How do you solve Maximize Number of Subsequences in a String on LeetCode?

Maximize Number of Subsequences in a String (LeetCode #2207) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Inserting characters can significantly change the number of subsequences.

What is the time complexity of Maximize Number of Subsequences in a String?

The optimal solution for Maximize Number of Subsequences in a String runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the text once to count occurrences, making it much more efficient than the brute force approach.

What is the brute force solution for Maximize Number of Subsequences in a String?

The brute force approach for Maximize Number of Subsequences in a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to insert each character of the pattern into every possible position in the text and counting how many times the pattern appears as a subsequence. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize Number of Subsequences in a String?

Maximize Number of Subsequences in a String uses the following concepts: String, Greedy, Prefix Sum. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Maximize Number of Subsequences in a String?

Explain your thought process clearly while coding. Consider edge cases, such as empty strings or patterns. Practice counting subsequences in simpler examples before tackling complex ones.

What are common mistakes when solving Maximize Number of Subsequences in a String?

Not considering all possible insertion points. Failing to count subsequences correctly after each insertion.

#2207

Maximize Number of Subsequences in a String

Medium

StringGreedyPrefix SumTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass to count occurrences of the pattern in the text while considering the effect of inserting each character of the pattern. This avoids the need for repeated counting and is much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Count occurrences of pattern[0] and pattern[1] in the text using a single pass.
2Step 2: For each position in the text, calculate how many times the pattern would occur if we inserted pattern[0] or pattern[1].
3Step 3: Return the maximum count from both insertions.

solution.py11 lines

1def max_subsequences(text, pattern):
2    count_a = 0
3    count_b = 0
4    total = 0
5    for char in text:
6        if char == pattern[0]:
7            total += count_b
8            count_a += 1
9        elif char == pattern[1]:
10            count_b += 1
11    return total + count_a + count_b

ℹ

Complexity note: The time complexity is O(n) because we only traverse the text once to count occurrences, making it much more efficient than the brute force approach.

1Inserting characters can significantly change the number of subsequences.
2Counting subsequences efficiently requires careful tracking of character occurrences.

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