How do you solve Distinct Echo Substrings on LeetCode?

Distinct Echo Substrings (LeetCode #1316) 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: Understanding how to generate and check substrings is crucial.

What is the brute force solution for Distinct Echo Substrings?

The brute force approach for Distinct Echo Substrings is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves generating all possible substrings of the given text and checking if each substring can be expressed as a concatenation of a smaller substring with itself. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Distinct Echo Substrings?

Distinct Echo Substrings uses the following concepts: String, Trie, Rolling Hash, Hash Function. The recommended patterns to study are: Hash Map, Rolling Hash.

What are the interview tips for Distinct Echo Substrings?

Always clarify the problem requirements before diving into coding. Discuss your thought process and potential optimizations with the interviewer. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Distinct Echo Substrings?

Not considering edge cases with very short strings. Overlooking the need to check for even-length substrings.

#1316

Distinct Echo Substrings

Hard

StringTrieRolling HashHash FunctionHash MapRolling Hash

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a rolling hash technique to efficiently check for distinct echo substrings. By hashing substrings, we can quickly determine if two substrings are equal without directly comparing them.

⚙️

Algorithm

3 steps

1Step 1: Use a rolling hash to generate hashes for all substrings of the input string.
2Step 2: For each substring, check if its length is even and if the first half matches the second half using the hash values.
3Step 3: Store unique valid hashes in a set to count distinct echo substrings.

solution.py19 lines

1def distinctEchoSubstrings(text):
2    MOD = 10**9 + 7
3    base = 31
4    n = len(text)
5    hashes = set()
6    power = [1] * (n + 1)
7    hash_value = 0
8
9    for i in range(n):
10        hash_value = (hash_value * base + (ord(text[i]) - ord('a') + 1)) % MOD
11        power[i + 1] = (power[i] * base) % MOD
12
13    for length in range(1, n // 2 + 1):
14        for start in range(n - 2 * length + 1):
15            hash1 = (hash_value - (hash_value - (ord(text[start]) - ord('a') + 1) * power[length]) % MOD + MOD) % MOD
16            hash2 = (hash_value - (hash_value - (ord(text[start + length]) - ord('a') + 1) * power[length]) % MOD + MOD) % MOD
17            if hash1 == hash2:
18                hashes.add(text[start:start + 2 * length])
19    return len(hashes)

ℹ

Complexity note: The time complexity is O(n) because we compute hash values in linear time and check for distinct substrings using a set. The space complexity is O(n) due to the storage of hash values.

1Understanding how to generate and check substrings is crucial.
2Rolling hash can significantly reduce the time complexity of substring comparison.

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