How do you solve Minimum Capacity Box on LeetCode?

Minimum Capacity Box (LeetCode #3861) 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: Iterate through the array to find the minimum capacity.

What is the time complexity of Minimum Capacity Box?

The optimal solution for Minimum Capacity Box runs in O(n) time and uses O(1) space. We only traverse the array once, leading to O(n) time complexity with constant space usage.

What is the brute force solution for Minimum Capacity Box?

The brute force approach for Minimum Capacity Box is Brute Force, with O(n) time complexity and O(1) space complexity. We check each box's capacity to find the minimum that can hold the item. This approach is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Capacity Box?

Minimum Capacity Box uses the following concepts: Array. The recommended patterns to study are: Array, Linear Search.

What are the interview tips for Minimum Capacity Box?

Explain your thought process clearly. Consider edge cases like empty arrays or all boxes being too small. Optimize your solution after the brute force approach.

What are common mistakes when solving Minimum Capacity Box?

Not handling the case where no box can store the item. Forgetting to update the index only on strictly smaller capacities.

#3861

Minimum Capacity Box

Easy

ArrayArrayLinear Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We efficiently find the minimum box that can hold the item by scanning through the list once, keeping track of the best candidate.

⚙️

Algorithm

3 steps

1Step 1: Initialize min_index to -1.
2Step 2: Iterate through each box in capacity.
3Step 3: If capacity[i] >= itemSize and (min_index == -1 or capacity[i] < capacity[min_index]), update min_index.

solution.py7 lines

1def min_capacity_box(capacity, itemSize):
2    min_index = -1
3    for i in range(len(capacity)):
4        if capacity[i] >= itemSize:
5            if min_index == -1 or capacity[i] < capacity[min_index]:
6                min_index = i
7    return min_index

ℹ

Complexity note: We only traverse the array once, leading to O(n) time complexity with constant space usage.

1Iterate through the array to find the minimum capacity.
2Keep track of the smallest index for ties.

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