# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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: Similarly, for bad shifts we have: Below is the implementation of above approach:

## C++

 `// C++ program for the above approach ` `#include ` `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 = 1; ` `        ``dp2 = 0; ` `    ``} ` `    ``else` `{ ` `        ``dp1 = 0; ` `        ``dp2 = 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   `

Output:

```1
```

Time Complexity: O(N)
Auxiliary Space: O(K) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : dewantipandeydp