3.3.2 String Searching (Z Algorithm)

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/*

Given a single string (needle) and a single text (haystack) to be searched, determine the first
position in which the needle occurs within the haystack in linear time using the Z algorithm. In
comparison, `std::string::find` does not guarantee a linear worst-case bound.

The Z array stores, for each position of a string, the length of the longest substring starting
there that matches a prefix of the whole string. It is computed in linear time by maintaining the
rightmost matched window and seeding each position from its mirror earlier in that window.

The `find_substring_z()` function below calls the Z algorithm on the concatenation of `needle` and
`haystack`, separated by a sentinel value that is guaranteed not to collide with any byte in either
input. Any position whose Z value reaches the needle's length marks an occurrence.

- `z_array(s)` constructs the Z array of sequence `s`. Each `z[i]` is the length of the longest
  prefix of `s` that also occurs starting at index `i`.
- `find_substring_z(haystack, needle)` returns the first position where `needle` occurs in
  `haystack`, or `std::string::npos` if it cannot be found. To return all matches, let the loop run
  and store the results instead of returning early.

Time Complexity:
- O(n) per call to `z_array(s)`, where $n$ is the length of `s`.
- O(n + m) per call to `find_substring_z(haystack, needle)`, where $n$ and $m$ are the lengths of
  `haystack` and `needle`.

Space Complexity:
- O(n) auxiliary for `z_array(s)`, where $n$ is the length of `s`.
- O(n + m) auxiliary for `find_substring_z(haystack, needle)`, where $n$ is the length of `haystack`
  and $m$ is the length of `needle`.

*/

#include <algorithm>
#include <cstddef>
#include <string>
#include <vector>
using std::string;

template<typename Seq>
std::vector<int> z_array(const Seq &s) {
  std::vector<int> z(s.size());
  for (int i = 1, l = 0, r = 0; i < static_cast<int>(z.size()); i++) {
    if (i <= r) {
      z[i] = std::min(r - i + 1, z[i - l]);
    }
    while (i + z[i] < static_cast<int>(z.size()) && s[z[i]] == s[i + z[i]]) {
      z[i]++;
    }
    if (r < i + z[i] - 1) {
      l = i;
      r = i + z[i] - 1;
    }
  }
  return z;
}

std::size_t find_substring_z(const string &haystack, const string &needle) {
  if (needle.empty()) {
    return 0;
  }
  std::vector<int> s;
  s.reserve(needle.size() + haystack.size() + 1);
  for (unsigned char c : needle) {
    s.push_back(c + 1);
  }
  s.push_back(0);
  for (unsigned char c : haystack) {
    s.push_back(c + 1);
  }
  auto z = z_array(s);
  int m = static_cast<int>(needle.size());
  for (int i = m + 1; i < static_cast<int>(z.size()); i++) {
    if (z[i] == m) {
      return i - m - 1;
    }
  }
  return string::npos;
}

/*** Example Usage ***/

#include <cassert>
using namespace std;

int main() {
  assert((z_array(string("aaaaa")) == vector<int>{0, 4, 3, 2, 1}));
  assert(find_substring_z("ABC ABCDAB ABCDABCDABDE", "ABCDABD") == 15);
  assert(find_substring_z("ABC ABCDAB ABCDABCDA", "ABCDABD") == string::npos);
  assert(find_substring_z("aaa", "aa") == 0);  // First overlapping occurrence.
  assert(find_substring_z("abc", "") == 0);
  return 0;
}