Problem Statement
The challenge presented requires determining the length of the longest repeating substring within a given string s. If s does not contain any repeating substrings, the function should return 0. A repeating substring can be described as a sequence of characters that appears more than once in the string, and the goal is to identify the length of the longest sequence of this nature.
Examples
Example 1
Input:
s = "abcd"
Output:
0
Explanation:
There is no repeating substring.
Example 2
Input:
s = "abbaba"
Output:
2
Explanation:
The longest repeating substrings are "ab" and "ba", each of which occurs twice.
Example 3
Input:
s = "aabcaabdaab"
Output:
3
Explanation:
The longest repeating substring is "aab", which occurs 3 times.
Constraints
1 <= s.length <= 2000sconsists of lowercase English letters.
Approach and Intuition
The problem of finding the longest repeating substring in a string combines elements of string manipulation and search techniques. From the provided examples and constraints, we can glean some insights and potential approaches:
- A Brute Force approach would involve generating all possible substrings and checking for repetitions, which would be computationally expensive given
scan be up to 2000 characters long. - Optimizations could include using binary search to find the "longest" feasible substring length, combined with some form of hashing or searching technique like the Rabin-Karp algorithm for efficient substring comparison.
- As evident from the examples:
- For
s = "abcd", there are no substrings that repeat, leading to a return of0. - For
s = "abbaba", the longest repeating substrings are "ab" and "ba", each of two characters in length, hence the function returns2. - In
s = "aabcaabdaab", the three-character substring "aab" repeats, and thus the function returns3as the length of the longest repeating substrings.
- For
- The constraints:
- The maximum length for
sis 2000, suggesting that while a solution must be efficient, particularly in terms of time complexity, there is some processing power available. - Only lowercase English letters will be present, simplifying the need to handle a broad charset, and allowing for more straightforward hashing or array index-based solutions.
- The maximum length for
Solutions
- C++
- Java
- Python
class Solution {
public:
int findLongestRepeatingSubstring(string s) {
int n = s.size();
vector<string> subs(n);
// Store all suffixes of the input string
for (int i = 0; i < n; i++) {
subs[i] = s.substr(i);
}
// Apply MSD radix sort to the suffixes
radixSort(subs);
int maxLen = 0;
// Find the longest common prefix between consecutive suffixes
for (int i = 1; i < n; i++) {
int cnt = 0;
// Compare characters of adjacent suffixes
while (cnt < min(subs[i].size(), subs[i - 1].size()) &&
subs[i][cnt] == subs[i - 1][cnt]) {
cnt++;
}
// Update the maximum prefix length found
maxLen = max(maxLen, cnt);
}
return maxLen; // Return the length of the longest repeating substring
}
private:
void radixSort(vector<string>& arr) {
vector<string> temp(arr.size());
stableSort(arr, 0, arr.size() - 1, 0, temp);
}
void stableSort(vector<string>& arr, int low, int high, int depth,
vector<string>& temp) {
if (low >= high) return;
vector<int> count(28, 0);
// Count characters at 'depth'
for (int i = low; i <= high; i++) {
count[charAt(arr[i], depth) + 1]++;
}
// Calculate positions
for (int i = 1; i < 28; i++) {
count[i] += count[i - 1];
}
// Distribute records
for (int i = low; i <= high; i++) {
temp[count[charAt(arr[i], depth)]++] = arr[i];
}
// Copy back to arr
for (int i = low; i <= high; i++) {
arr[i] = temp[i - low];
}
// Recursive subarray sorts
for (int i = 0; i < 27; i++) {
stableSort(arr, low + count[i], low + count[i + 1] - 1, depth + 1, temp);
}
}
int charAt(const string& s, int d) {
if (d >= s.size()) return 0;
return s[d] - 'a' + 1;
}
};
The solution for finding the longest repeating substring in C++ leverages suffix sorting and comparison. Here's a quick breakdown of the process:
- Suffix Array Creation: Starts by generating all suffixes of the given string and storing them in an array.
- Sorting: Uses a most significant digit (MSD) radix sort algorithm to lexicographically sort the suffixes of the string.
- Common Prefix Length Measurement: Iterates over the sorted suffixes to identify the length of the longest common prefix between each pair of consecutive suffixes.
- Result Calculation: Captures and returns the maximum length of these common prefixes as the length of the longest repeating substring.
Detailed Steps within the Functions:
findLongestRepeatingSubstring:
- Initializes an array to store the suffixes.
- Sorts the suffixes using the radix sort method.
- Calculates the longest repeating substring by comparing adjacent sorted suffixes.
radixSort:
- Utilizes the stableSort function recursively which organizes the suffixes by comparing characters incrementally, ensuring stability and lexicographic order.
stableSort:
- Counts occurrences of each character at the current depth in all suffixes using an auxiliary array.
- Sorts the suffixes based on these counts and recursively sorts the resultant buckets, increasing the depth each time to achieve complete sorting.
This approach efficiently handles the substring search by sorting and then simply comparing adjacent elements in the sorted array, rather than engaging in a more naive or brute force comparison of all possible substrings.
class Solution {
public int longestRepeatedSubstring(String str) {
int length = str.length();
String[] suffixes = new String[length];
// Generating suffix array
for (int i = 0; i < length; i++) {
suffixes[i] = str.substring(i);
}
// Implementing MSD Radix Sort on suffixes
radixSortMSD(suffixes);
int maxLcpLength = 0;
// Identifying longest common prefix among sorted suffixes
for (int i = 1; i < length; i++) {
int lcp = 0;
while (
lcp < Math.min(suffixes[i].length(), suffixes[i - 1].length()) &&
suffixes[i].charAt(lcp) == suffixes[i - 1].charAt(lcp)
) {
lcp++;
}
maxLcpLength = Math.max(maxLcpLength, lcp);
}
return maxLcpLength;
}
// Method to apply MSD Radix Sort on suffix array
private void radixSortMSD(String[] array) {
msdSort(array, 0, array.length - 1, 0, new String[array.length]);
}
// Helper sort method for MSD radix sorting
private void msdSort(String[] array, int lo, int hi, int d, String[] temp) {
if (lo >= hi) return;
int[] count = new int[28];
for (int i = lo; i <= hi; i++) {
count[charPosition(array[i], d) + 1]++;
}
for (int i = 1; i < 28; i++) {
count[i] += count[i - 1];
}
for (int i = lo; i <= hi; i++) {
temp[count[charPosition(array[i], d)]++] = array[i];
}
for (int i = lo; i <= hi; i++) {
array[i] = temp[i - lo];
}
for (int i = 0; i < 27; i++) {
msdSort(array, lo + count[i], lo + count[i + 1] - 1, d + 1, temp);
}
}
// Fetch character position or return 0 for bounds
private int charPosition(String s, int index) {
if (index >= s.length()) return 0;
return s.charAt(index) - 'a' + 1;
}
}
In solving the problem of finding the length of the longest repeating substring in Java, the approach involves constructing a suffix array and implementing radix sort algorithms. Here's the breakdown:
- Start by extracting each possible suffix from the given input string and developing a suffix array from these suffixes.
- Employ the most significant digit (MSD) radix sort to organize the suffixes lexicographically. This sort handles sorting based on multiple characters of the strings.
- Once sorted, identify the longest common prefix (LCP) among consecutive entries in the sorted suffix list. This reveals the longest repeating substring length by comparing adjacent suffixes and counting maximal identical initial segments.
- Efficiently manage sorting using recursive helper methods that handle array partitions based on character positions, utilizing temporary arrays and counting distributions for sorting characters at specific positions.
By following these methods, the radix sort effectively groups substrings facilitating the efficient discovery of the longest recurring sequence within the string. The overall logic and flow remain clear, ensuring optimal performance and readability of the solution.
class Solution:
def findLongestRepeatedSubstring(self, input_str: str) -> int:
n = len(input_str)
substrings = []
# Construct array with all possible substrings
for start in range(n):
substrings.append(input_str[start:])
# Sort these substrings efficiently
self._sort_substrings(substrings)
longest = 0
# Determine longest repeated substring by comparing adjacent
for idx in range(1, n):
temp_length = 0
# Compare character by character
while (
temp_length < min(len(substrings[idx]), len(substrings[idx - 1]))
and substrings[idx][temp_length] == substrings[idx - 1][temp_length]
):
temp_length += 1
# Record the length of longest match
longest = max(longest, temp_length)
return longest
def _sort_substrings(self, array: List[str]) -> None:
temporary = ["" for _ in array]
self._msd_sort_helper(array, 0, len(array) - 1, 0, temporary)
def _msd_sort_helper(
self, array: List[str], start: int, end: int, char_index: int, temp: List[str]
) -> None:
if start >= end:
return
# Initialize character bucket sizes (incl. pseudo-EOF at index 0)
buckets = [0] * 28
# Count frequency of each character at the current depth
for idx in range(start, end + 1):
buckets[self._char_idx(array[idx], char_index) + 1] += 1
# Accumulate bucket sizes to get positions
for i in range(1, 28):
buckets[i] += buckets[i - 1]
# Move items into place in the temp array
for idx in range(start, end + 1):
position = self._char_idx(array[idx], char_index)
temp[buckets[position]] = array[idx]
buckets[position] += 1
# Copy sorted items back into original array
for idx in range(start, end + 1):
array[idx] = temp[idx - start]
# Recursively sort individual buckets
for j in range(27):
self._msd_sort_helper(
array, start + buckets[j], start + buckets[j + 1] - 1, char_index + 1, temp
)
def _char_idx(self, s: str, index: int) -> int:
# 0 for out-of-bounds (acts as string end marker)
if index >= len(s):
return 0
return ord(s[index]) - ord("a") + 1
The Python3 program you're working with aims to identify the longest repeating substring within a given string by utilizing a sophisticated method based on comparison of suffixes and sorting. Follow these insights to understand the solution:
The central task of the program is to process an input string and identify the length of the longest repeating substring present in it. The main class
Solutionhas two methods to accomplish this:findLongestRepeatedSubstringand_sort_substrings.In
findLongestRepeatedSubstring, an arraysubstringsis created to hold all possible subsequent substrings of the input. These substrings are constructed in a loop and then sorted by invoking_sort_substrings.The
_sort_substringsmethod uses most significant digit (MSD) sorting, well-suited for string sorting, employing a temporary array and a character-index mapping.After sorting, adjacent substrings are compared character by character to ascertain the length of the highest common prefix, which would indicate a repeating sequence. The longest such sequence length is tracked and finally returned by the program.
The detailed sorting mechanism involves first counting characters using a bucket array that effectively segregates substrings based on their current character. It essentially rearranges substrings according to character ranks at the indexed position, and then sorts within these categorizations recursively, deepening to successively higher character positions.
Understanding this solution leverages knowledge of suffix arrays, string sorting, and recursive subdivision of data, which enables efficient identification of repeating substrings amid large or complex datasets. This approach guarantees that the solution not only finds but also confirms the maximal repeating sequences in the string by direct comparison of ordered substring candidates.
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