Problem Description

In this problem, we have a string word, which is indexed starting from 0, and an integer k. The goal is to determine the minimum time to bring word back to its original state through a series of operations performed every second. Each second, we need to:

1. Remove the first k characters (sub-string) from word.
2. Append any k characters at the end of word.

The added twist is that the characters you add don't have to be the ones you just removed. The requirement is to perform both operations simultaneously, and we need to find the minimum number of seconds necessary to revert word to its initial state.

Intuition

The intuition behind the solution relies on discovering a pattern in the transformation of the string word. At each operation, we take a block of k characters from the front and append k characters at the back. To find when the word returns to its original state, we should look for a repetition in the string that matches the start of the string.

The key observation is that if after i operations (where i is a multiple of k), the part of the string after k*i characters matches the original string starting from the 0th index up to n-k*i, where n is the length of word, then the word has come back to its original state. This would mean every k*i characters in the string can be cycled through removals and additions to achieve the initial state of the string.

To implement this idea, we iterate through the string, checking at each multiple of k if the postfixed string word[k*i:] is the same as the start of the original string word[:-k*i]. If we find a match, then dividing the matching index by k gives us the total operations (seconds) taken to get back to the starting state. If we don't find any such match, then the smallest number of operations needed would involve going through the entire string once, hence (n + k - 1) // k operations.

Solution Approach

The implementation of the solution is simple and direct. It follows the steps outlined in the Reference Solution Approach, utilizing a single loop and basic string comparison to find the cycle that brings the word back to its initial state.

No complex data structures are needed for this approach. The main algorithm pattern in place here is enumeration, which is a straightforward technique to iterate through the possibilities until the solution is found.

The process of the algorithm is as follows:

1. Calculate the length of the string word and store it in a variable n.
2. Use a for loop to iterate over the string, starting from k and continuing in increments of k until the end of the string n.
3. Inside the loop, check if the substring of word from the current index i to the end (word[i:]) matches the substring from the start of word to the n - i index (word[:-i]). This represents that after removing and appending k characters i // k times, word has returned to its initial state.
4. If the condition in step 3 is true, immediately return i // k as the minimum time.
5. If the loop completes without finding a match, then the word does not revert to its initial state until we have gone through all its characters at least once. Thus we return (n + k - 1) // k, which accounts for the case where the last set of characters to revert might not be a full k characters in length.

The return (n + k - 1) // k line handles the edge case that when we're left with fewer than k characters at the end, we still need an extra operation to complete the cycle.

This procedure finds the minimum number of operations necessary to restore word to its original state, by systematically checking for cycles in the string formed by the series of operations defined in the problem.

Example Walkthrough

Let's consider a simple example to illustrate the solution approach using a string word = "abcabc" and k = 3.

Step by step process:

1. We calculate the length n of word, which in this case is 6.
2. Now, k is 3, so we will be taking and appending blocks of 3 characters at each operation. We start iterating with a for loop beginning from k (3 in this case) and increment by k with each iteration.
3. At each iteration i (which is a multiple of k), we compare the substring of word from index i to the end, word[i:], with the substring from the start of word to the index n-i, word[:n-i].
   • At i = 3, we take the substring word[i:] which is "abc", and compare it with word[:n-i], which is also "abc". They match, indicating that after 1 operation (removing "abc" from the start and appending "abc" to the end), the word will return to its original state.
4. Since we have found a match, we return i // k, which is 3 // 3 or 1 in this case.

Therefore, in this example, it takes a minimum of 1 operation to bring word back to its original state.

No further iterations are necessary because we found the cycle length in this example during the first iteration. If the word did not revert to its initial state by the end of the iteration through its length, we would have used the formula (n + k - 1) // k to determine the minimum number of operations required.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n^2). In the worst case, the for loop runs for n/k iterations, and in each iteration, it performs a substring operation and string comparison with a complexity of O(n) leading to a total of O(n * (n/k)) which simplifies to O(n^2) because k is a constant and does not grow with n. Substring operations involve creating new strings which take O(n) time each.

The space complexity is O(n) primarily due to the substring operation within the loop that potentially creates a new string of size n at each iteration. Although only one substring is kept in memory at a time, its size could be as large as the original word, which means it directly scales with the input size n.

Learn more about how to find time and space complexity quickly using problem constraints.