# 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++ 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[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 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 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# 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 `

Output:
```1
```

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

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