How do you solve Maximum Candies You Can Get from Boxes on LeetCode?

Maximum Candies You Can Get from Boxes (LeetCode #1298) 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: Using BFS allows us to explore all reachable boxes efficiently.

What is the time complexity of Maximum Candies You Can Get from Boxes?

The optimal solution for Maximum Candies You Can Get from Boxes runs in O(n) time and uses O(n) space. This complexity arises because we process each box once and manage keys and opened boxes efficiently.

What is the brute force solution for Maximum Candies You Can Get from Boxes?

The brute force approach for Maximum Candies You Can Get from Boxes is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will explore each box sequentially, collecting candies and opening boxes as we go. This method is straightforward but inefficient because it may require checking each box multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Candies You Can Get from Boxes?

Maximum Candies You Can Get from Boxes uses the following concepts: Array, Breadth-First Search, Graph Theory. The recommended patterns to study are: Breadth-First Search (BFS), Graph Traversal.

What are the interview tips for Maximum Candies You Can Get from Boxes?

Clearly explain your thought process while coding, especially how you manage state. Consider edge cases, such as all boxes being closed or all boxes being open. Practice BFS problems to become comfortable with queue management.

What are common mistakes when solving Maximum Candies You Can Get from Boxes?

Not managing the state of keys and opened boxes properly, leading to missed opportunities. Failing to use a queue for BFS, which can complicate the traversal logic.

#1298

Maximum Candies You Can Get from Boxes

Hard

ArrayBreadth-First SearchGraph TheoryBreadth-First Search (BFS)Graph Traversal

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)

Using BFS allows us to efficiently explore all boxes without revisiting them unnecessarily. We can track opened boxes and keys dynamically, ensuring we maximize candy collection.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue with the initial boxes and a set for opened boxes.
2Step 2: While the queue is not empty, process each box: collect candies, add keys, and enqueue contained boxes.
3Step 3: After processing a box, check if any keys can open new boxes and enqueue them.

solution.py23 lines

1def maxCandies(status, candies, keys, containedBoxes, initialBoxes):
2    from collections import deque
3    queue = deque(initialBoxes)
4    total_candies = 0
5    keys_set = set()
6    opened_boxes = set(initialBoxes)
7
8    while queue:
9        box = queue.popleft()
10        if status[box]:
11            total_candies += candies[box]
12            for key in keys[box]:
13                keys_set.add(key)
14            for new_box in containedBoxes[box]:
15                if new_box not in opened_boxes:
16                    opened_boxes.add(new_box)
17                    queue.append(new_box)
18            for key in list(keys_set):
19                if key not in opened_boxes:
20                    opened_boxes.add(key)
21                    queue.append(key)
22                    keys_set.remove(key)
23    return total_candies

ℹ

Complexity note: This complexity arises because we process each box once and manage keys and opened boxes efficiently.

1Using BFS allows us to explore all reachable boxes efficiently.
2Tracking keys and opened boxes dynamically prevents unnecessary revisits.

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