# 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

• /**
* @param {string} s
* @param {number} power
* @param {number} modulo
* @param {number} k
* @param {number} hashValue
* @return {string}
*/
var subStrHash = function (s, power, modulo, k, hashValue) {
power = BigInt(power);
modulo = BigInt(modulo);
hashValue = BigInt(hashValue);
const n = s.length;
let pk = 1n;
let ac = 0n;
// 倒序滑动窗口
for (let i = n - 1; i > n - 1 - k; i--) {
ac = (ac * power + getCode(s, i)) % modulo;
pk = (pk * power) % modulo;
}
let ans = -1;
if (ac == hashValue) {
ans = n - k;
}
for (let i = n - 1 - k; i >= 0; i--) {
let pre = (getCode(s, i + k) * pk) % modulo;
ac = (ac * power + getCode(s, i) - pre + modulo) % modulo;
if (ac == hashValue) {
ans = i;
}
}
return ans == -1 ? '' : s.substring(ans, ans + k);
};

function getCode(str, index) {
return BigInt(str.charCodeAt(index) - 'a'.charCodeAt(0) + 1);
}


• class Solution {
public String subStrHash(String s, int power, int modulo, int k, int hashValue) {
long h = 0, p = 1;
int n = s.length();
for (int i = n - 1; i >= n - k; --i) {
int val = s.charAt(i) - 'a' + 1;
h = ((h * power % modulo) + val) % modulo;
if (i != n - k) {
p = p * power % modulo;
}
}
int j = n - k;
for (int i = n - k - 1; i >= 0; --i) {
int pre = s.charAt(i + k) - 'a' + 1;
int cur = s.charAt(i) - 'a' + 1;
h = ((h - pre * p % modulo + modulo) * power % modulo + cur) % modulo;
if (h == hashValue) {
j = i;
}
}
return s.substring(j, j + k);
}
}

• class Solution {
public:
string subStrHash(string s, int power, int modulo, int k, int hashValue) {
long long h = 0, p = 1;
int n = s.size();
for (int i = n - 1; i >= n - k; --i) {
int val = s[i] - 'a' + 1;
h = ((h * power % modulo) + val) % modulo;
if (i != n - k) {
p = p * power % modulo;
}
}
int j = n - k;
for (int i = n - k - 1; i >= 0; --i) {
int pre = s[i + k] - 'a' + 1;
int cur = s[i] - 'a' + 1;
h = ((h - pre * p % modulo + modulo) * power % modulo + cur) % modulo;
if (h == hashValue) {
j = i;
}
}
return s.substr(j, k);
}
};

• class Solution:
def subStrHash(
self, s: str, power: int, modulo: int, k: int, hashValue: int
) -> str:
h, n = 0, len(s)
p = 1
for i in range(n - 1, n - 1 - k, -1):
val = ord(s[i]) - ord("a") + 1
h = ((h * power) + val) % modulo
if i != n - k:
p = p * power % modulo
j = n - k
for i in range(n - 1 - k, -1, -1):
pre = ord(s[i + k]) - ord("a") + 1
cur = ord(s[i]) - ord("a") + 1
h = ((h - pre * p) * power + cur) % modulo
if h == hashValue:
j = i
return s[j : j + k]


• func subStrHash(s string, power int, modulo int, k int, hashValue int) string {
h, p := 0, 1
n := len(s)
for i := n - 1; i >= n-k; i-- {
val := int(s[i] - 'a' + 1)
h = (h*power%modulo + val) % modulo
if i != n-k {
p = p * power % modulo
}
}
j := n - k
for i := n - k - 1; i >= 0; i-- {
pre := int(s[i+k] - 'a' + 1)
cur := int(s[i] - 'a' + 1)
h = ((h-pre*p%modulo+modulo)*power%modulo + cur) % modulo
if h == hashValue {
j = i
}
}
return s[j : j+k]
}

• function subStrHash(
s: string,
power: number,
modulo: number,
k: number,
hashValue: number,
): string {
let h: bigint = BigInt(0),
p: bigint = BigInt(1);
const n: number = s.length;
const mod = BigInt(modulo);
for (let i: number = n - 1; i >= n - k; --i) {
const val: bigint = BigInt(s.charCodeAt(i) - 'a'.charCodeAt(0) + 1);
h = (((h * BigInt(power)) % mod) + val) % mod;
if (i !== n - k) {
p = (p * BigInt(power)) % mod;
}
}
let j: number = n - k;
for (let i: number = n - k - 1; i >= 0; --i) {
const pre: bigint = BigInt(s.charCodeAt(i + k) - 'a'.charCodeAt(0) + 1);
const cur: bigint = BigInt(s.charCodeAt(i) - 'a'.charCodeAt(0) + 1);
h = ((((h - ((pre * p) % mod) + mod) * BigInt(power)) % mod) + cur) % mod;
if (Number(h) === hashValue) {
j = i;
}
}
return s.substring(j, j + k);
}