How do you solve Match Substring After Replacement on LeetCode?

Match Substring After Replacement (LeetCode #2301) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transformations can be represented as a mapping for quick lookups.

What is the brute force solution for Match Substring After Replacement?

The brute force approach for Match Substring After Replacement is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of `s` that has the same length as `sub`. For each substring, we will see if we can transform `sub` into that substring using the provided mappings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Match Substring After Replacement?

Match Substring After Replacement uses the following concepts: Array, Hash Table, String, String Matching. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Match Substring After Replacement?

Clarify the problem requirements and constraints before diving into coding. Think about edge cases and how mappings might affect them. Explain your thought process and approach clearly during the interview.

What are common mistakes when solving Match Substring After Replacement?

Not considering all possible transformations. Failing to handle cases where characters cannot be transformed.

#2301

Match Substring After Replacement

Hard

ArrayHash TableStringString MatchingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a sliding window technique combined with a mapping to quickly check if `sub` can be transformed into any substring of `s`. This reduces the number of comparisons and leverages the mappings effectively.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping dictionary from the mappings array for quick lookups.
2Step 2: Use a sliding window of size equal to `sub` to traverse `s`.
3Step 3: For each character in the current window, check if it matches the corresponding character in `sub` or can be transformed using the mapping.

solution.py15 lines

1def canMatchSubstring(s, sub, mappings):
2    mapping_dict = {}
3    for old, new in mappings:
4        mapping_dict.setdefault(old, []).append(new)
5    n, m = len(s), len(sub)
6    for i in range(n - m + 1):
7        if canTransform(sub, s[i:i + m], mapping_dict):
8            return True
9    return False
10
11def canTransform(sub, target, mapping_dict):
12    for i in range(len(sub)):
13        if sub[i] != target[i] and target[i] not in mapping_dict.get(sub[i], []):
14            return False
15    return True

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string `s` once with a sliding window, and the space complexity is O(n) due to the mapping dictionary that stores possible transformations.

1Transformations can be represented as a mapping for quick lookups.
2Using a sliding window helps in efficiently checking substrings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.