.out.print(countWays(S, T, K)); 
  
// This code is contributed by Pratima Pandey  

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Output:

1

Time Complexity: O(N)
Auxiliary Space: O(K)

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