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3303. Find the Occurrence of First Almost Equal Substring 👍

  • Time: $O(|\texttt{s}| + |\texttt{pattern}|)$
  • Space: $O(|\texttt{s}| + |\texttt{pattern}|)$
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class Solution {
 public:
  int minStartingIndex(string s, string pattern) {
    const vector<int> z1 = zFunction(pattern + s);
    const vector<int> z2 = zFunction(reversed(pattern) + reversed(s));

    // Match s[i..i + len(pattern) - 1] with `pattern` from both the prefix and
    // the suffix.
    for (int i = 0; i <= s.length() - pattern.length(); ++i)
      if (z1[pattern.length() + i] + z2[s.length() - i] >= pattern.length() - 1)
        return i;

    return -1;
  }

 private:
  // 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;
  }

  string reversed(const string& s) {
    return {s.rbegin(), s.rend()};
  }
};

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class Solution {
  public int minStartingIndex(String s, String pattern) {
    int[] z1 = zFunction(new StringBuilder(pattern).append(s).toString());
    int[] z2 = zFunction(new StringBuilder(pattern)
                             .reverse() //
                             .append(new StringBuilder(s).reverse())
                             .toString());

    // Match s[i..i + len(pattern) - 1] with `pattern` from both the prefix and
    // the suffix.
    for (int i = 0; i <= s.length() - pattern.length(); ++i)
      if (z1[pattern.length() + i] + z2[s.length() - i] >= pattern.length() - 1)
        return i;

    return -1;
  }

  // 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;
  }

  private String reversed(String s) {
    return new StringBuilder(s).reverse().toString();
  }
}

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class Solution:
  def minStartingIndex(self, s: str, pattern: str) -> int:
    z1 = self._zFunction(pattern + s)
    z2 = self._zFunction(pattern[::-1] + s[::-1])

    # Match s[i..i + len(pattern) - 1] with `pattern` from both the prefix and
    # the suffix.
    for i in range(len(s) - len(pattern) + 1):
      if z1[len(pattern) + i] + z2[len(s) - i] >= len(pattern) - 1:
        return i

    return -1

  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