Optimal Partition of String

Published: Sep 21, 2022

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()
            seen.add(c)
        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
      seen.add(c)
    end
    count
end

Complexities

• Time: O(n)
• Space: O(n)