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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.

• Java
• C++
• Python
• Go
• TypeScript

• 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
  }
  

• function maxUniqueSplit(s: string): number {
      const n = s.length;
      const st = new Set<string>();
      let ans = 0;
      const dfs = (i: number): void => {
          if (st.size + n - i <= ans) {
              return;
          }
          if (i >= n) {
              ans = Math.max(ans, st.size);
              return;
          }
          for (let j = i + 1; j <= n; ++j) {
              const t = s.slice(i, j);
              if (!st.has(t)) {
                  st.add(t);
                  dfs(j);
                  st.delete(t);
              }
          }
      };
      dfs(0);
      return ans;
  }
  
  

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