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

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  • Difficulty Level : Hard
  • Last Updated : 05 May, 2021

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:
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:
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++




// 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;
}


Java




// 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


Python3




# Python3 program for the above approach
mod = 1000000007
 
# Function to count number of ways
# to convert string S to string T by
# performing K cyclic shifts
def countWays(s, t, k):
     
    # Calculate length of string
    n = len(s)
     
    # a is no. of good cyclic shifts
    # b is no. of bad cyclic shifts
    a = 0
    b = 0
     
    # Iterate in string
    for i in range(n):
        p = s[i : n - i + 1] + s[: i + 1]
         
        # Precompute the number of good
        # and bad cyclic shifts
        if(p == t):
            a += 1
        else:
            b += 1
             
    # Initialize two dp arrays
    # dp1[i] to store the no of ways to
    # get to a goof shift in i moves
     
    # dp2[i] to store the no of ways to
    # get to a bad shift in i moves
    dp1 = [0] * (k + 1)
    dp2 = [0] * (k + 1)
     
    if(s == t):
        dp1[0] = 1
        dp2[0] = 0
    else:
        dp1[0] = 0
        dp2[0] = 1
         
    # Calculate good and bad shifts    
    for i in range(1, k + 1):
        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
 
# Given Strings
S = 'ab'
T = 'ab'
 
# Given K shifts required
K = 2
 
# Function call
print(countWays(S, T, K))
 
# This code is contributed by Arjun Saini


C#




// C# program for the above approach
using System;
 
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
    Console.Write(countWays(S, T, K));
}
}
 
// This code is contributed by rutvik_56


Javascript




<script>
 
// JavaScript program for the above approach
 
let mod = 10000000007;
 
// Function to count number of ways to
// convert string S to string T by
// performing K cyclic shifts
function countWays(s, t, k)
{
     
    // Calculate length of string
    let n = s.length;
 
    // 'a' is no of good cyclic shifts
    // 'b' is no of bad cyclic shifts
    let a = 0, b = 0;
 
    // Iterate in the string
    for(let i = 0; i < n; i++)
    {
    let 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
    let dp1 = Array.from({length: k+1}, (_, i) => 0);
    let dp2 = Array.from({length: k+1}, (_, i) => 0);
 
    if (s == t)
    {
        dp1[0] = 1;
        dp2[0] = 0;
    }
    else
    {
        dp1[0] = 0;
        dp2[0] = 1;
    }
 
    // Calculate good and bad shifts
    for(let 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
 
    // Given Strings
    let S = "ab", T = "ab";
 
    // Given K shifts required
    let K = 2;
 
    // Function Call
    document.write(countWays(S, T, K));
 
</script>


Output: 

1

 

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


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