Formatted question description: https://leetcode.ca/all/2156.html
2156. Find Substring With Given Hash Value (Medium)
The hash of a 0indexed 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[k1]) * p^{k1}) 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 nonempty 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 * 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.
Similar Questions:
Solution 1. Rolling Hash
In RabinKarp algorithm, the hash function is “reversed” hash(s, p, m) = (val(s[0]) * p^(k1) + val(s[1]) * p^(k2) + ... + val(s[k1]) * p^0) mod m
.
During the contest, I felt that if I generate the hash from left to right, I need to do division and it might cause trouble with modulo.
To be safe, I reversed the string and calculated the hashes using the RabinKarp algorithm.

// OJ: https://leetcode.com/problems/findsubstringwithgivenhashvalue/ // Time: O(N) // Space: O(N) which can be reduced to `O(1)` if we really calculate from right to left class Solution { public: string subStrHash(string s, int p, int mod, int k, int target) { long h = 0, N = s.size(), pp = 1; // `pp` = p^k vector<long> hash(N); string r(rbegin(s), rend(s)); for (int i = 0; i < N; ++i) { if (i < k) pp = pp * p % mod; h = (h * p + (r[i]  'a' + 1)) % mod; // push r[i] into the window if (i  k >= 0) { // pop r[ik] out of the window h = (h  (r[i  k]  'a' + 1) * pp % mod + mod) % mod; } if (i >= k  1) hash[i] = h; } reverse(begin(hash), end(hash)); for (int i = 0; i < N; ++i) { if (hash[i] == target) return s.substr(i, k); // hash[i] is the hash of `s[i .. (i+k1)]` } return ""; } };

// Todo

# 2156. Find Substring With Given Hash Value # https://leetcode.com/problems/findsubstringwithgivenhashvalue class Solution: def subStrHash(self, s: str, power: int, M: int, k: int, hashValue: int) > str: def val(x): return ord(x)  ord('a') + 1 curr = 0 b = 1 res = n = len(s) for i in range(n  1, 1, 1): curr = (curr * power + val(s[i])) % M if i + k >= n: b = b * power % M else: curr = (curr  val(s[i + k]) * b) % M if curr == hashValue: res = i return s[res: res + k]
FAQ
1. Why could division + modulo be problematic?
Because when we do division to a moduloed number, we might get the wrong answer.
Assume mod = 100
, a hash h
’s real value is 123
and we want to divide it by 3
. So the expected result should be 41
. But in fact we need to do 123 % 100 / 3
which incorrectly returns 7
.
2. Why do we need division if we calculate from left to right?
Assume s = "abcd", k = 3
.
If we calculate from left to right, hash("abc") = a * p^0 + b * p^1 + c * p^2
. To get hash("bcd")
, we need to subtract a
then divide by p
then add d * p^2
. Here comes the division.
If we calculate from right to left, we can leverage the RabinKarp algorithm which only requires multiplication. We get hash("dcb") = d * p^2 + c * p^1 + b * p^0
first. To get hash("abc")
, we just need to multiply by p
then add a
then subtract d * p^3
.
Discuss
https://leetcode.com/problems/findsubstringwithgivenhashvalue/discuss/1730114/