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

Description

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.

Solutions

DFS.

  • class Solution {
    private Set<String> vis = new HashSet<>();
    private int ans = 1;
    private String s;

    public int maxUniqueSplit(String s) {
        this.s = s;
        dfs(0, 0);
        return ans;
    }

    private void dfs(int i, int t) {
        if (i >= s.length()) {
            ans = Math.max(ans, t);
            return;
        }
        for (int j = i + 1; j <= s.length(); ++j) {
            String x = s.substring(i, j);
            if (vis.add(x)) {
                dfs(j, t + 1);
                vis.remove(x);
            }
        }
    }
}

  • class Solution {
public:
    unordered_set<string> vis;
    string s;
    int ans = 1;

    int maxUniqueSplit(string s) {
        this->s = s;
        dfs(0, 0);
        return ans;
    }

    void dfs(int i, int t) {
        if (i >= s.size()) {
            ans = max(ans, t);
            return;
        }
        for (int j = i + 1; j <= s.size(); ++j) {
            string x = s.substr(i, j - i);
            if (!vis.count(x)) {
                vis.insert(x);
                dfs(j, t + 1);
                vis.erase(x);
            }
        }
    }
};

  • class Solution:
    def maxUniqueSplit(self, s: str) -> int:
        def dfs(i, t):
            if i >= len(s):
                nonlocal ans
                ans = max(ans, t)
                return
            for j in range(i + 1, len(s) + 1):
                if s[i:j] not in vis:
                    vis.add(s[i:j])
                    dfs(j, t + 1)
                    vis.remove(s[i:j])

        vis = set()
        ans = 1
        dfs(0, 0)
        return ans


  • func maxUniqueSplit(s string) int {
	ans := 1
	vis := map[string]bool{}

	var dfs func(i, t int)
	dfs = func(i, t int) {
		if i >= len(s) {
			ans = max(ans, t)
			return
		}
		for j := i + 1; j <= len(s); j++ {
			x := s[i:j]
			if !vis[x] {
				vis[x] = true
				dfs(j, t+1)
				vis[x] = false
			}
		}
	}
	dfs(0, 0)
	return ans
}

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