3003. Maximize the Number of Partitions After Operations

Description

You are given a string s and an integer k.

First, you are allowed to change at most one index in s to another lowercase English letter.

After that, do the following partitioning operation until s is empty:

• Choose the longest prefix of s containing at most k distinct characters.
• Delete the prefix from s and increase the number of partitions by one. The remaining characters (if any) in s maintain their initial order.

Return an integer denoting the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.

 

Example 1:

Input: s = "accca", k = 2

Output: 3

Explanation:

The optimal way is to change s[2] to something other than a and c, for example, b. then it becomes "acbca".

Then we perform the operations:

1. The longest prefix containing at most 2 distinct characters is "ac", we remove it and s becomes "bca".
2. Now The longest prefix containing at most 2 distinct characters is "bc", so we remove it and s becomes "a".
3. Finally, we remove "a" and s becomes empty, so the procedure ends.

Doing the operations, the string is divided into 3 partitions, so the answer is 3.

Example 2:

Input: s = "aabaab", k = 3

Output: 1

Explanation:

Initially s contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.

Example 3:

Input: s = "xxyz", k = 1

Output: 4

Explanation:

The optimal way is to change s[0] or s[1] to something other than characters in s, for example, to change s[0] to w.

Then s becomes "wxyz", which consists of 4 distinct characters, so as k is 1, it will divide into 4 partitions.

 

Constraints:

• 1 <= s.length <= 104
• s consists only of lowercase English letters.
• 1 <= k <= 26

Solutions

Solution 1

Python3JavaC++GoTypeScript

class Solution:
    def maxPartitionsAfterOperations(self, s: str, k: int) -> int:
        @cache
        def dfs(i: int, cur: int, t: int) -> int:
            if i >= n:
                return 1
            v = 1 << (ord(s[i]) - ord("a"))
            nxt = cur | v
            if nxt.bit_count() > k:
                ans = dfs(i + 1, v, t) + 1
            else:
                ans = dfs(i + 1, nxt, t)
            if t:
                for j in range(26):
                    nxt = cur | (1 << j)
                    if nxt.bit_count() > k:
                        ans = max(ans, dfs(i + 1, 1 << j, 0) + 1)
                    else:
                        ans = max(ans, dfs(i + 1, nxt, 0))
            return ans

        n = len(s)
        return dfs(0, 0, 1)

class Solution {
    private Map<List<Integer>, Integer> f = new HashMap<>();
    private String s;
    private int k;

    public int maxPartitionsAfterOperations(String s, int k) {
        this.s = s;
        this.k = k;
        return dfs(0, 0, 1);
    }

    private int dfs(int i, int cur, int t) {
        if (i >= s.length()) {
            return 1;
        }
        var key = List.of(i, cur, t);
        if (f.containsKey(key)) {
            return f.get(key);
        }
        int v = 1 << (s.charAt(i) - 'a');
        int nxt = cur | v;
        int ans = Integer.bitCount(nxt) > k ? dfs(i + 1, v, t) + 1 : dfs(i + 1, nxt, t);
        if (t > 0) {
            for (int j = 0; j < 26; ++j) {
                nxt = cur | (1 << j);
                if (Integer.bitCount(nxt) > k) {
                    ans = Math.max(ans, dfs(i + 1, 1 << j, 0) + 1);
                } else {
                    ans = Math.max(ans, dfs(i + 1, nxt, 0));
                }
            }
        }
        f.put(key, ans);
        return ans;
    }
}

class Solution {
public:
    int maxPartitionsAfterOperations(string s, int k) {
        int n = s.size();
        unordered_map<long long, int> f;
        function<int(int, int, int)> dfs = [