2156. Find Substring With Given Hash Value

Problem

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]) * 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 * 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.

Solution

class Solution {
    public String subStrHash(String s, int power, int modulo, int k, int hashValue) {
        long mul1 = 1;
        int times = k - 1;
        while (times-- > 0) {
            mul1 = mul1 * power % modulo;
        }
        int index = -1;
        long hash = 0;
        int end = s.length() - 1;
        for (int i = s.length() - 1; i >= 0; i--) {
            int val = s.charAt(i) - 96;
            hash = (hash * power % modulo + val) % modulo;
            if (end - i + 1 == k) {
                if (hash == hashValue) {
                    index = i;
                }
                hash = (hash - (s.charAt(end) - 96) * mul1 % modulo + modulo) % modulo;
                end--;
            }
        }
        return s.substring(index, index + k);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).