# Optimal Partition of String

Published: Sep 21, 2022

Medium String Hash Table Greedy

## Problem 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.

Constraints:

• `1 <= s.length <= 10**5`
• `s` consists of only English lowercase letters.

https://leetcode.com/problems/optimal-partition-of-string/

## Examples

``````Example 1
Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a"), ("ab","a","ca","ba"), and ("ab","ac","ab","a").
``````
``````Exammle 2
Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").
``````

## How to Solve

To solve string partition problems, hash table or set is used to save a previous occurrence. If the problem asks a length, it needs hash table. However, this problem asks just a number of groups, so set is a good data structure.

Use a set to save the previous occurrence of a same character. If the same character appears, count up and clear the set. In the end, if the set is not empty, count up.

## Solution

``````class OptimalPartitionString {
private:
int count = 1;
set<char> seen;
public:
int partitionString(string s) {
for (char& c : s) {
if (seen.count(c)) {
count++;
seen.clear();
}
seen.emplace(c);
}
return count;
}
};
``````
``````
``````
``````
``````
``````class OptimalPartitionString:
def partitionString(self, s: str) -> int:
count, seen = 0, set()
for c in s:
if c in seen:
count += 1
seen.clear()
return count + 1 if seen else count
``````
``````# @param {String} s
# @return {Integer}
def partition_string(s)
count, seen = 1, Set.new
s.each_char do |c|
if seen.include?(c)
count += 1
seen.clear
end
• Time: `O(n)`
• Space: `O(n)`