LeetCode 1593: Split a String Into the Max Number of Unique Substrings

Explanation

To solve this problem, we can use backtracking to try all possible ways of splitting the string into unique substrings. We can start by iterating through the string and at each position, we can check if the substring formed by the current position and all characters before it is unique. If it is unique, we can consider it as a valid split and recursively try to split the remaining part of the string. We keep track of the maximum number of unique substrings we have found so far.

Algorithm

1. Start with an empty set to store the unique substrings and a variable to keep track of the maximum number of unique substrings found.
2. Implement a backtracking function that takes the input string, current index, set of unique substrings, and the maximum number of unique substrings found so far as parameters.
3. In the backtracking function, iterate through the string starting from the current index.
4. Check if the substring formed by the current index and all characters before it is not already present in the set of unique substrings.
5. If it is unique, add it to the set, update the maximum number of unique substrings found so far, and recursively call the function with the updated parameters.
6. Return the maximum number of unique substrings found.

Time Complexity

The time complexity of this approach is exponential as we are exploring all possible splits of the string.

Space Complexity

The space complexity is also exponential due to the recursion stack.

class Solution {
    public int maxUniqueSplit(String s) {
        return backtrack(s, new HashSet<>(), 0, 0);
    }
    
    private int backtrack(String s, Set<String> uniqueSubstrings, int index, int maxUnique) {
        if (index == s.length()) {
            return maxUnique;
        }
        
        int currentMax = 0;
        for (int i = index + 1; i <= s.length(); i++) {
            String substring = s.substring(index, i);
            if (!uniqueSubstrings.contains(substring)) {
                uniqueSubstrings.add(substring);
                currentMax = Math.max(currentMax, backtrack(s, uniqueSubstrings, i, maxUnique + 1));
                uniqueSubstrings.remove(substring);
            }
        }
        
        return currentMax;
    }
}