How do you solve Find the Most Competitive Subsequence on LeetCode?

Find the Most Competitive Subsequence (LeetCode #1673) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The most competitive subsequence is determined by the smallest elements appearing first.

What is the brute force solution for Find the Most Competitive Subsequence?

The brute force approach for Find the Most Competitive Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and then selecting the one that is most competitive. This method is straightforward but inefficient due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Most Competitive Subsequence?

Find the Most Competitive Subsequence uses the following concepts: Array, Stack, Greedy, Monotonic Stack. The recommended patterns to study are: Greedy algorithms, Monotonic stack.

What are the interview tips for Find the Most Competitive Subsequence?

Always clarify the problem requirements and constraints before diving into coding. Discuss your thought process and potential approaches with the interviewer. Practice writing clean and efficient code, as readability can be just as important as functionality.

What are common mistakes when solving Find the Most Competitive Subsequence?

Failing to understand the importance of maintaining the order of elements in the subsequence. Not recognizing when to pop elements from the stack, leading to suboptimal subsequences.

#1673

Find the Most Competitive Subsequence

Medium

ArrayStackGreedyMonotonic StackGreedy algorithmsMonotonic stack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a stack to maintain the most competitive subsequence while iterating through the array. This method ensures that we always keep the smallest possible elements and can efficiently discard larger elements that would make the subsequence less competitive.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack and calculate how many elements we can remove from nums.
2Step 2: Iterate through each number in nums, and for each number, check if it can replace the top of the stack to maintain competitiveness.
3Step 3: If the stack size is less than k, push the current number onto the stack.
4Step 4: Return the first k elements from the stack.

solution.py13 lines

1# Full working Python code
2from collections import deque
3
4def mostCompetitive(nums, k):
5    stack = []
6    to_remove = len(nums) - k
7    for num in nums:
8        while to_remove > 0 and stack and stack[-1] > num:
9            stack.pop()
10            to_remove -= 1
11        stack.append(num)
12    return stack[:k]
13

ℹ

Complexity note: The time complexity is O(n) because we process each element of the array once. The space complexity is O(n) due to the stack that may store up to n elements in the worst case.

1The most competitive subsequence is determined by the smallest elements appearing first.
2Using a stack allows us to efficiently manage the elements we want to keep.

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