What is the brute force solution for Existence of a Substring in a String and Its Reverse?

The brute force approach for Existence of a Substring in a String and Its Reverse is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible substring of length 2 in the original string and see if it exists in the reversed string. This is simple but not efficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Existence of a Substring in a String and Its Reverse?

Existence of a Substring in a String and Its Reverse uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Existence of a Substring in a String and Its Reverse?

Always think about the simplest solution first, but be ready to discuss its limitations. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice writing code on a whiteboard or in a shared document to simulate real interview conditions.

What are common mistakes when solving Existence of a Substring in a String and Its Reverse?

Not considering edge cases, such as strings with less than 2 characters. Failing to recognize that checking for existence in a string can be optimized using a set.

#3083

Existence of a Substring in a String and Its Reverse

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of checking each substring against the reversed string, we can use a set to store all substrings of length 2 from the original string and check for matches in the reversed string.

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Algorithm

3 steps

1Step 1: Create a set to store all substrings of length 2 from the original string s.
2Step 2: Loop through the reversed string and check if any substring of length 2 exists in the set.
3Step 3: If a match is found, return true. If no matches are found after checking all, return false.

solution.py10 lines

1# Full working Python code
2s = 'leetcode'
3substrings = {s[i:i+2] for i in range(len(s) - 1)}
4reverse_s = s[::-1]
5for i in range(len(reverse_s) - 1):
6    if reverse_s[i:i+2] in substrings:
7        print(True)
8        break
9else:
10    print(False)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string a couple of times (once for creating the set and once for checking the reversed string). The space complexity is O(n) due to the storage of substrings in the set.

1Using a set allows for O(1) average time complexity for lookups, making the optimal solution much faster than the brute force approach.
2Understanding how to manipulate strings and utilize data structures like sets can greatly improve efficiency.

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