Distinct Echo Substrings - Leetcode Solution

Problem Statement:

Given a string s, return the number of distinct substrings of s that are echoed.

A substring is echoed if it appears exactly twice in s. For example, the substring "abc" is echoed in the string "abcabc", but not in the string "abcdabcd".

Solution:

To solve this problem, we can start by iterating over all possible substrings of the given string s, and check if the substring is repeated elsewhere in the string (i.e., it is echoed). To avoid checking the same substring multiple times, we can keep track of the substrings that we've seen before in a set.

Let's start by defining a function called count_echoes, which takes a string as input and returns the number of distinct echoed substrings in the string. Here's the code for it:

def count_echoes(s): echoes = set() for i in range(len(s)): for j in range(i+1, len(s)): if s[i:j] == s[j:j+(j-i)]: echoes.add(s[i:j]) return len(echoes)

This function works by iterating over all possible pairs of indices (i, j) such that i < j, and checking if the substring s[i:j] is repeated elsewhere in the string. If it is, we add it to the set of echoes. Finally, we return the length of this set, which gives us the number of distinct echoed substrings in the string.

Now that we have this function, we can use it to solve the main problem. We just need to iterate over all possible substrings of the given string s, and call count_echoes on each one to get the number of echoed substrings. We can do this using a nested loop as follows:

def distinctEchoSubstrings(s): echoes = set() for i in range(len(s)): for j in range(i+1, len(s)+1): echoes.update(count_echoes(s[i:j])) return len(echoes)

This function works by iterating over all possible pairs of indices (i, j) such that i < j, and calling count_echoes on the substring s[i:j]. We then add the number of echoed substrings in this substring to a set of all echoes. Finally, we return the length of this set, which gives us the number of distinct echoed substrings in the original string s.

Time Complexity: O(n^2 * m), where n is the length of the input string s, and m is the length of the longest echoed substring. Since we need to iterate over all possible substrings of s, and then for each substring, we need to scan the remaining string to check if it is echoed, the total time complexity is O(n^2 * m).

Space Complexity: O(m), where m is the length of the longest echoed substring. Since we only need to store the echoes in a set, and there can be at most m distinct echoed substrings (one for each possible length), the space complexity is O(m).

Overall, this solution is not the most efficient one out there, but it is a simple and intuitive approach that works well for small inputs. For larger inputs, more sophisticated algorithms such as the Z algorithm or a suffix tree can be used to improve the time complexity.

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