This algorithm is based on the concept of hashing, so if you are not familiar with string hashing, refer to the string hashing article.

This algorithm was authored by Rabin and Karp in 1987.

Problem: Given two strings - a pattern $s$ and a text $t$, determine if the pattern appears in the text and if it does, enumerate all its occurrences in $O(|s| + |t|)$ time.

Algorithm: Calculate the hash for the pattern $s$. Calculate hash values for all the prefixes of the text $t$. Now, we can compare a substring of length $|s|$ with $s$ in constant time using the calculated hashes. So, compare each substring of length $|s|$ with the pattern. This will take a total of $O(|t|)$ time. Hence the final complexity of the algorithm is $O(|t| + |s|)$: $O(|s|)$ is required for calculating the hash of the pattern and $O(|t|)$ for comparing each substring of length $|s|$ with the pattern.

```
vector<int> rabin_karp(string const& s, string const& t) {
const int p = 31;
const int m = 1e9 + 9;
int S = s.size(), T = t.size();
vector<long long> p_pow(max(S, T));
p_pow[0] = 1;
for (int i = 1; i < (int)p_pow.size(); i++)
p_pow[i] = (p_pow[i-1] * p) % m;
vector<long long> h(T + 1, 0);
for (int i = 0; i < T; i++)
h[i+1] = (h[i] + (t[i] - 'a' + 1) * p_pow[i]) % m;
long long h_s = 0;
for (int i = 0; i < S; i++)
h_s = (h_s + (s[i] - 'a' + 1) * p_pow[i]) % m;
vector<int> occurences;
for (int i = 0; i + S - 1 < T; i++) {
long long cur_h = (h[i+S] + m - h[i]) % m;
if (cur_h == h_s * p_pow[i] % m)
occurences.push_back(i);
}
return occurences;
}
```