Formatted question description: https://leetcode.ca/all/1593.html

1593. Split a String Into the Max Number of Unique Substrings (Medium)

Given a string s, return the maximum number of unique substrings that the given string can be split into.

You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique.

A substring is a contiguous sequence of characters within a string.

 

Example 1:

Input: s = "ababccc"
Output: 5
Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times.

Example 2:

Input: s = "aba"
Output: 2
Explanation: One way to split maximally is ['a', 'ba'].

Example 3:

Input: s = "aa"
Output: 1
Explanation: It is impossible to split the string any further.

 

Constraints:

  • 1 <= s.length <= 16

  • s contains only lower case English letters.

Related Topics:
Backtracking

Solution 1. Backtrack

// OJ: https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/
// Time: O(2^N)
// Space: O(N)
class Solution {
    unordered_set<string> m;
    int ans = 0;
    void dfs(string &s, int i) {
        if (i == s.size()) {
            ans = max(ans, (int)m.size());
            return;
        }
        for (int j = i; j < s.size(); ++j) {
            string sub = s.substr(i, j - i + 1);
            if (m.count(sub)) continue;
            m.insert(sub);
            dfs(s, j + 1);
            m.erase(sub);
        }
    }
public:
    int maxUniqueSplit(string s) {
        dfs(s, 0);
        return ans;
    }
};

Java

  • class Solution {
        public int maxUniqueSplit(String s) {
            int length = s.length();
            if (length == 1)
                return 1;
            int maxSplit = 1;
            int total = 1 << (length - 1);
            for (int i = 1; i < total; i++) {
                boolean[] split = new boolean[length - 1];
                int mask = i;
                int splitCount = 1;
                for (int j = 0; j < length - 1 && mask > 0; j++) {
                    if (mask % 2 == 1) {
                        split[j] = true;
                        splitCount++;
                    }
                    mask /= 2;
                }
                if (splitCount > maxSplit && checkUnique(s, split, splitCount))
                    maxSplit = splitCount;
            }
            return maxSplit;
        }
    
        public boolean checkUnique(String s, boolean[] split, int splitCount) {
            Set<String> set = new HashSet<String>();
            int start = 0;
            int length = s.length();
            for (int i = 0; i < length - 1; i++) {
                if (split[i]) {
                    set.add(s.substring(start, i + 1));
                    start = i + 1;
                }
            }
            if (start < length)
                set.add(s.substring(start));
            return set.size() == splitCount;
        }
    }
    
  • // OJ: https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/
    // Time: O(2^N)
    // Space: O(N)
    class Solution {
        unordered_set<string> m;
        int ans = 0;
        void dfs(string &s, int i) {
            if (i == s.size()) {
                ans = max(ans, (int)m.size());
                return;
            }
            for (int j = i; j < s.size(); ++j) {
                string sub = s.substr(i, j - i + 1);
                if (m.count(sub)) continue;
                m.insert(sub);
                dfs(s, j + 1);
                m.erase(sub);
            }
        }
    public:
        int maxUniqueSplit(string s) {
            dfs(s, 0);
            return ans;
        }
    };
    
  • # 1593. Split a String Into the Max Number of Unique Substrings
    # https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/
    
    class Solution:
        def maxUniqueSplit(self, S: str) -> int:
            def backtrack(S, seen):
                res = 0
                for i in range(1,len(S)+1):
                    s = S[:i]
                    if s not in seen:
                        seen.add(s)
                        res = max(res, 1 + backtrack(S[i:], seen))
                        seen.remove(s)
                
                return res
            
            seen = set()
            return backtrack(S, seen)
        
            
                    
            
    

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