rolling hash Archives - Huahua's Tech Road

The hash of a 0-indexed string s of length k, given integers p and m, is computed using the following function:

hash(s, p, m) = (val(s[0]) * p⁰ + val(s[1]) * p¹ + ... + val(s[k-1]) * p^k-1) mod m.

Where val(s[i]) represents the index of s[i] in the alphabet from val('a') = 1 to val('z') = 26.

You are given a string s and the integers power, modulo, k, and hashValue. Return sub, the first substring of s of length k such that hash(sub, power, modulo) == hashValue.

The test cases will be generated such that an answer always exists.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "leetcode", power = 7, modulo = 20, k = 2, hashValue = 0
Output: "ee"
Explanation: The hash of "ee" can be computed to be hash("ee", 7, 20) = (5 * 1 + 5 * 7) mod 20 = 40 mod 20 = 0. 
"ee" is the first substring of length 2 with hashValue 0. Hence, we return "ee".

Example 2:

Input: s = "fbxzaad", power = 31, modulo = 100, k = 3, hashValue = 32
Output: "fbx"
Explanation: The hash of "fbx" can be computed to be hash("fbx", 31, 100) = (6 * 1 + 2 * 31 + 24 * 31²) mod 100 = 23132 mod 100 = 32. 
The hash of "bxz" can be computed to be hash("bxz", 31, 100) = (2 * 1 + 24 * 31 + 26 * 31²) mod 100 = 25732 mod 100 = 32. 
"fbx" is the first substring of length 3 with hashValue 32. Hence, we return "fbx".
Note that "bxz" also has a hash of 32 but it appears later than "fbx".

Constraints:

1 <= k <= s.length <= 2 * 10⁴
1 <= power, modulo <= 10⁹
0 <= hashValue < modulo
s consists of lowercase English letters only.
The test cases are generated such that an answer always exists.

Solution: Sliding window

hash = (((hash – (s[i+k] * p^k-1) % mod + mod) * p) + (s[i] * p⁰)) % mod

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string subStrHash(string s, int power, int modulo, int k, int hashValue) {

const int n = s.length();

long p = 1;

for (int i = 1; i < k; ++i)

p = (p * power) % modulo;

long ans = 0;

for (long i = n - 1, cur = 0; i >= 0; --i) {

if (i + k < n)

cur = ((cur - (s[i + k] - 'a' + 1) * p) % modulo + modulo) % modulo;

cur = (cur * power + (s[i] - 'a' + 1)) % modulo;

if (i + k <= n && cur == hashValue)

ans = i;

}

return s.substr(ans, k);

}

};

Input: s = "level" Output: "l" Explanation: s contains 4 prefix excluding itself ("l", "le", "lev", "leve"), and suffix ("l", "el", "vel", "evel"). The largest prefix which is also suffix is given by "l".

Solution: Rolling Hash

Time complexity: O(n) / worst case: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string longestPrefix(string_view s) {

const int kMod = 16769023;

const int kBase = 128;

const int n = s.length();

int p = 0;

int q = 0;

int a = 1;

int ans = 0;

for (int i = 1; i < n; ++i) {

p = (p + a * s[i - 1]) % kMod;

q = (q * kBase + s[n - i]) % kMod;

a = (a * kBase) % kMod;

if (p == q && s.substr(0, i) == s.substr(n - i))

ans = i;

}

return string(s.substr(0, ans));

}

};

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