Z Algorithm
O(n)z[i] is the length of the longest common prefix of s and s[i..]. Same jobs as the prefix function (matching, borders, periods) with a shape some find easier to reason about. For matching, run it on pattern + '#' + text and look for z values equal to the pattern length.
// use it on
CSES: Finding Borders ↗
A border is a prefix that is also a suffix: position i is a border exactly when i + z[i] equals the string length.
// the code
// z[i] = longest common prefix of s and s.substr(i)
vector<int> z_function(const string& s) {
int n = s.size();
vector<int> z(n, 0);
z[0] = n;
for (int i = 1, l = 0, r = 0; 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;
}