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3031. Minimum Time to Revert Word to Initial State II 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  // Same as 3029. Minimum Time to Revert Word to Initial State I
  int minimumTimeToInitialState(string word, int k) {
    const int n = word.length();
    const int maxOps = (n - 1) / k + 1;
    const vector<int> z = zFunction(word);
    for (int ans = 1; ans < maxOps; ++ans)
      if (z[ans * k] >= n - ans * k)
        return ans;
    return maxOps;
  }

  // Returns the z array, where z[i] is the length of the longest prefix of
  // s[i..n) which is also a prefix of s.
  //
  // https://cp-algorithms.com/string/z-function.html#implementation
  vector<int> zFunction(const string& s) {
    const int n = s.length();
    vector<int> z(n);
    int l = 0;
    int r = 0;
    for (int i = 1; i < n; ++i) {
      if (i < r)
        z[i] = min(r - i, z[i - l]);
      while (i + z[i] < n && s[z[i]] == s[i + z[i]])
        ++z[i];
      if (i + z[i] > r) {
        l = i;
        r = i + z[i];
      }
    }
    return z;
  }
};

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class Solution {
  // Same as 3029. Minimum Time to Revert Word to Initial State I
  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;
  }

  // Returns the z array, where z[i] is the length of the longest prefix of
  // s[i..n) which is also a prefix of s.
  //
  // https://cp-algorithms.com/string/z-function.html#implementation
  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;
  }
}

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class Solution:
  # Same as 3029. Minimum Time to Revert Word to Initial State I
  def minimumTimeToInitialState(self, word: str, k: int) -> int:
    n = len(word)
    maxOps = (n - 1) // k + 1
    z = self._zFunction(word)

    for ans in range(1, maxOps):
      if z[ans * k] >= n - ans * k:
        return ans

    return maxOps

  def _zFunction(self, s: str) -> list[int]:
    """
    Returns the z array, where z[i] is the length of the longest prefix of
    s[i..n) which is also a prefix of s.

    https://cp-algorithms.com/string/z-function.html#implementation
    """
    n = len(s)
    z = [0] * n
    l = 0
    r = 0
    for i in range(1, n):
      if i < r:
        z[i] = min(r - i, z[i - l])
      while i + z[i] < n and s[z[i]] == s[i + z[i]]:
        z[i] += 1
      if i + z[i] > r:
        l = i
        r = i + z[i]
    return z