2156. Find Substring With Given Hash Value

Hard

381

374

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`

`k`

`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 = 0Output:"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 = 32Output:"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**.

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