String Hashing

Tutorial

Implementation - Single Base

As mentioned in the articles above, there is no need to calculate modular inverses.

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#include <vector>
#include <string>

using namespace std;

class HashedString {
	private:
		// change M and P if you want
		static const long long M = 1e9 + 9;
		static const long long P = 9973;

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import java.util.*;

public class HashedString {
	// Change M and P if you want
	public static final long M = (long) 1e9 + 9;
	public static final long P = 9973;

	// pow[i] contains P^i % M
	private static ArrayList<Long> pow = new ArrayList<>();

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class HashedString:
	# Change M and P if you want
	M = int(1e9) + 9
	P = 9973

	# pow[i] contains P^i % M
	_pow = [1]

	def __init__(self, s: str):
		while len(self._pow) < len(s):

This implementation calculates

The hash of any particular substring $S[a : b]$ is then calculated as

using prefix sums. This is nice because the highest power of $P$ in that polynomial will always be $P^{b - a}$.

Since $10^9 + 9$ is prime, the probability of collision when using this hash is at most $\frac{N}{10^9 + 9} < 10^{-4}$, by the Schwarz-Zippel lemma. This means that if you select any two different strings of length at most $N$ and a random base modulo $10^9 + 9$ (e.g. $9973$ in the code), the probability that they hash to the same value is at most $10^{-4}$.

Implementation - Multiple Bases

It's generally a good idea to use two randomized bases rather than just one to decrease the probability that two random strings hash to the same value.

Solution - Searching For Strings

One Hash

Time Complexity: $\mathcal O((|N| + |H|) \cdot \Sigma)$, where $\Sigma$ is the size of the alphabet.

We'll use a sliding window over $H$ to find the "matches" with $N$.

Since we don't care about relative order when comparing two substrings, we can store frequency tables of the characters in the current window and in $N$. When we slide the window, at most two values in that table change. To compare two substrings, we simply compare the 26 values in each table.

If we only needed to count the number of matches, then the above alone would suffice (in fact, IOI 2006 Writing is just that). However, we need to count the distinct permutations of $N$ in $H$, so we need to be a bit more clever.

One way to solve this is by storing the polynomial hashes of each match in a hashset, since we expect different permutations to have different polynomial hashes. The answer would simply be the size of that hashset at the end.

Since the test data for this particular problem is very strong, we will probably get hash collisions with only one hash. To remedy this, we use two hashes for each match - this significantly decreases the probability of collisions.

Using the base $9973$ with the two modulos $10^9 + 9$ and $10^9 + 7$ works for this problem. (Note that using two bases with the same modulo works too.)

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#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll P = 9973, M1 = 1e9 + 9, M2 = 1e9 + 7;

int freq_target[26], freq_curr[26];
string n, h;

int main() {

Two Hashes

Time Complexity: $\mathcal O((|N| + |H|) \log M)$

An alternative solution without frequency tables would be to hash the substrings that we're trying to match. Since order doesn't matter, we need to modify our hash function slightly.

In particular, instead of computing the polynomial hash of the substrings, compute the product $(P + s_1)(P + s_2) \cdots (P + s_k) \bmod M$ as the hash (again, using two modulos). This hash is nice because the relative order of the letters doesn't matter, as multiplication is commutative.

Since this hash requires the modular inverse, there's an extra $\log M$ factor in the time complexity.

Alternative hashes (e.g. computing the sum $(P + s_1)^2 + (P + s_2)^2 + \dots + (P + s_k)^2 \bmod M$) also work for other hashing problems, but the test cases are too strong for that to pass here.

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#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll P = 9973, M1 = 1e9 + 9, M2 = 1e9 + 7;

ll inv(ll base, ll MOD) {
	ll ans = 1, expo = MOD - 2;
	while (expo) {
		if (expo & 1) ans = ans * base % MOD;

Problems