How do you solve Maximum Spending After Buying Items on LeetCode?

Maximum Spending After Buying Items (LeetCode #2931) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a max-heap allows efficient retrieval of the most expensive item available.

What is the time complexity of Maximum Spending After Buying Items?

The optimal solution for Maximum Spending After Buying Items runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the heap operations, where n is the total number of items across all shops.

What is the brute force solution for Maximum Spending After Buying Items?

The brute force approach for Maximum Spending After Buying Items is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through each day and selecting the rightmost available item from any shop, calculating the cost based on the day. This method is straightforward but inefficient as it checks all items repeatedly. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Spending After Buying Items?

Maximum Spending After Buying Items uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue), Matrix. The recommended patterns to study are: Heap, Greedy.

What are the interview tips for Maximum Spending After Buying Items?

Always clarify the problem statement and constraints. Discuss your thought process and approach before jumping into coding. Consider edge cases and test your solution with them.

What are common mistakes when solving Maximum Spending After Buying Items?

Not considering the rightmost item correctly. Failing to track which items have been bought.

#2931

Maximum Spending After Buying Items

Hard

ArrayGreedySortingHeap (Priority Queue)MatrixHeapGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal approach uses a max-heap to efficiently track and retrieve the rightmost available items across all shops. This allows us to always buy the most expensive item available for each day, maximizing the total spending.

⚙️

Algorithm

3 steps

1Step 1: Initialize a max-heap to store items along with their calculated costs for each day.
2Step 2: For each shop, push the rightmost item into the heap with its cost for day 1.
3Step 3: For each day, pop the item with the maximum cost from the heap, add it to the total spending, and push the next rightmost item from the same shop into the heap.

solution.py19 lines

1# Full working Python code
2import heapq
3
4values = [[8,5,2],[6,4,1],[9,7,3]]
5
6def maxSpendingOptimal(values):
7    total_spending = 0
8    m, n = len(values), len(values[0])
9    max_heap = []
10    for i in range(m):
11        heapq.heappush(max_heap, (-values[i][n-1], i, n-1))
12    for day in range(1, m * n + 1):
13        cost, shop, item = heapq.heappop(max_heap)
14        total_spending += -cost * day
15        if item > 0:
16            heapq.heappush(max_heap, (-values[shop][item - 1], shop, item - 1))
17    return total_spending
18
19print(maxSpendingOptimal(values))

ℹ

Complexity note: The complexity is O(n log n) due to the heap operations, where n is the total number of items across all shops.

1Using a max-heap allows efficient retrieval of the most expensive item available.
2The greedy approach ensures that we maximize spending each day.

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