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

• 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:

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[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

 ``

Output:

`1`

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

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