2405. Optimal Partition of String

Description

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 <= 105
• s consists of only English lowercase letters.

Solutions

Solution 1: Greedy

According to the problem, each substring should be as long as possible and contain unique characters. We just need to partition greedily.

During the process, we can use a hash table to record all characters in the current substring, with a space complexity of $O(n)$; or we can use a number to record characters using bitwise operations, with a space complexity of $O(1)$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$.

• class Solution {
public int partitionString(String s) {
Set<Character> ss = new HashSet<>();
int ans = 1;
for (char c : s.toCharArray()) {
if (ss.contains(c)) {
++ans;
ss.clear();
}
}
return ans;
}
}

• class Solution {
public:
int partitionString(string s) {
unordered_set<char> ss;
int ans = 1;
for (char c : s) {
if (ss.count(c)) {
++ans;
ss.clear();
}
ss.insert(c);
}
return ans;
}
};

• class Solution:
def partitionString(self, s: str) -> int:
ss = set()
ans = 1
for c in s:
if c in ss:
ans += 1
ss = set()
return ans


• func partitionString(s string) int {
ss := map[rune]bool{}
ans := 1
for _, c := range s {
if ss[c] {
ans++
ss = map[rune]bool{}
}
ss[c] = true
}
return ans
}

• function partitionString(s: string): number {
const set = new Set();
let res = 1;
for (const c of s) {
if (set.has(c)) {
res++;
set.clear();
}
}
return res;
}


• use std::collections::HashSet;
impl Solution {
pub fn partition_string(s: String) -> i32 {
let mut set = HashSet::new();
let mut res = 1;
for c in s.as_bytes().iter() {
if set.contains(c) {
res += 1;
set.clear();
}
set.insert(c);
}
res
}
}