I have a doubt about the implementation of the Rabin-Karp Algorithm given on CP Algorithms.com
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.
Solution:
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;
}
In the last comparison:
if (cur_h == h_s * p_pow[i] % m)
occurences.push_back(i);
Why do we need to multiply h_s by p_pow[i], shouldn't it be just if(cur_h == h_s)
Please help.
Read this paragraph on cp-algorithms.
Because in h_s starting power is 0 but in intermidiate hash starting power is not 0 and there is an extra power of pow[i] in each term so to make comparable we multiply initial hash with pow[i]