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]) * 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.

 

Solutions

Solution: Rolling Hash

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {string} s
 * @param {number} power
 * @param {number} modulo
 * @param {number} k
 * @param {number} hashValue
 * @return {string}
 */
const subStrHash = function (s, power, modulo, k, hashValue) {
  const BASE_POWER = BigInt(power);
  const TARGET_HASH = BigInt(hashValue);
  const MOD = BigInt(modulo);
  const BASE_CODE = 'a'.charCodeAt(0) - 1;
  const n = s.length;
  let hash = 0n;
  let pow = 1n;
  let start = 0;

  for (let index = n - 1; index >= 0; index--) {
    const val = BigInt(s[index].charCodeAt(0) - BASE_CODE);

    hash = (hash * BASE_POWER + val) % MOD;

    if (index + k < n) {
      const removeVal = BigInt(s[index + k].charCodeAt(0) - BASE_CODE);

      hash = (((hash - removeVal * pow) % MOD) + MOD) % MOD;
    } else {
      pow = (pow * BASE_POWER) % MOD;
    }

    if (hash === TARGET_HASH) {
      start = index;
    }
  }

  return s.slice(start, start + k);
};