3137. Minimum Number of Operations to Make Word K-Periodic
Description
You are given a string word
of size n
, and an integer k
such that k
divides n
.
In one operation, you can pick any two indices i
and j
, that are divisible by k
, then replace the substring of length k
starting at i
with the substring of length k
starting at j
. That is, replace the substring word[i..i + k - 1]
with the substring word[j..j + k - 1]
.
Return the minimum number of operations required to make word
k-periodic.
We say that word
is k-periodic if there is some string s
of length k
such that word
can be obtained by concatenating s
an arbitrary number of times. For example, if word == “ababab”
, then word
is 2-periodic for s = "ab"
.
Example 1:
Input: word = "leetcodeleet", k = 4
Output: 1
Explanation:
We can obtain a 4-periodic string by picking i = 4 and j = 0. After this operation, word becomes equal to "leetleetleet".
Example 2:
Input: word = "leetcoleet", k = 2
Output: 3
Explanation:
We can obtain a 2-periodic string by applying the operations in the table below.
i | j | word |
---|---|---|
0 | 2 | etetcoleet |
4 | 0 | etetetleet |
6 | 0 | etetetetet |
Constraints:
1 <= n == word.length <= 105
1 <= k <= word.length
k
dividesword.length
.word
consists only of lowercase English letters.
Solutions
Solution 1: Counting
We can divide the string word
into substrings of length $k$, then count the occurrence of each substring, and finally return $n/k$ minus the count of the most frequently occurring substring.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string word
.
1 2 3 4 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|