What is the time complexity of Minimum Substring Partition of Equal Character Frequency?

The optimal solution for Minimum Substring Partition of Equal Character Frequency runs in O(n) time and uses O(n) space. The time complexity is O(n) because we make a single pass through the string, updating frequencies and checking partitions efficiently.

What is the brute force solution for Minimum Substring Partition of Equal Character Frequency?

The brute force approach for Minimum Substring Partition of Equal Character Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible substring partition to see if they are balanced. This method is straightforward but inefficient, as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Substring Partition of Equal Character Frequency?

Minimum Substring Partition of Equal Character Frequency uses the following concepts: Hash Table, String, Dynamic Programming, Counting. The recommended patterns to study are: Hash Map, Dynamic Programming.

What are the interview tips for Minimum Substring Partition of Equal Character Frequency?

Start with a brute force solution to demonstrate understanding. Clearly explain how you optimize your solution. Practice dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Minimum Substring Partition of Equal Character Frequency?

Not checking all possible partitions in the brute force approach. Failing to efficiently update character frequencies in the optimal solution.

#3144

Minimum Substring Partition of Equal Character Frequency

Medium

Hash TableStringDynamic ProgrammingCountingHash MapDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to efficiently calculate the minimum number of partitions by leveraging previously computed results. It checks character frequencies in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Initialize a frequency array to count occurrences of each character.
2Step 2: Use dynamic programming to store the minimum partitions required for each prefix of the string.
3Step 3: For each character, update the frequency array and check if the current substring is balanced.
4Step 4: Update the minimum partitions based on valid partitions found.

solution.py17 lines

1def min_partitions(s):
2    from collections import defaultdict
3    n = len(s)
4    dp = [float('inf')] * n
5    freq = defaultdict(int)
6    for i in range(n):
7        freq[s[i]] += 1
8        if len(set(freq.values())) == 1:
9            dp[i] = 1
10        else:
11            for j in range(i):
12                if len(set(freq.values())) == 1:
13                    dp[i] = min(dp[i], dp[j] + 1)
14        freq[s[i]] -= 1
15        if freq[s[i]] == 0:
16            del freq[s[i]]
17    return dp[-1]

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string, updating frequencies and checking partitions efficiently.

1A balanced substring has equal frequency of all characters.
2Dynamic programming can optimize the solution by storing results of subproblems.

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