Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself (i.e. it can be written as a + a where a is some string).
Example 1:
Input: text = "abcabcabc" Output: 3 Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".
Example 2:
Input: text = "leetcodeleetcode" Output: 2 Explanation: The 2 substrings are "ee" and "leetcodeleetcode".
Constraints:
1 <= text.length <= 2000text has only lowercase English letters.This approach involves utilizing a sliding window to iterate over possible substrings and employing hashing to detect if a substring is a repeated concatenation of a smaller substring. The advantage of hashing is that it allows comparison of substrings in O(1) time, making the solution efficient.
This Python function calculates the number of distinct echo substrings by iterating over possible substring lengths up to half of the text length. It uses a sliding window to compare each substring of length 'length' with the succeeding substring of the same length. If they are equal, it stores the combined substring in a set to ensure uniqueness.
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Time Complexity: O(n^2), where n is the length of the text. Space Complexity: O(n), for storing hash values.
This approach uses Dynamic Programming (DP) to keep track of repeated substring patterns. A DP table can help store results of previous computations, thus optimizing the check for repeated substrings by re-using calculated solutions.
The Python implementation uses a 2D DP table where dp[i][j] stores the length of the longest common suffix of text[i:] and text[j:]. It stores unique echo substrings in a set. The function performs checks to ensure that only valid echo substrings are stored.
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C++
JavaScript
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C
Time Complexity: O(n^2). Space Complexity: O(n^2) due to the DP table.
| Approach | Complexity |
|---|---|
| Sliding Window with Hashing | Time Complexity: O(n^2), where n is the length of the text. Space Complexity: O(n), for storing hash values. |
| Dynamic Programming | Time Complexity: O(n^2). Space Complexity: O(n^2) due to the DP table. |
Minimum Window Substring - Airbnb Interview Question - Leetcode 76 • NeetCode • 385,124 views views
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