LeetCode in Kotlin

2156. Find Substring With Given Hash Value

Hard

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 * 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

class Solution {
    fun subStrHash(s: String, power: Int, modulo: Int, k: Int, hashValue: Int): String {
        var mul1: Long = 1
        var times = k - 1
        while (times-- > 0) {
            mul1 = mul1 * power % modulo
        }
        var index = -1
        var hash: Long = 0
        var end = s.length - 1
        for (i in s.length - 1 downTo 0) {
            val `val` = s[i].code - 96
            hash = (hash * power % modulo + `val`) % modulo
            if (end - i + 1 == k) {
                if (hash == hashValue.toLong()) {
                    index = i
                }
                hash = (hash - (s[end].code - 96) * mul1 % modulo + modulo) % modulo
                end--
            }
        }
        return s.substring(index, index + k)
    }
}

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