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

Post a Comment

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 ...
Post a Comment
Read more

Dynamic Programming - Easy Level - Question 1

Dynamic Programming - Easy Level - Question 1 Leetcode 1646  Get Maximum in Generated Array You are given an integer n. An array nums of length n + 1 is generated in the following way: nums[0] = 0 nums[1] = 1 nums[2 * i] = nums[i] when 2 <= 2 * i <= n nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n Return the maximum integer in the array nums​​​. Constraints: 0 <= n <= 100 Analysis: This question is quick straightforward: the state and transitional formula are given; the initialization is also given. So we can just ready the code to iterate all the states and find the maximum. See the code below: class Solution { public: int getMaximumGenerated(int n) { int res = 0; if(n<2) return n; vector<int> f(n+1, 0); f[1] = 1; for(int i=2; i<=n; ++i) { if(i&1) f[i] = f[i/2] + f[i/2+1]; else f[i] = f[i/2]; // cout<<i<<" "<<f[i]<<endl; ...
Post a Comment
Read more