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.
Input: S = “ab”, T = “ab”, K = 2
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
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:
- Precompute the number of good(denoted by a) and bad(denoted by b) cyclic shifts.
- 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.
- For transition, we are only concerned about previous state i.e., (i – 1)th state and the answer to this question is dp1[K].
- 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)
- 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:
Time Complexity: O(N)
Auxiliary Space: O(K)
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Improved By : dewantipandeydp