How do you solve Minimum Discards to Balance Inventory on LeetCode?

Minimum Discards to Balance Inventory (LeetCode #3679) 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: Sliding window allows efficient management of counts.

What is the time complexity of Minimum Discards to Balance Inventory?

The optimal solution for Minimum Discards to Balance Inventory runs in O(n) time and uses O(n) space. Each arrival is processed once, maintaining a sliding window leads to O(n).

What is the brute force solution for Minimum Discards to Balance Inventory?

The brute force approach for Minimum Discards to Balance Inventory is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible window for each arrival and count occurrences. Discard arrivals that exceed the limit in their respective windows. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Discards to Balance Inventory?

Minimum Discards to Balance Inventory uses the following concepts: Array, Hash Table, Sliding Window, Simulation, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Discards to Balance Inventory?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Minimum Discards to Balance Inventory?

Not maintaining the window size correctly. Forgetting to update counts when discarding.

#3679

Minimum Discards to Balance Inventory

Medium

ArrayHash TableSliding WindowSimulationCountingHash 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)

Use a sliding window to efficiently track item counts and discard excess items without re-evaluating the entire window.

⚙️

Algorithm

3 steps

1Step 1: Initialize a count map and two pointers for the sliding window.
2Step 2: For each arrival, update the count and check if it exceeds m; if so, increment discard count.
3Step 3: Move the left pointer to maintain the window size and adjust counts accordingly.

solution.py12 lines

1def min_discards(arrivals, w, m):
2    from collections import defaultdict
3    discard_count = 0
4    counts = defaultdict(int)
5    left = 0
6    for right in range(len(arrivals)):
7        counts[arrivals[right]] += 1
8        while counts[arrivals[right]] > m:
9            counts[arrivals[left]] -= 1
10            left += 1
11            discard_count += 1
12    return discard_count

ℹ

Complexity note: Each arrival is processed once, maintaining a sliding window leads to O(n).

1Sliding window allows efficient management of counts.
2Using a hash map simplifies counting occurrences.

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