Problem Description

You are given a 2D integer matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that:

• The number of elements taken from the i^th row of grid does not exceed limits[i].

Return the maximum sum.

Intuition

To solve this problem, we need to select elements across the matrix grid such that their sum is maximized, while adhering to the limitations laid out by limits and k. The constraints imply that we cannot exceed taking more than limits[i] elements from the i^th row, and overall, we can't take more than k elements in total.

The intuition here is to leverage a greedy approach with the help of a priority queue (min-heap):

1. Iterate through each row of the matrix. For each row:

   • Sort the elements to easily access the largest values.
   • Choose the largest elements up to the limit specified for that row.

2. Utilize a min-heap to keep track of the largest k elements from the sorted selections across all rows:

   • Push each selected element into the min-heap.
   • If the heap grows beyond size k, remove the smallest element. This ensures that the heap always contains the largest possible elements we've seen thus far.

3. Once all rows are processed, the sum of elements within the min-heap will yield the desired maximum sum, observing the constraints given by limits and k.

By structuring the solution in this way, we ensure we're always prioritizing the largest elements for our sum while respecting the given constraints.

Learn more about Greedy, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The provided solution uses a greedy approach combined with a priority queue (min-heap) to determine the maximum sum of selected elements.

1. Priority Queue (Min-Heap): This data structure will help maintain the largest k elements efficiently. The operation of pushing and popping from a heap enables us to continually keep track of the most valuable elements while discarding less useful ones.

2. Algorithm Steps:

   • Initialize an empty priority queue pq.
   • For each row in the grid, paired with its corresponding value from limits:
     • Sort the row to bring larger numbers to one end, making it easy to select the largest values.
     • Select elements: From the sorted row, take up to limit elements and push each into the priority queue:

       for nums, limit in zip(grid, limits):
           nums.sort()
           for _ in range(limit):
               heappush(pq, nums.pop())
               if len(pq) > k:
                   heappop(pq)

     • Maintain size: Make sure the total number of elements in pq does not exceed k. If it does, remove the smallest element (the top of the min-heap).

3. Compute Maximum Sum: Once all rows are processed, sum all the elements in the priority queue. Since the queue contains only the largest possible selections of size k, this will be our maximum sum:

   return sum(pq)

This approach efficiently selects and sums the maximum possible elements while respecting the constraints on the number of elements selectable from each row and the total elements selected.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Example Input:

• grid = [[5, 3, 4], [8, 1], [6, 9, 2]]
• limits = [2, 1, 1]
• k = 3

Step-by-Step Approach:

1. Initialize the Priority Queue:

   • Start with an empty min-heap pq = [].

2. Process Each Row:

   • Row 1: [5, 3, 4] with limit 2

     • Sort the row to get [3, 4, 5].
     • Select the two largest values, 4 and 5:
       • Push 4 to pq: pq = [4]
       • Push 5 to pq: pq = [4, 5]

   • Row 2: [8, 1] with limit 1

     • Sort the row to get [1, 8].
     • Select the single largest value, 8:
       • Push 8 to pq: pq = [4, 5, 8]

   • Row 3: [6, 9, 2] with limit 1

     • Sort the row to get [2, 6, 9].
     • Select the single largest value, 9:
       • Push 9 to pq: pq = [4, 5, 8, 9]
       • The heap exceeds size k, so remove the smallest element, 4: pq = [5, 9, 8]

3. Compute the Maximum Sum:

   • The elements in the min-heap are [5, 9, 8].
   • The maximum sum is the sum of these elements: 5 + 9 + 8 = 22.

Thus, following the outlined strategy and constraints, the maximum sum we can achieve is 22.

Solution Implementation

Time and Space Complexity

The code iterates over each row of the grid, which has n rows and m columns. For each row, the contents are sorted in O(m \log m). For each of the limit elements in the row, the code performs a heappush operation, which is O(\log k). If the size of pq exceeds k, a heappop operation is also executed, which is O(\log k). Thus, for each row, the cumulative operations take O(m \log m + \text{limit} \times \log k).

Considering limit can be up to m, this implies the worst-case time complexity for a single row is O(m \log m + m \log k). Therefore, for all n rows, the time complexity becomes O(n \times m \times (\log m + \log k)).

The space complexity is dictated by the size of the heap pq, which at most contains k elements. Therefore, the space complexity is O(k).

Learn more about how to find time and space complexity quickly.