2405 - Optimal Partition of String (Medium)

Problem Statement

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.

Approach 1: Greedy

Written by @wingkwong

class Solution {
public:
    // the idea is to keep each partition as long as possible 
    // so that we could have the minimum number of substrings
    int partitionString(string s) {
        // the minimum number of substring is at least 1
        // e.g. "a"
        int ans = 1;
        // cnt is used to count the frequency of each character
        vector<int> cnt(26);
        // for each character
        for (auto& c : s) {
            // we check if it exists before
            // if so, then we should create a new partition
            // because no letter appears in a single substring more than once
            if (cnt[c - 'a']) {
                // reset the counter
                cnt = vector<int>(26);
                // create a new partition
                ans++;
            }
            // increase the frequency of the current character by 1
            cnt[c - 'a']++;
        } 
        return ans;
    }
};