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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;
}
};

• 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 splitCount = 1;
for (int j = 0; j < length - 1 && mask > 0; j++) {
if (mask % 2 == 1) {
split[j] = true;
splitCount++;
}
}
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]) {
start = i + 1;
}
}
if (start < length)
return set.size() == splitCount;
}
}

############

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);
dfs(j, t + 1);
vis.remove(x);
}
}
}
}

• // 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;
}
};

• 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:
dfs(j, t + 1)
vis.remove(s[i:j])

vis = set()
ans = 1
dfs(0, 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:
res = max(res, 1 + backtrack(S[i:], seen))
seen.remove(s)

return res

seen = set()
return backtrack(S, seen)


• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}