Split a String Into the Max Number of Unique Substrings in Java
The problem of "Split a String Into the Max Number of Unique Substrings" is challenging in string manipulation and algorithm design. It involves partitioning a string into substrings, with each substring appearing only once. The above problem is relevant in text processing, data compression, and string parsing.
Example 1:
Input: s = "ababccc"
Output: 5
Explanation: One method for dividing most effectively is ['a', 'b,' 'ab,' 'c,' 'cc']. Dividing like ['a', 'b,' 'a', 'b,' 'c,' 'cc'] is invalid because 'a' and 'b' appear multiple times.
Example 2:
Input: s = “aba”
Output: 2
Explanation: ['a', 'ba'] is one option for splitting maximally.
Using Backtracking
The code initializes a `UniqueSplitSolution` class and defines an input string `exampleString.` It invokes the `maxUniqueSplit` method with the input string `exampleString,` iterating over all possible substrings, checking if they're not already visited, adding them to the visited set, making a recursive call, and removing them for backtracking. The recursion continues until the start index reaches the end of the input string, updating the `maxSplitCount.`
Here's an implementation of spliting a string into the max number of unique substrings using backtracking :
FileName: UniqueSplitSolution.java
import java.util.HashSet; public class UniqueSplitSolution { int maxSplitCount; String inputString; public int maxUniqueSplit(String s) { maxSplitCount = 0; this.inputString = s; generateUniqueSplits(0, new HashSet()); return maxSplitCount; } // Helper method to generate unique splits recursively private void generateUniqueSplits(int startIndex, HashSet visited) { // Base case: If the startIndex reaches the end of the string if (startIndex == inputString.length()) { // Update maxSplitCount with the maximum of current unique splits or maxSplitCount maxSplitCount = Math.max(visited.size(), maxSplitCount); return; } for (int endIndex = startIndex + 1; endIndex <= inputString.length(); endIndex++) { // Extract substring from startIndex to endIndex String newSubstring = inputString.substring(startIndex, endIndex); // Check if the substring is not already visited if (!visited.contains(newSubstring)) { visited.add(newSubstring); // Add the substring to visited set // Recursively call generateUniqueSplits with updated endIndex and visit the set generateUniqueSplits(endIndex, visited); visited.remove(newSubstring); // Remove the substring for backtracking } } } public static void main(String[] args) { UniqueSplitSolution uniqueSplitSolution = new UniqueSplitSolution(); // Create an instance of UniqueSplitSolution String exampleString = "ababccc"; System.out.println(uniqueSplitSolution.maxUniqueSplit(exampleString)); } }
Output:
5
Complexity analysis: The code's time complexity is O(n^3) due to the number of unique substrings generated using the `generateUniqueSplits` method. The time complexity is O(n^3) as it iterates through each starting and ending index. The space complexity is O(n^2) due to the `visited` set and recursion stack. The `visited` set can store all unique substrings of the input string with a maximum size of O(2^n). The recursion depth depends on the input string length and the number of recursive calls. Other variables like `maxSplitCount` and `inputString` occupy constant space.
Using Greedy Approach
The greedy approach is an algorithm that splits an input string into the maximum number of unique substrings. It starts by initializing a set called `uniqueSubstrings,` then calls the `greedySplit` method. The method takes two parameters: the current substring `s` to be split and the set of unique substrings encountered so far. It iterates over all possible substrings of `s,` checking if it is not already present in the set. If it is, it is added to the set, and the method recursively calls itself with the remaining part of the string. The method backtracks after exploring all possibilities, ensuring all valid combinations are considered.
Here's an implementation of spliting a string into the max number of unique substrings using the greedy approach :
FileName: UniqueSplitSolution1.java
import java.util.HashSet; import java.util.Set; public class UniqueSplitSolution1 { // Main method to find the maximum number of unique splits public int findMaxUniqueSplit(String s) { SetuniqueSubstrings = new HashSet<>(); return greedySplit(s, uniqueSubstrings); } // Greedy approach to split the string and find the maximum number of unique splits private int greedySplit(String s, Set uniqueSubstrings) { if (s.isEmpty()) { return uniqueSubstrings.size(); } int maxSplit = 0; for (int i = 1; i <= s.length(); i++) { String substring = s.substring(0, i); // If the substring hasn't been seen before, add it to the set of unique substrings if (!uniqueSubstrings.contains(substring)) { uniqueSubstrings.add(substring); // Recursively call greedySplit with the remaining part of the string int splitCount = greedySplit(s.substring(i), uniqueSubstrings); // Update maxSplit with the maximum split count found so far maxSplit = Math.max(maxSplit, splitCount); // Backtrack: remove the current substring to explore other possibilities uniqueSubstrings.remove(substring); } } return maxSplit; } public static void main(String[] args) { UniqueSplitSolution1 solution = new UniqueSplitSolution1(); String exampleString = "ababccc"; System.out.println(solution.findMaxUniqueSplit(exampleString)); } }
Output:
5
Complexity analysis: The code's complexity analysis considers time and space complexities. The `greedySplit` method calls recursively for each substring, resulting in a time complexity of O(2^N), where N is the input string's length. However, memoization can be reduced to O(N^2). The space complexity depends on the space used by the set `uniqueSubstrings` and the recursion stack, with the worst case space complexity being O(N) due to the depth of the recursion stack.