2156. Find Substring With Given Hash Value

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2156. Find Substring With Given Hash Value

Description

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]) * p^0 + val(s[1]) * p^1 + ... + 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:

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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:

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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 * 31^2) mod 100 = 23132 mod 100 = 32.
The hash of "bxz" can be computed to be hash("bxz", 31, 100) = (2 * 1 + 24 * 31 + 26 * 31^2) 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 * 10^4
  • 1 <= power, modulo <= 10^9
  • 0 <= hashValue < modulo
  • s consists of lowercase English letters only.
  • The test cases are generated such that an answer always exists .

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    string subStrHash(string s, int power, int modulo, int k, int hashValue) {
        int n = s.size();
        long pow= 1;
        for (int i = 1; i < k; i++) {
            pow = pow * power % modulo;
        }
        int left = n - 1;
        long cur = 0;
        string res;
        while (left >= 0) {
            cur = (cur * power + s[left] - 'a' + 1) % modulo;
            if (n - left >= k) {
                if (cur == hashValue) {
                    res = s.substr(left, k);
                }
                int r = s[left + k - 1] - 'a' + 1;
                cur = ((cur - r * pow) % modulo + modulo) % modulo;
            }
            left--;
        }
        return res;
    }
};