What is the time complexity of Smallest K-Length Subsequence With Occurrences of a Letter?

The optimal solution for Smallest K-Length Subsequence With Occurrences of a Letter runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse the string once and use a stack to maintain the necessary characters.

What is the brute force solution for Smallest K-Length Subsequence With Occurrences of a Letter?

The brute force approach for Smallest K-Length Subsequence With Occurrences of a Letter is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of length k and checking each one for the required conditions. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest K-Length Subsequence With Occurrences of a Letter?

Smallest K-Length Subsequence With Occurrences of a Letter uses the following concepts: String, Stack, Greedy, Monotonic Stack. The recommended patterns to study are: Stack, Greedy Algorithms.

What are the interview tips for Smallest K-Length Subsequence With Occurrences of a Letter?

Always clarify the constraints and edge cases before jumping into coding. Think about how to maintain order and conditions simultaneously, especially with stacks. Practice explaining your thought process as you code; it helps solidify your understanding.

What are common mistakes when solving Smallest K-Length Subsequence With Occurrences of a Letter?

Failing to account for the remaining characters when deciding to pop from the stack. Not ensuring that the required letter is included in the final result.

#2030

Smallest K-Length Subsequence With Occurrences of a Letter

Hard

StringStackGreedyMonotonic StackStackGreedy Algorithms

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 a stack to build the desired subsequence while ensuring that we maintain the lexicographical order and the required number of occurrences of the specified letter.

⚙️

Algorithm

5 steps

1Step 1: Initialize a stack to build the result and count the occurrences of the letter in the string.
2Step 2: Iterate through each character in the string s.
3Step 3: For each character, decide whether to push it onto the stack or pop from the stack based on the current size, remaining characters, and required occurrences of the letter.
4Step 4: Ensure that the stack contains the required letter at least 'repetition' times by the end of the iteration.
5Step 5: Construct the final result from the stack.

solution.py15 lines

1def smallestKLengthSubsequence(s, k, letter, repetition):
2    stack = []
3    letter_count = s.count(letter)
4    for i, char in enumerate(s):
5        while (stack and stack[-1] > char and 
6               len(stack) + len(s) - i > k and 
7               (stack[-1] != letter or letter_count > repetition)):
8            stack.pop()
9        if len(stack) < k:
10            if char == letter:
11                stack.append(char)
12                letter_count -= 1
13            elif k - len(stack) > repetition:
14                stack.append(char)
15    return ''.join(stack[:k])

ℹ

Complexity note: The complexity is O(n) because we traverse the string once and use a stack to maintain the necessary characters.

1Using a stack helps maintain the order of characters while allowing for efficient popping of larger characters.
2Counting occurrences of the required letter upfront allows for better decision-making during the stack operations.

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