What is the brute force solution for Maximize the Number of Partitions After Operations?

The brute force approach for Maximize the Number of Partitions After Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible single character change in the string and calculating the maximum number of partitions for each modified string. This method is straightforward but inefficient due to the nested operations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize the Number of Partitions After Operations?

Maximize the Number of Partitions After Operations uses the following concepts: String, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Sliding Window, Hash Map.

What are the interview tips for Maximize the Number of Partitions After Operations?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice writing clean and efficient code.

What are common mistakes when solving Maximize the Number of Partitions After Operations?

Not considering all possible character changes. Failing to reset state variables when counting partitions.

#3003

Maximize the Number of Partitions After Operations

Hard

StringDynamic ProgrammingBit ManipulationBitmaskSliding WindowHash Map

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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 approach uses prefix sums to precompute the number of partitions for each substring and leverages this information to efficiently determine the maximum partitions after a single character change. This significantly reduces redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Precompute the number of partitions for each prefix of the string without any changes using a sliding window approach.
2Step 2: Store the results in a prefix array where each index represents the number of partitions for the substring from the start to that index.
3Step 3: For each character in the string, consider changing it to every other character and calculate the maximum partitions using the precomputed prefix array.

solution.py36 lines

1def maxPartitions(s, k):
2    n = len(s)
3    pref = [0] * n
4    distinct = {}
5    left = 0
6    count = 0
7
8    for right in range(n):
9        distinct[s[right]] = distinct.get(s[right], 0) + 1
10        while len(distinct) > k:
11            distinct[s[left]] -= 1
12            if distinct[s[left]] == 0:
13                del distinct[s[left]]
14            left += 1
15        count += 1
16        pref[right] = count
17
18    max_partitions = count
19    for i in range(n):
20        original_char = s[i]
21        for c in 'abcdefghijklmnopqrstuvwxyz':
22            if c != original_char:
23                modified_s = s[:i] + c + s[i+1:]
24                count = 0
25                distinct = {}
26                left = 0
27                for right in range(n):
28                    distinct[modified_s[right]] = distinct.get(modified_s[right], 0) + 1
29                    while len(distinct) > k:
30                        distinct[modified_s[left]] -= 1
31                        if distinct[modified_s[left]] == 0:
32                            del distinct[modified_s[left]]
33                        left += 1
34                    count += 1
35                max_partitions = max(max_partitions, count)
36    return max_partitions

ℹ

Complexity note: The time complexity is O(n) for the initial partitioning and O(n) for each character change, leading to an overall complexity of O(n) due to the efficient use of prefix sums and sliding window.

1Understanding how to count distinct characters efficiently is crucial.
2Precomputation can save time in repetitive calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.