How do you solve Minimum Array Length After Pair Removals on LeetCode?

Minimum Array Length After Pair Removals (LeetCode #2856) 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: The optimal solution focuses on counting occurrences to minimize operations.

What is the brute force solution for Minimum Array Length After Pair Removals?

The brute force approach for Minimum Array Length After Pair Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible pairs of elements in the array to see if they can be removed. This method is straightforward but inefficient, as it checks every combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Array Length After Pair Removals?

Minimum Array Length After Pair Removals uses the following concepts: Array, Hash Table, Two Pointers, Binary Search, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Array Length After Pair Removals?

Always consider edge cases, such as arrays with all identical elements. Explain your thought process clearly while coding to demonstrate your understanding. Practice identifying patterns in problems, as they often lead to optimal solutions.

What are common mistakes when solving Minimum Array Length After Pair Removals?

Overcomplicating the problem by trying to remove elements in a non-systematic way. Not recognizing that the array is sorted, which can simplify the approach.

#2856

Minimum Array Length After Pair Removals

Medium

ArrayHash TableTwo PointersBinary SearchGreedyCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the fact that we can remove pairs of the smallest and largest elements. By counting the occurrences of the smallest element and the largest element, we can determine how many pairs we can remove.

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Algorithm

4 steps

1Step 1: Count the occurrences of the smallest and largest elements in the array.
2Step 2: Determine the number of pairs that can be removed, which is the minimum of the counts of the smallest and largest elements.
3Step 3: Calculate the remaining length of the array after removing these pairs.
4Step 4: Return the remaining length.

solution.py10 lines

1# Full working Python code
2
3def min_length_after_removals(nums):
4    min_count = nums.count(nums[0])
5    max_count = nums.count(nums[-1])
6    pairs = min(min_count, max_count)
7    return len(nums) - 2 * pairs
8
9# Example usage
10print(min_length_after_removals([1, 2, 3, 4]))  # Output: 0

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once to count the occurrences of the smallest and largest elements. This is efficient compared to the brute-force approach.

1The optimal solution focuses on counting occurrences to minimize operations.
2Understanding the properties of sorted arrays helps in deriving efficient solutions.

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