Problem Description

Given a string s, you need to partition it into one or more balanced substrings. For example, if s == "ababcc" then ("abab", "c", "c"), ("ab", "abc", "c"), and ("ababcc") are all valid partitions, but ("a", "bab", "cc"), ("aba", "bc", "c"), and ("ab", "abcc") are not. The unbalanced substrings are bolded.

Return the minimum number of substrings that you can partition s into.

Note: A balanced string is a string where each character in the string occurs the same number of times.

Intuition

To solve the problem of partitioning a string into the minimum number of balanced substrings, we need to determine the correct points in the string where each partition should occur. A substring is considered balanced if all characters occur with the same frequency.

The approach involves designing a recursive function dfs(i) that calculates the minimum number of substrings starting from the index i. The key idea is to explore potential cuts at each position and use memoization to save previously computed results to avoid redundant calculations.

1. Initialization: Define two hash tables cnt and freq. cnt will track the frequency of each character in the current segment, while freq will count how many characters have the same frequency, allowing us to quickly check if all characters in the current substring have the same frequency.

2. Iterative Exploration: For each character position j starting from i, update the character counts in cnt and accordingly adjust freq. If at any point freq contains only one entry, it indicates that all characters in the current substring have uniform frequency, making it balanced. At this point, we have a potential point to partition the string.

3. Recursive Evaluation: When a balanced substring is identified ending at j, calculate the number of partitions by calling dfs(j + 1) which determines the partitions for the rest of the string. The result for the current position is 1 + dfs(j + 1).

4. Minimize Partitions: Continue evaluating each possible ending j and maintain the smallest number of partitions found.

5. Memoization: Use memoization to store results of dfs(i) calls to optimize calculations and prevent re-computation of results for the same state.

By following this approach, we ensure that the solution is computed efficiently, minimizing the number of balanced substrings into which s is partitioned.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution to partition the string into the minimum number of balanced substrings is implemented using a recursive approach with memoization, also known as depth-first search (DFS) combined with caching. Here's a detailed walk-through:

1. Recursive Function (dfs): We define a function dfs(i) which calculates the minimum number of balanced substrings starting from the index i. The base case for this function is i >= n, where n is the length of the string s. In this case, all characters have been processed, so it returns 0.

2. Data Structures:

   • cnt: A dictionary that tracks the frequency of each character in the current substring being considered.
   • freq: A dictionary that keeps track of the frequency of these character frequencies. For example, if three characters each occur twice in the current substring, freq[2] == 3.

3. Iterative Exploration: For each starting index i, iterate over potential end positions j, ranging from i to n-1. This iteration considers each possible substring starting from i:

   • Update character frequency in cnt and subsequently adjust freq to reflect these updates. If an existing frequency is altered, decrement its count in freq and update the new frequency count.

4. Check for Balanced Substring: After updating cnt and freq, check if freq contains only one entry. If it does, it implies that all characters in the current substring have the same frequency, hence the substring is balanced.

5. Recursive Evaluation: For a balanced substring ending at j, recursively call dfs(j + 1) to determine balanced substring partitions for the remainder of the string s. The result is 1 + dfs(j + 1).

6. Minimize Substrings: Track and update the minimum number of balanced substrings obtained through all potential partitions ending at each j from the current i.

7. Dynamic Programming with Memoization: Use memoization to store previously computed results for dfs(i), thereby avoiding repeated calculations and enhancing efficiency. This is efficiently handled by using the @cache decorator in Python, which automatically memoizes the function.

The implementation ensures optimal efficiency by leveraging a combination of recursive exploration with memoization, allowing the solution to efficiently navigate and compute the minimum partitions across various potential substrings.

Example Walkthrough

Let's walk through a simple example using the string s = "aabbcc". The goal is to partition s into the minimum number of balanced substrings, where each character in every substring appears with the same frequency.

Assume the helper function dfs(i) returns the minimum number of balanced substrings starting from index i.

1. Initialization:

   • We start at index i = 0 with empty dictionaries cnt and freq.

2. Iterative Exploration:

   • Index j = 0:

     • cnt updates to {'a': 1}.
     • freq updates to {1: 1} (one character with frequency 1).
     • Substring s[0:1] = "a" is balanced, making a potential partition. Calculate 1 + dfs(1).

   • Index j = 1:

     • cnt updates to {'a': 2}.
     • freq updates to {2: 1} (one character with frequency 2).
     • Substring s[0:2] = "aa" is balanced, making another potential partition. Calculate 1 + dfs(2).

   • Index j = 2:

     • cnt becomes {'a': 2, 'b': 1}.
     • freq becomes {2: 1, 1: 1} (mixed frequencies, not balanced yet).

   • Index j = 3:

     • cnt updates to {'a': 2, 'b': 2}.
     • freq updates to {2: 2} (all characters have frequency 2).
     • Substring s[0:4] = "aabb" is balanced. Calculate 1 + dfs(4).

   • Index j = 4:

     • cnt becomes {'a': 2, 'b': 2, 'c': 1}.
     • freq becomes {2: 2, 1: 1}.

   • Index j = 5:

     • cnt updates to {'a': 2, 'b': 2, 'c': 2}.
     • freq updates to {2: 3} (all characters have frequency 2).
     • Substring s[0:6] = "aabbcc" is balanced. Calculate 1 + dfs(6).

3. Recursive Evaluation:

   • Evaluate the results of each balanced substring to determine which yields the minimum partitions.
   • For each balanced substring (index j where substring ends), call dfs(j + 1) recursively to determine partitions for the remaining string.

4. Memoization:

   • Store results of subproblems (dfs(i)) using memoization to avoid recalculations and achieve efficiency.

In this example, you can achieve the minimum number of balanced substrings partitioning as ("aabbcc"), giving us a result of 1 by having the entire string as a balanced substring.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n^2). This is because the main work is done within the nested loop structure where for each starting index i, we attempt to extend the end index j for all subsequent indices, thus forming a double loop which results in O(n^2) operations.

The space complexity is O(n * |\Sigma|), where |\Sigma| is the size of the character set. Given the use of memoization via the @cache decorator and dictionaries (cnt and freq) to keep track of character counts, the space complexity scales by both the length of the string n and the character set |\Sigma|, which is 26 for alphabetic characters in this context.

Learn more about how to find time and space complexity quickly.