Leetcode #2156: Find Substring With Given Hash Value
In this guide, we solve Leetcode #2156 Find Substring With Given Hash Value in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Sliding Window, Hash Function, Rolling Hash
Intuition
We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.
Expanding and shrinking the window lets us maintain validity without restarting the scan.
Approach
Grow the window with a right pointer, and shrink from the left only when the constraint is violated.
Track the best window as you go to keep the solution linear.
Steps:
- Expand the right end of the window.
- While invalid, move the left end to restore constraints.
- Update the best window found.
Example
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".
Python Solution
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]
Complexity
The time complexity is , where is the length of the string. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.