Problem Description

The problem presents us with a two-dimensional grid filled with integers and a list of queries. Each query represents an attempt to accumulate points by moving through the grid starting from the top-left cell. The rule to accumulate points is straightforward: you can only move to an adjacent cell (up, down, left, or right) if the query's value is strictly greater than the current cell's value, earning one point the first time you visit a cell. If it's not greater, you stop moving and can’t earn any points from that cell. The process for each query is independent of the others, and you can visit the same cell multiple times with different queries, potentially earning points each time.

The ultimate goal is to determine, for each query, the maximum number of points one can earn following the above rules.

Flowchart Walkthrough

First, let's analyze the problem using the Flowchart. Here's a step-by-step breakdown:

Is it a graph?

• Yes: Although not explicitly structured as a traditional graph, the grid in the provided problem can be viewed as a graph where each cell potentially connects to adjacent cells.

Is it a tree?

• No: The grid doesn't inherently form a single branching structure typical of trees. It potentially involves multiple branching possibilities from each cell.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem discusses maximizing points through grid queries and not processing tasks or dependencies as in DAGs.

Is the problem related to shortest paths?

• No: The main objective is to maximize points, not to find the shortest path in the grid.

Does the problem involve connectivity?

• Yes: There is an implicit need to determine the connections between points to ensure optimal movement or picking strategy throughout the grid.

Is the graph weighted?

• No information on weights is implied in the query of maximizing points from the grid, suggesting an unweighted approach.

Conclusion: The problem aligns well with using BFS according to the characteristics of connectivity in an unweighted graph, as indicated by the flowchart. Thus, BFS should be utilized to explore potential paths or contiguous sections of the grid efficiently to maximize the points collected.

Intuition

Approaching the solution to this problem, we observe that the sequential nature of movements in the grid resembles a breadth-first traversal, where from any given cell, we can move in any of the four cardinal directions as long as the value of the query allows it.

One naive approach would be to start at the top-left cell and perform a breadth-first search for each query independently, but this would be inefficient, particularly when the grid and the number of queries are large. A better approach is needed to optimize the search across multiple queries.

The core intuition for an optimized approach lies in two observations:

1. As we process the queries in ascending order, we can leverage the work from previous queries. Once a cell becomes reachable (i.e., its value is less than the current query value), it remains reachable for all subsequent queries.
2. We can maintain a priority queue to track cells in the order of their values. This ensures that we only visit cells that are reachable for the current query value and avoid re-processing the entire grid unnecessarily.

Therefore, the steps we follow to arrive at the solution are:

• First, sort the queries alongside their original indices. This allows us to process them in ascending order and maintain the ability to return answers in the order they were originally asked.
• Use a priority queue (min-heap) to hold grid cells prioritized by their value along with their coordinates. We start by adding only the top-left cell to this queue.
• Initialize a counter to track the number of distinct cells visited during the process.
• For each query, processed in order, we:
  • Keep extracting cells from the priority queue as long as their value is less than the current query value. Visiting a cell for the first time under a specific query earns a point (as long as the cell's value is less than the query value).
  • For each cell visited, we check its adjacent cells and if any of them are unvisited and reachable, we add them to the priority queue.
• The counter value after processing a specific query corresponds to the maximum points for that query. We store this in the corresponding index in the answer array.
• Return the answer array after processing all queries. The use of a priority queue and the early termination condition for cells values that do not meet the threshold optimize this approach.

With this approach, the solution scales better with the size of the grid and the number of queries, providing an efficient way to calculate the maximum number of points for each query.

Learn more about Breadth-First Search, Union Find, Two Pointers, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The provided code uses a few classic data structures and algorithms to implement an efficient solution to the problem:

• Priority Queue (Min-Heap): The heapq module in Python (used as heappop and heappush in the code) provides an implementation of the priority queue data structure. The priority queue is essential to manage the cells in the order of their grid values, ensuring that we always process the smallest-valued cell that qualifies for the current query.

• Breadth-First Search (BFS): This search pattern is used because we explore the grid from the top-left working outwards, considering cells one step away (adjacent cells) from any given cell. Whenever a cell's value is less than the current query value, we can move to its neighbors.

• List Comprehension and Tuple Packing: The code uses list comprehension to create lists (like vis and ans) and tuple packing to store the multiple pieces of data together for each cell (such as grid value, x-coordinate, and y-coordinate).

• Boolean Matrix for Visited Cells: A matrix vis of the same dimensions as the grid is initialized to False and used to mark whether or not a cell has been visited during the traversal for the present query value. Once a cell is visited, it is marked True.

• Sorting the Queries: The queries are sorted to ensure that we process them in ascending order. This is beneficial as once a cell becomes reachable for a certain query value, it remains reachable for all subsequent queries.

• Counter: A variable cnt is used to keep track of how many points we can get for the current query. Every time we pop a cell from the priority queue that satisfies the query condition, we increment cnt.

In the code, we translate these concepts into implementation steps as follows:

1. Initialize the dimensions m and n for the grid, sort the queries with their original index, and create an answer list ans initialized with zeros to hold the result.
2. Prepare the priority queue q by pushing the top-left cell onto it.
3. Initialize the cnt counter to 0 and a vis matrix to False indicating all cells are initially unvisited.
4. Start iterating over the sorted queries. For each (v, k) pair (value and index from the sorted queries):
   • While the priority queue is not empty and the top cell value is less than v:
     • Pop the cell from the priority queue, increment cnt, and check its adjacent cells.
     • For each adjacent cell (x, y) that is within the grid boundaries and unvisited, push it onto the priority queue and mark it as visited.
   • After all reachable cells for this query value have been visited and added, update the answer array at the index k with the current counter value cnt.
5. Once all queries have been processed, return the ans array.

The algorithm effectively uses the earlier processes and stored state (priority queue and visited matrix) to reduce redundant computations as it progresses through each query, making the overall approach efficient for this type of grid traversal and accumulation problem.

Example Walkthrough

Let's consider a small example to illustrate the solution approach described above. Imagine we have a grid and a list of queries as follows:

Grid:

4 3 2
5 3 1

Queries: [3, 4, 6]

Now let's walk through this example following the solution steps:

1. Initialize the grid dimensions (m = 2, n = 3), sort the queries alongside their original indices, and prepare an answer list ans to hold the results. After sorting, the queries and their indices are: [(3, 0), (4, 1), (6, 2)].

2. Initialize the priority queue q with the top-left cell (4, 0, 0) where 4 is the value and (0, 0) are its coordinates.

3. Initialize the counter (cnt) to 0 and the visited matrix (vis) with all values as False.

4. Start iterating through the sorted queries. For the first query (3, 0):

   • The top of the priority queue has a cell value 4, which is not less than the query value 3, so we don't extract any cell from the queue for this query. Thus, the counter stays at 0, and ans[0] is set to 0.

5. For the second query (4, 1):

   • Now, the top of the priority queue is (4, 0, 0), which is equal to the query value 4, so again, we do not visit new cells, leaving counter 0. Hence, ans[1] stays 0.

6. For the last query (6, 2):

   • The top of the priority queue is (4, 0, 0) and 4 is now less than the query value 6. We pop this cell from the queue, increment cnt to 1, and mark it as visited.
   • We explore adjacent cells [5, 3] and [3, 3]. Both are less than 6, so they get pushed to the queue, and cnt increases to 3.
   • Continue to pop cells from the priority queue as long as their value is less than 6, but all remaining cells have already been visited, so we stop.
   • Therefore, ans[2] is set to 3.

7. The final ans array after processing all queries is [0, 0, 3].

This walkthrough demonstrates the application of the provided solution approach on a simplified version of the given problem, showcasing how sorting queries and using a priority queue can efficiently determine the maximum points earned for each query.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code depends on several factors:

• Sorting the queries: The sort operation has a complexity of O(Q log Q), where Q is the number of queries.
• Priority queue operations: The main loop iterates potentially over all cells in the grid, performing heappush and heappop operations. Insertion and removal from a heap have complexity O(log K), where K is the number of elements in the heap.
• Checking neighboring cells: For each cell, it checks at most 4 neighbors, adding a constant factor.

Assuming that in the worst case each cell is visited and added to the priority queue, the complexity becomes O(MN log(MN)). Therefore, combining this with the sort, we get a time complexity of O(Q log Q + MN log(MN)), where M and N are the dimensions of the grid.

Space Complexity

The space complexity is determined by:

• The heap/priority queue: In the worst case, it can hold all cells in the grid, resulting in O(MN).
• The visited grid: An M x N boolean grid for visited cells also results in O(MN).
• The answer list: Storing the result for each query gives O(Q).

Therefore, the total space complexity is O(MN + Q), with MN for the visited matrix and the priority queue, and Q being the size of the answer list.

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