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Given a string `s`

, partition the string into one or more **substrings** such that the characters in each substring are **unique**. That is, no letter appears in a single substring more than **once**.

Return *the minimum number of substrings in such a partition.*

Note that each character should belong to exactly one substring in a partition.

**Example 1:**

**Input:** s = "abacaba"
**Output:** 4
**Explanation:**
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.

**Example 2:**

**Input:** s = "ssssss"
**Output:** 6
**Explanation:
**The only valid partition is ("s","s","s","s","s","s").

**Constraints:**

`1 <= s.length <= 10`

^{5}`s`

consists of only English lowercase letters.

```
class Solution {
public:
bool checkBit(int &flag, int &n){
return flag & (1<<n);
}
void setBit(int &flag, int &n){
flag = flag | (1<<n);
}
int partitionString(string s) {
int flag = 0;
int i = 0, ans = 1;
while(i < s.size()){
int n = s[i] - 'a';
if( checkBit(flag, n) ) {
flag = 0;
ans++;
}
setBit(flag, n);
i++;
}
return ans;
}
};
```

```
class Solution {
public int partitionString(String s) {
int ans = 1;
HashSet<Character> st = new HashSet<>();
for(int i=0;i<s.length();i++){
// Insert Till we find duplicate element.
if(!st.contains(s.charAt(i))){
st.add(s.charAt(i));
}
else{
// If we found duplicate char then increment count and clear set and start with new set.
ans++;
st.clear();
st.add(s.charAt(i));
}
}
return ans;
}
}
```

```
class Solution:
def partitionString(self, s: str) -> int:
def isUnique(s):
return len(set(s)) == len(s)
left = 0
res = 0
for right in range(1,len(s)+1):
if not isUnique(s[left:right]):
res += 1
left = right-1
return res+1 # don't forget there is still one partition we didn't add into res
```

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