What is the time complexity of Maximum Number of Subsequences After One Inserting?

The optimal solution for Maximum Number of Subsequences After One Inserting runs in O(n) time and uses O(n) space. We only traverse the string a few times, leading to linear time complexity. Space is used for counting arrays.

What is the brute force solution for Maximum Number of Subsequences After One Inserting?

The brute force approach for Maximum Number of Subsequences After One Inserting is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible strings by inserting one letter at every position and count the 'LCT' subsequences for each. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Subsequences After One Inserting?

Maximum Number of Subsequences After One Inserting uses the following concepts: String, Dynamic Programming, Greedy, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Subsequences After One Inserting?

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice coding without an IDE to simulate interview conditions.

What are common mistakes when solving Maximum Number of Subsequences After One Inserting?

Forgetting to handle edge cases with short strings. Not updating counts correctly during insertion.

#3628

Maximum Number of Subsequences After One Inserting

Medium

StringDynamic ProgrammingGreedyPrefix SumHash MapArray

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)

We can precompute counts of 'L', 'LC', 'T', and 'CT' at each position to efficiently calculate the maximum subsequences after one insertion.

⚙️

Algorithm

3 steps

1Step 1: Precompute arrays for counts of 'L', 'LC', 'T', and 'CT' at each index.
2Step 2: Calculate the base count of 'LCT' subsequences without insertion.
3Step 3: For each position, compute potential counts by inserting 'L', 'C', or 'T' and find the maximum.

solution.py14 lines

1def maxSubsequences(s):
2    n = len(s)
3    preL, preLC, sufT, sufCT = [0] * (n + 1), [0] * (n + 1), [0] * (n + 1), [0] * (n + 1)
4    for i in range(n):
5        preL[i + 1] = preL[i] + (s[i] == 'L')
6        preLC[i + 1] = preLC[i] + (s[i] == 'C') * preL[i + 1]
7    for i in range(n - 1, -1, -1):
8        sufT[i] = sufT[i + 1] + (s[i] == 'T')
9        sufCT[i] = sufCT[i + 1] + (s[i] == 'C') * sufT[i + 1]
10    base = sum(preLC[i] * sufT[i] for i in range(n + 1))
11    max_count = base
12    for i in range(n + 1):
13        max_count = max(max_count, preLC[i] * (sufT[i] + 1) + (preL[i] + 1) * sufCT[i])
14    return max_count

ℹ

Complexity note: We only traverse the string a few times, leading to linear time complexity. Space is used for counting arrays.

1Precomputation of counts allows efficient subsequence calculation.
2Inserting at strategic positions maximizes subsequences.

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