Solution Explanation: Finding Substring with Given Hash Value

This problem requires finding the first substring of length k within a given string s that matches a target hashValue. The hash function is defined as:

hash(s, p, m) = (val(s[0]) * p<sup>0</sup> + val(s[1]) * p<sup>1</sup> + ... + val(s[k-1]) * p<sup>k-1</sup>) mod m

where val(s[i]) is the alphabetical index of the character (a=1, b=2,... z=26), p is the power, and m is the modulo.

A naive approach would calculate the hash for every substring of length k, which has a time complexity of O(n*k), where n is the length of the string. However, we can optimize this using a sliding window and rolling hash technique.

Optimized Approach (Sliding Window and Rolling Hash)

The core idea is to efficiently update the hash value as the window slides along the string, avoiding redundant calculations. We process the string from right to left to simplify the hash update.

1. Initial Hash: Calculate the hash of the last k characters of the string.

2. Sliding Window: Iterate through the string from right to left (index i from n-k-1 to 0).

3. Hash Update: For each iteration:

   • Remove the contribution of the character leaving the window (at i+k).
   • Add the contribution of the character entering the window (at i).
   • This update is done efficiently using modular arithmetic.

4. Check and Return: If the updated hash matches hashValue, the substring starting at index i is the solution.

Time and Space Complexity

• Time Complexity: O(n), where n is the length of the string. We iterate through the string once. The hash update within the loop is constant time.
• Space Complexity: O(1). We use a constant amount of extra space to store the current hash and other variables.

Code Implementation (Python)

class Solution:
    def subStrHash(self, s, power, modulo, k, hashValue):
        h, n = 0, len(s)
        p = 1  # power^0 = 1. We use this to avoid repetitive calculation of powers.
        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 # update p for next iteration
 
        j = n - k # Initialize j as the start index of the last substring
        for i in range(n - 1 - k, -1, -1):
            pre = ord(s[i + k]) - ord("a") + 1 #char leaving the window
            cur = ord(s[i]) - ord("a") + 1 #char entering the window
            h = ((h - (pre * p) % modulo + modulo) * power + cur) % modulo # efficient hash update
            if h == hashValue:
                j = i
        return s[j : j + k]

The code in other languages (Java, C++, Go, TypeScript, JavaScript) follows a similar structure and logic, adapting syntax and data types as needed. The core algorithm remains the same, offering efficient computation by leveraging rolling hash and sliding window techniques. The use of BigInt in Javascript and Typescript is necessary to handle potentially large hash values to avoid integer overflow.