What is the optimal solution for the Minimum Substring Partition of Equal Character Frequency problem?

The optimal solution involves using dynamic programming with state transitions and hash tables to track character frequencies efficiently.

How does dynamic programming help solve the Minimum Substring Partition problem?

Dynamic programming tracks the minimum partitions up to each index, allowing you to decide whether a substring can be balanced by checking frequency distributions.

Can this problem be solved without dynamic programming?

While dynamic programming is the most efficient approach, trying brute-force methods or greedy solutions would lead to inefficient and incorrect results for larger inputs.

What role does the hash table play in the solution?

A hash table is used to store and update the character frequencies as you traverse the string, helping to quickly check if a substring is balanced.

How can I improve the performance of my solution to this problem?

Focus on optimizing the dynamic programming state transitions and using a sliding window technique with the hash table to reduce unnecessary recalculations.

#3144

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

分割字符频率相等的最少子字符串

给你一个字符串 s ，你需要将它分割成一个或者更多的平衡子字符串。比方说， s == "ababcc" 那么 ("abab", "c", "c") ， ("ab", "abc", "c") 和 ("ababcc") 都是合法分割，但是 ("a", "bab" , "cc") ， ( "aba" …

哈希表字符串动态规划计数

题目描述

给你一个字符串 s ，你需要将它分割成一个或者更多的平衡子字符串。比方说，s == "ababcc" 那么 ("abab", "c", "c") ，("ab", "abc", "c") 和 ("ababcc") 都是合法分割，但是 ("a", "bab", "cc") ，("aba", "bc", "c") 和 ("ab", "abcc") 不是，不平衡的子字符串用粗体表示。

请你返回 s 最少能分割成多少个平衡子字符串。

注意：一个平衡字符串指的是字符串中所有字符出现的次数都相同。

示例 1：

输入：s = "fabccddg"

输出：3

解释：

我们可以将 s 分割成 3 个子字符串：("fab, "ccdd", "g") 或者 ("fabc", "cd", "dg") 。

示例 2：

输入：s = "abababaccddb"

输出：2

解释：

我们可以将 s 分割成 2 个子字符串：("abab", "abaccddb") 。

提示：

1 <= s.length <= 1000
s 只包含小写英文字母。

lightbulb

解题思路

方法一：记忆化搜索 + 哈希表

我们设计一个函数 $\textit{dfs}(i)$ ，表示从字符串 $s[i]$ 开始分割的最少子字符串数量。那么答案就是 $\textit{dfs}(0)$ 。

函数 $\textit{dfs}(i)$ 的计算过程如下：

如果 $i \geq n$ ，表示已经处理完了所有字符，返回 $0$ 。

否则，我们维护一个哈希表 $\textit{cnt}$ ，表示当前子字符串中每个字符出现的次数。另外，我们还维护一个哈希表 $\textit{freq}$ ，表示每个字符出现的次数的频率。

然后我们枚举 $j$ 从 $i$ 到 $n-1$ ，表示当前子字符串的结束位置。对于每个 $j$ ，我们更新 $\textit{cnt}$ 和 $\textit{freq}$ ，然后判断 $\textit{freq}$ 的大小是否为 $1$ ，如果是的话，我们可以从 $j+1$ 开始分割，此时答案为 $1 + \textit{dfs}(j+1)$ ，我们取所有 $j$ 中答案的最小值作为函数的返回值。

为了避免重复计算，我们使用记忆化搜索。

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class Solution:
    def minimumSubstringsInPartition(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            cnt = defaultdict(int)
            freq = defaultdict(int)
            ans = n - i
            for j in range(i, n):
                if cnt[s[j]]:
                    freq[cnt[s[j]]] -= 1
                    if not freq[cnt[s[j]]]:
                        freq.pop(cnt[s[j]])
                cnt[s[j]] += 1
                freq[cnt[s[j]]] += 1
                if len(freq) == 1 and (t := 1 + dfs(j + 1)) < ans:
                    ans = t
            return ans

        n = len(s)
        return dfs(0)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate should demonstrate an understanding of dynamic programming and its applications to partitioning problems.
question_mark
Look for understanding of how state transitions in DP work, especially how to handle the frequency distributions.
question_mark
Check if the candidate optimizes the solution using hash tables to manage character counts effectively.

warning

常见陷阱

外企场景

error
Failing to recognize that the problem requires checking character frequency distributions across partitions.
error
Misunderstanding the DP state transition, particularly how to track and update dp[i] values correctly.
error
Inefficient handling of the string with brute-force approaches that don't make use of dynamic programming or sliding windows.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Implementing this problem with a greedy approach may miss valid partitions and lead to incorrect results.
arrow_right_alt
A brute-force solution can become inefficient for longer strings, especially when dealing with up to 1000 characters.
arrow_right_alt
Trying to solve without using dynamic programming will lead to higher time complexities and possible redundant calculations.

help

常见问题

外企场景

继续练习

#3335 字符串转换后的长度 I #3337 字符串转换后的长度 II #3389 使字符频率相等的最少操作次数