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:
Solutions#
Solution 1#
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| 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
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| 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);
}
}
}
}
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| 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);
}
}
}
};
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| 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
}
|