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

• C++
• Java
• JavaScript
• Python
• Ruby

• 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)