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

// 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);
  }
};

The post 花花酱 LeetCode 2156. Find Substring With Given Hash Value appeared first on Huahua's Tech Road.

Given a string s. Return the longest happy prefix of s .

Return an empty string if no such prefix exists.

Example 1:

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".

Example 2:

Input: s = "ababab"
Output: "abab"
Explanation: "abab" is the largest prefix which is also suffix. They can overlap in the original string.

Example 3:

Input: s = "leetcodeleet"
Output: "leet"

Example 4:

Input: s = "a"
Output: ""

Constraints:

• 1 <= s.length <= 10^5
• s contains only lowercase English letters.

Solution: Rolling Hash

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

// 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));
  }
};

The post 花花酱 LeetCode 1392. Longest Happy Prefix appeared first on Huahua's Tech Road.