#3144

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Substring Partition of Equal Character Frequency

Partition a string into substrings with equal character frequencies using dynamic programming and state transitions.

hash table string dynamic programming counting

Problem Statement

You are given a string s, and your task is to partition it into one or more substrings. Each substring should be balanced, meaning all characters in the substring appear the same number of times. The goal is to minimize the number of partitions. For example, 'ababcc' can be partitioned into valid combinations such as ('abab', 'c', 'c').

The challenge is to return the minimum number of partitions such that each substring has an equal frequency for all characters. A balanced substring is defined as one where the character frequencies match. For instance, in 'fabccddg', the solution can be 3 partitions like ('fab', 'ccdd', 'g').

Examples

Example 1

Input: s = "fabccddg"

Output: 3

We can partition the string s into 3 substrings in one of the following ways: ("fab, "ccdd", "g") , or ("fabc", "cd", "dg") .

Example 2

Input: s = "abababaccddb"

Output: 2

We can partition the string s into 2 substrings like so: ("abab", "abaccddb") .

Constraints

1 <= s.length <= 1000
s consists only of English lowercase letters.

Solution Approach

State Transition Dynamic Programming

Use dynamic programming (DP) to calculate the minimum partitions. Let dp[i] be the minimum number of partitions for the prefix of the string ending at index i + 1. The core idea is to update the DP state based on the frequency distribution of characters. Check if the frequency distribution from the start to any point can form a valid balanced substring.

Hash Table to Track Frequencies

Use a hash table to maintain the frequency of characters as you iterate through the string. This allows efficient checking for whether the frequencies are balanced, aiding in making decisions about valid substring partitions. As you process the string, store intermediate results for substring partition possibilities.

Iterate and Update with Sliding Window

To efficiently find the solution, use a sliding window approach to move through the string and keep track of character frequencies within that window. At each step, check if the frequencies are equal. When they are, you can consider partitioning and updating your DP table for the minimum result.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach used, but it generally involves iterating over the string and updating the DP table, leading to an O(n^2) solution in the worst case. The space complexity depends on the storage of the frequency table and DP states, which is O(n).

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Memoized Search + Hash Table

We design a function $\textit{dfs}(i)$, which represents the minimum number of substrings starting from $s[i]$. The answer is $\textit{dfs}(0)$.

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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)

Solution 2: Memoized Search (Optimization)

We can optimize Solution 1 by not maintaining the $\textit{freq}$ hash table. Instead, we only need to maintain a hash table $\textit{cnt}$, which represents the frequency of each character in the current substring. Additionally, we maintain two variables $k$ and $m$ to represent the number of distinct characters in the current substring and the maximum frequency of any character, respectively. For a substring $s[i..j]$, if $j-i+1 = m \times k$, then this substring is a balanced substring.

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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)

Solution 3: Dynamic Programming

We can convert the memoized search into dynamic programming. Define the state $f[i]$ as the minimum number of substrings required to partition the first $i$ characters. Initially, $f[0] = 0$, and the rest $f[i] = +\infty$ or $f[i] = n$.

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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)

Continue Practicing

#3335 total characters in string after transformations i #3337 total characters in string after transformations ii #3389 minimum operations to make character frequencies equal