Minimum Time to Revert Word to Initial State II - Practice Coding | SlaveCode
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3031. Minimum Time to Revert Word to Initial State II
Hard
50 Points
String
Rolling Hash
String Matching
Hash Function
You are given a 0-indexed string word and an integer k.
At every second, you must perform the following operations:
Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second.
Return the minimum time greater than zero required for word to revert to its initial state.
Examples
Example 1
Input: word = "abacaba", k = 3
Output: 2
Explanation: At the 1st second, we remove characters "aba" from the prefix of word, and add characters "bac" to the end of word. Thus, word becomes equal to "cababac".
At the 2nd second, we remove characters "cab" from the prefix of word, and add "aba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.
Example 2
Input: word = "abacaba", k = 4
Output: 1
Explanation: At the 1st second, we remove characters "abac" from the prefix of word, and add characters "caba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.
Example 3
Input: word = "abcbabcd", k = 2
Output: 4
Explanation: At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.
After 4 seconds, word becomes equal to "abcbabcd" and reverts to its initial state.
It can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.
Constraints
1 <= word.length <= 106
1 <= k <= word.length
word consists only of lowercase English letters.
3031. Minimum Time to Revert Word to Initial State II
Hard
50 Points
String
Rolling Hash
String Matching
Hash Function
You are given a 0-indexed string word and an integer k.
At every second, you must perform the following operations:
Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second.
Return the minimum time greater than zero required for word to revert to its initial state.
Examples
Example 1
Input: word = "abacaba", k = 3
Output: 2
Explanation: At the 1st second, we remove characters "aba" from the prefix of word, and add characters "bac" to the end of word. Thus, word becomes equal to "cababac".
At the 2nd second, we remove characters "cab" from the prefix of word, and add "aba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.
Example 2
Input: word = "abacaba", k = 4
Output: 1
Explanation: At the 1st second, we remove characters "abac" from the prefix of word, and add characters "caba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.
Example 3
Input: word = "abcbabcd", k = 2
Output: 4
Explanation: At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.
After 4 seconds, word becomes equal to "abcbabcd" and reverts to its initial state.
It can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.