Leetcode #1638: Count Substrings That Differ by One Character
In this guide, we solve Leetcode #1638 Count Substrings That Differ by One Character 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
Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Hash Table, String, Dynamic Programming, Enumeration
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "aba", t = "baba"
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
The underlined portions are the substrings that are chosen from s and t.
Python Solution
class Solution:
def countSubstrings(self, s: str, t: str) -> int:
ans = 0
m, n = len(s), len(t)
for i, a in enumerate(s):
for j, b in enumerate(t):
if a != b:
l = r = 0
while i > l and j > l and s[i - l - 1] == t[j - l - 1]:
l += 1
while (
i + r + 1 < m and j + r + 1 < n and s[i + r + 1] == t[j + r + 1]
):
r += 1
ans += (l + 1) * (r + 1)
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
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.