2503. Maximum Number of Points From Grid Queries
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:
- 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.
- 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 asheappop
andheappush
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
andans
) 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 toFalse
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 markedTrue
. -
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 incrementcnt
.
In the code, we translate these concepts into implementation steps as follows:
- Initialize the dimensions
m
andn
for thegrid
, sort thequeries
with their original index, and create an answer listans
initialized with zeros to hold the result. - Prepare the priority queue
q
by pushing the top-left cell onto it. - Initialize the
cnt
counter to 0 and avis
matrix toFalse
indicating all cells are initially unvisited. - 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.
- Pop the cell from the priority queue, increment
- 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 valuecnt
.
- While the priority queue is not empty and the top cell value is less than
- 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.
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Start EvaluatorExample 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:
-
Initialize the grid dimensions (
m = 2
,n = 3
), sort the queries alongside their original indices, and prepare an answer listans
to hold the results. After sorting, the queries and their indices are:[(3, 0), (4, 1), (6, 2)]
. -
Initialize the priority queue
q
with the top-left cell(4, 0, 0)
where4
is the value and(0, 0)
are its coordinates. -
Initialize the counter (
cnt
) to0
and the visited matrix (vis
) with all values asFalse
. -
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 value3
, so we don't extract any cell from the queue for this query. Thus, the counter stays at0
, andans[0]
is set to0
.
- The top of the priority queue has a cell value
-
For the second query
(4, 1)
:- Now, the top of the priority queue is
(4, 0, 0)
, which is equal to the query value4
, so again, we do not visit new cells, leaving counter0
. Hence,ans[1]
stays0
.
- Now, the top of the priority queue is
-
For the last query
(6, 2)
:- The top of the priority queue is
(4, 0, 0)
and4
is now less than the query value6
. We pop this cell from the queue, incrementcnt
to1
, and mark it as visited. - We explore adjacent cells
[5, 3]
and[3, 3]
. Both are less than6
, so they get pushed to the queue, andcnt
increases to3
. - 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 to3
.
- The top of the priority queue is
-
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
1from heapq import heappop, heappush
2from itertools import pairwise
3
4class Solution:
5 def maxPoints(self, points_grid: List[List[int]], queries: List[int]) -> List[int]]:
6 # Get the dimensions of the grid
7 rows, cols = len(points_grid), len(points_grid[0])
8
9 # Sort the queries along with their indices for efficient access
10 sorted_queries = sorted((value, index) for index, value in enumerate(queries))
11
12 # Initialize the answer list to have the same size as queries with default 0 values
13 results = [0] * len(sorted_queries)
14
15 # Initialize the priority queue with the starting point (0,0)
16 priority_queue = [(points_grid[0][0], 0, 0)]
17
18 # Counter to keep track of how many points have been visited
19 points_count = 0
20
21 # Initialize the visited matrix to track visited cells in the grid
22 visited = [[False] * cols for _ in range(rows)]
23 visited[0][0] = True
24
25 # Iterate over the sorted queries
26 for value, index in sorted_queries:
27 # Process cells from the priority queue until the cell’s value is less than the query value
28 while priority_queue and priority_queue[0][0] < value:
29 # Pop the smallest value cell from the priority queue
30 _, current_row, current_col = heappop(priority_queue)
31
32 # Increment the visited points counter
33 points_count += 1
34
35 # Use the pairwise utility to iterate over the adjacent cells
36 for delta_row, delta_col in pairwise((-1, 0, 1, 0, -1)):
37 # Calculate the adjacent cell's row and column
38 adjacent_row, adjacent_col = current_row + delta_row, current_col + delta_col
39
40 # Check if the adjacent cell is inside the grid and not visited
41 if 0 <= adjacent_row < rows and 0 <= adjacent_col < cols and not visited[adjacent_row][adjacent_col]:
42 # Push the adjacent cell into the priority queue
43 heappush(priority_queue, (points_grid[adjacent_row][adjacent_col], adjacent_row, adjacent_col))
44
45 # Mark the adjacent cell as visited
46 visited[adjacent_row][adjacent_col] = True
47
48 # After processing the priority queue, assign the number of points visited to the results
49 results[index] = points_count
50
51 # Return the results list with the number of points visited for each query
52 return results
53
1import java.util.Arrays;
2import java.util.PriorityQueue;
3
4class Solution {
5
6 public int[] maxPoints(int[][] grid, int[] queries) {
7 int queryLength = queries.length;
8 // Array to store value and index of each query
9 int[][] sortedQueries = new int[queryLength][2];
10 // Fill the array with queries and their original indexes
11 for (int i = 0; i < queryLength; ++i) {
12 sortedQueries[i] = new int[] { queries[i], i };
13 }
14 // Sort queries based on their values
15 Arrays.sort(sortedQueries, (a, b) -> a[0] - b[0]);
16
17 // Initialize the answer array to store results for each query
18 int[] answers = new int[queryLength];
19 // Get the dimensions of the grid
20 int rows = grid.length;
21 int columns = grid[0].length;
22 // Keep track of visited cells
23 boolean[][] visited = new boolean[rows][columns];
24 // Mark the start cell as visited
25 visited[0][0] = true;
26
27 // Priority queue to store cell values with coordinates, sorted by value
28 PriorityQueue<int[]> queue = new PriorityQueue<>((a, b) -> a[0] - b[0]);
29 // Offer the first cell to the priority queue
30 queue.offer(new int[] { grid[0][0], 0, 0 });
31
32 // Array to help navigate through neighbors of a cell
33 int[] directions = new int[] { -1, 0, 1, 0, -1 };
34 // Counter to keep the count of cells processed
35 int count = 0;
36
37 // Process each sorted query
38 for (int[] query : sortedQueries) {
39 int queryValue = query[0];
40 int queryIndex = query[1];
41
42 // Poll cells with value less than the current query value
43 while (!queue.isEmpty() && queue.peek()[0] < queryValue) {
44 int[] cell = queue.poll();
45 // Increment the count whenever a cell is polled
46 ++count;
47
48 // Navigate to neighboring cells
49 for (int i = 0; i < 4; ++i) {
50 int newRow = cell[1] + directions[i];
51 int newColumn = cell[2] + directions[i + 1];
52
53 // If the neighbor is within bounds and not visited yet
54 if (newRow >= 0 && newRow < rows
55 && newColumn >= 0 && newColumn < columns
56 && !visited[newRow][newColumn]) {
57 // Mark as visited
58 visited[newRow][newColumn] = true;
59 // Add to queue with values
60 queue.offer(new int[] { grid[newRow][newColumn], newRow, newColumn });
61 }
62 }
63 }
64 // Save counter value in the answer array corresponding to the original query index
65 answers[queryIndex] = count;
66 }
67 // Return the answer array with counts for each query
68 return answers;
69 }
70}
71
1#include <vector>
2#include <queue>
3#include <algorithm>
4#include <cstring>
5using namespace std;
6
7class Solution {
8public:
9 // Directions array for moving up, right, down, left in a grid
10 const int directions[5] = {-1, 0, 1, 0, -1};
11
12 // This function will return the max points from the top-left corner to the cells
13 // value greater than or equal to the given queries on a grid
14 vector<int> maxPoints(vector<vector<int>>& grid, vector<int>& queries) {
15 int querySize = queries.size();
16 vector<pair<int, int>> sortedQueries(querySize);
17
18 // Attach each query value to its original index
19 for (int i = 0; i < querySize; ++i) {
20 sortedQueries[i] = {queries[i], i};
21 }
22 // Sort the queries in ascending order
23 sort(sortedQueries.begin(), sortedQueries.end());
24
25 // Initialize the answer vector to store results for the original queries order
26 vector<int> answer(querySize);
27 int rows = grid.size(), cols = grid[0].size();
28 bool visited[rows][cols];
29 // Set visited status of all cells to false
30 memset(visited, 0, sizeof visited);
31
32 // Start from the top-left corner
33 visited[0][0] = true;
34 priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> minHeap;
35 minHeap.push({grid[0][0], 0, 0});
36
37 int exploredCount = 0; // Number of cells explored that satisfy a given query
38
39 for (auto& queryPair : sortedQueries) {
40 int requiredValue = queryPair.first; // The minimum required value for this query
41 int originalIndex = queryPair.second; // The original index of this query
42
43 // Explore cells until we meet the condition of the current query
44 while (!minHeap.empty() && get<0>(minHeap.top()) < requiredValue) {
45 auto [value, row, col] = minHeap.top();
46 minHeap.pop();
47 exploredCount++; // Increment the count of explored cells
48
49 // Explore adjacent cells
50 for (int i = 0; i < 4; ++i) {
51 int newRow = row + directions[i], newCol = col + directions[i + 1];
52 // Check for valid and unvisited adjacent cells
53 if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && !visited[newRow][newCol]) {
54 visited[newRow][newCol] = true;
55 minHeap.push({grid[newRow][newCol], newRow, newCol});
56 }
57 }
58 }
59 // Record the number of cells explored that satisfy the query's condition
60 answer[originalIndex] = exploredCount;
61 }
62
63 // Return the result in the order of the original queries
64 return answer;
65 }
66};
67
1type Grid = number[][];
2type Query = number[];
3type HeapElement = [number, number, number]; // Typing for clarity: [value, row, col]
4
5// Directions array for moving up, right, down, and left in a grid
6const directions: number[] = [-1, 0, 1, 0, -1];
7
8// Function that returns the max points from the top-left corner to the cells
9// with a value greater than or equal to the given queries on a grid
10function maxPoints(grid: Grid, queries: Query): number[] {
11 const querySize: number = queries.length;
12 let sortedQueries: Array<[number, number]> = new Array(querySize);
13
14 // Attach each query value to its original index
15 for (let i = 0; i < querySize; ++i) {
16 sortedQueries[i] = [queries[i], i];
17 }
18
19 // Sort the queries in ascending order
20 sortedQueries.sort((a, b) => a[0] - b[0]);
21
22 // Initialize the answer array to store results for the original queries order
23 let answers: number[] = new Array(querySize).fill(0);
24 const rows: number = grid.length,
25 cols: number = grid[0].length;
26
27 // Array to check if a cell has been visited
28 let visited: boolean[][] = Array.from({ length: rows }, () => Array(cols).fill(false));
29
30 // Start from the top-left corner and mark it as visited
31 visited[0][0] = true;
32
33 // Priority queue (using a min-heap) for BFS
34 let minHeap: HeapElement[] = [];
35 minHeap.push([grid[0][0], 0, 0]);
36
37 // Function to insert elements into the min-heap, maintaining sorted order
38 function pushToMinHeap(element: HeapElement) {
39 minHeap.push(element);
40 minHeap.sort(([valueA], [valueB]) => valueB - valueA); // sort to keep the smallest element at the end
41 }
42
43 let exploredCount: number = 0; // Number of cells explored that satisfy a given query
44
45 // Function to extract the smallest element from the min-heap
46 function popFromMinHeap(): HeapElement {
47 return <HeapElement>minHeap.pop(); // Explicit type assertion for clarity
48 }
49
50 for (let [requiredValue, originalIndex] of sortedQueries) {
51 // Explore cells until we meet the condition of the current query
52 while (minHeap.length > 0 && minHeap[minHeap.length - 1][0] < requiredValue) {
53 let [value, row, col] = popFromMinHeap();
54 exploredCount++; // Increment the count of explored cells
55
56 // Explore adjacent cells
57 for (let i = 0; i < 4; ++i) {
58 let newRow = row + directions[i],
59 newCol = col + directions[i + 1];
60
61 // Check for valid and unvisited adjacent cells
62 if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && !visited[newRow][newCol]) {
63 visited[newRow][newCol] = true;
64 pushToMinHeap([grid[newRow][newCol], newRow, newCol]);
65 }
66 }
67 }
68
69 // Record the number of cells explored that satisfy the query's condition
70 answers[originalIndex] = exploredCount;
71 }
72
73 // Return the result in the order of the original queries
74 return answers;
75}
76
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)
, whereQ
is the number of queries. - Priority queue operations: The main loop iterates potentially over all cells in the grid, performing
heappush
andheappop
operations. Insertion and removal from a heap have complexityO(log K)
, whereK
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 inO(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.
In a binary min heap, the maximum element can be found in:
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