Minimum Time to Revert Word to Initial State I

Ah, the classic dilemma of wanting to revert a word to its initial state. It’s like trying to un-bake a cake after realizing you forgot the eggs. You know, the moment when you think, “Why did I even start this?” Well, welcome to the world of LeetCode, where we tackle the problem of reverting a word to its original form with the least amount of effort—because who has time for that?

Language Options

If you’re feeling adventurous and want to explore solutions in other languages, check these out:

Code Solution

Here’s the Java solution that will save you from the existential crisis of reverting words:


class Solution {
  public int minimumTimeToInitialState(String word, int k) {
    final int n = word.length();
    final int maxOps = (n - 1) / k + 1;
    final int[] z = zFunction(word);
    for (int ans = 1; ans < maxOps; ++ans)
      if (z[ans * k] >= n - ans * k)
        return ans;
    return maxOps;
  }

  private int[] zFunction(final String s) {
    final int n = s.length();
    int[] z = new int[n];
    int l = 0;
    int r = 0;
    for (int i = 1; i < n; ++i) {
      if (i < r)
        z[i] = Math.min(r - i, z[i - l]);
      while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
        ++z[i];
      if (i + z[i] > r) {
        l = i;
        r = i + z[i];
      }
    }
    return z;
  }
}

Approach

The approach here is as elegant as a cat walking on a keyboard. We utilize the Z-function, which helps us find the longest prefix of the string that matches a substring. The algorithm calculates how many operations are needed to revert the word back to its original state by checking how many segments of size k can be reverted in one go. If the Z-value at a certain position indicates that the remaining substring can be reverted, we return the number of operations needed.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n), where n is the length of the word.
Space Complexity	O(n) for storing the Z-array.

Real-World Example

Imagine you’re trying to clean your room, but every time you put something back in its place, your cat decides to knock it over. You can only pick up a few items at a time (let’s say k items). The goal is to figure out how many rounds of cleaning you need to do before your room is back to its pristine state. This problem is essentially about minimizing those rounds of cleaning—because who has time for that?