How do you solve Count Substrings That Differ by One Character on LeetCode?

Count Substrings That Differ by One Character (LeetCode #1638) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m + n² * m) time complexity and O(n * m) space complexity. Key insight: Using a HashMap to store substrings allows for quick lookups.

What is the brute force solution for Count Substrings That Differ by One Character?

The brute force approach for Count Substrings That Differ by One Character is Brute Force, with O(n² * m) time complexity and O(1) space complexity. We will generate all possible substrings of `s`, modify each substring by changing one character, and check if the modified substring exists in `t`. This method is straightforward but inefficient due to the number of comparisons. The optimal Optimal Solution approach improves this to O(n * m + n² * m).

What algorithm or data structure is used to solve Count Substrings That Differ by One Character?

Count Substrings That Differ by One Character uses the following concepts: Hash Table, String, Dynamic Programming, Enumeration. The recommended patterns to study are: Hash Map, Enumeration.

What are the interview tips for Count Substrings That Differ by One Character?

Explain your thought process clearly. Discuss time complexity and space complexity as you go. Consider edge cases, such as empty strings or single character strings.

What are common mistakes when solving Count Substrings That Differ by One Character?

Not considering all character replacements. Forgetting to check if the modified substring exists in `t`.

#1638

Count Substrings That Differ by One Character

Medium

Hash TableStringDynamic ProgrammingEnumerationHash MapEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * m + n² * m)Space O(n * m)

Instead of generating all substrings, we can use a HashMap to store all substrings of `t` and then iterate through `s` to check for valid modifications. This reduces the number of checks needed significantly.

⚙️

Algorithm

3 steps

1Step 1: Store all substrings of `t` in a HashMap with their counts.
2Step 2: Iterate through all substrings of `s` and for each substring, generate all possible modified versions by changing one character.
3Step 3: For each modified substring, check if it exists in the HashMap and add the count to the result.

solution.py16 lines

1def countSubstrings(s, t):
2    from collections import defaultdict
3    count_map = defaultdict(int)
4    for i in range(len(t)):
5        for j in range(i + 1, len(t) + 1):
6            count_map[t[i:j]] += 1
7    count = 0
8    for i in range(len(s)):
9        for j in range(i + 1, len(s) + 1):
10            substr = s[i:j]
11            for k in range(len(substr)):
12                for c in 'abcdefghijklmnopqrstuvwxyz':
13                    if c != substr[k]:
14                        modified = substr[:k] + c + substr[k + 1:]
15                        count += count_map[modified]
16    return count

ℹ

Complexity note: The complexity arises from storing all substrings of `t` (O(n * m)) and then checking each substring of `s` (O(n² * m)). However, this is significantly more efficient than the brute force method due to reduced checks.

1Using a HashMap to store substrings allows for quick lookups.
2Generating all substrings can be expensive; optimizing checks is crucial.

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