Skip to main content

Rolling Hash

Rolling hash is one common trick used to increase efficiency of substring comparisons by compressing (or hashing) a string into a integer. After this step, we can compare two strings directly without comparing each chars. So the efficiency can be increased from O(N) to O(1).

So how to implement the rolling hash?

First we need to choose a base for the expansion and a modulo to mod. The basic formula is (suppose the window is n, and the rolling direction is from left to right),

HashVal = (A1*p^(n-1) + A2*p^(n-2) + ... + An-1*p^1 + An*p^0)%mod

where HashVal is the hash value, Ai is the ith element, p is the base, and mod is the modulo.

To avoid collision as much as we can, p and modulo usually need to be large prime numbers.

One corner case is that the base order in the above formula cannot be reversed. Or to be more clear, if the rolling direction is from left to right in an array, the first element should be in the highest order of the base, or times p^(n-1), and the last element times p^0. Then when rolling to the next element, we just need to do two steps:

1. continue to adding the next element as previously

HashVal = (HashVal * p % mod + An+1 * p^0) % mod

2. remove the first element (since it is out of the window n now)

HashVal = (HashVal - A1 * p^n + mod) % mod

If we use a reversed order of the base (in this case, A1 would time p^0 and An * p^(n-1)), we need to divide p instead of times when rolling to the next element, which will lead errors in the module step. More details can be found in the first example in the Question List below.

Question List


Comments

Popular posts from this blog

Brute Force - Question 2

2105. Watering Plants II Alice and Bob want to water n plants in their garden. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at x = i. Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way: Alice waters the plants in order from left to right, starting from the 0th plant. Bob waters the plants in order from right to left, starting from the (n - 1)th plant. They begin watering the plants simultaneously. It takes the same amount of time to water each plant regardless of how much water it needs. Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant. In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice ...

Graph Question - Hard Level - Question 1

2246. Longest Path With Different Adjacent Characters You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1. You are also given a string s of length n, where s[i] is the character assigned to node i. Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them. Constraints: n == parent.length == s.length 1 <= n <= 10^5 0 <= parent[i] <= n - 1 for all i >= 1 parent[0] == -1 parent represents a valid tree. s consists of only lowercase English letters. Analysis For this question, we need to construct a graph, or more specifically, a n-nary tree. How to construct the graph/tree? What we only know is the parent array. So we need to go through the parent array and constr...