LeetCode #2503 — HARD

Maximum Number of Points From Grid Queries

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an m x n integer matrix grid and an array queries of size k.

Find an array answer of size k such that for each integer queries[i] you start in the top left cell of the matrix and repeat the following process:

If queries[i] is strictly greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any adjacent cell in all 4 directions: up, down, left, and right.
Otherwise, you do not get any points, and you end this process.

After the process, answer[i] is the maximum number of points you can get. Note that for each query you are allowed to visit the same cell multiple times.

Return the resulting array answer.

Example 1:

Input: grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]
Output: [5,8,1]
Explanation: The diagrams above show which cells we visit to get points for each query.

Example 2:

Input: grid = [[5,2,1],[1,1,2]], queries = [3]
Output: [0]
Explanation: We can not get any points because the value of the top left cell is already greater than or equal to 3.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 1000
4 <= m * n <= 10⁵
k == queries.length
1 <= k <= 10⁴
1 <= grid[i][j], queries[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an m x n integer matrix grid and an array queries of size k. Find an array answer of size k such that for each integer queries[i] you start in the top left cell of the matrix and repeat the following process: If queries[i] is strictly greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any adjacent cell in all 4 directions: up, down, left, and right. Otherwise, you do not get any points, and you end this process. After the process, answer[i] is the maximum number of points you can get. Note that for each query you are allowed to visit the same cell multiple times. Return the resulting array answer.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Union-Find

Example 1

[[1,2,3],[2,5,7],[3,5,1]]
[5,6,2]

Example 2

[[5,2,1],[1,1,2]]
[3]

Core Insight

What unlocks the optimal approach

The queries are all given to you beforehand so you can answer them in any order you want.
Sort the queries knowing their original order to be able to build the answer array.
Run a BFS on the graph and answer the queries in increasing order.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
class Solution {
    public int[] maxPoints(int[][] grid, int[] queries) {
        int k = queries.length;
        int[][] qs = new int[k][2];
        for (int i = 0; i < k; ++i) {
            qs[i] = new int[] {queries[i], i};
        }
        Arrays.sort(qs, (a, b) -> a[0] - b[0]);
        int[] ans = new int[k];
        int m = grid.length, n = grid[0].length;
        boolean[][] vis = new boolean[m][n];
        vis[0][0] = true;
        PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        q.offer(new int[] {grid[0][0], 0, 0});
        int[] dirs = new int[] {-1, 0, 1, 0, -1};
        int cnt = 0;
        for (var e : qs) {
            int v = e[0];
            k = e[1];
            while (!q.isEmpty() && q.peek()[0] < v) {
                var p = q.poll();
                ++cnt;
                for (int h = 0; h < 4; ++h) {
                    int x = p[1] + dirs[h], y = p[2] + dirs[h + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
                        vis[x][y] = true;
                        q.offer(new int[] {grid[x][y], x, y});
                    }
                }
            }
            ans[k] = cnt;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
func maxPoints(grid [][]int, queries []int) []int {
	k := len(queries)
	qs := make([]pair, k)
	for i, v := range queries {
		qs[i] = pair{v, i}
	}
	sort.Slice(qs, func(i, j int) bool { return qs[i].v < qs[j].v })
	ans := make([]int, k)
	m, n := len(grid), len(grid[0])
	q := hp{}
	heap.Push(&q, tuple{grid[0][0], 0, 0})
	dirs := []int{-1, 0, 1, 0, -1}
	vis := map[int]bool{0: true}
	cnt := 0
	for _, e := range qs {
		v := e.v
		k = e.i
		for len(q) > 0 && q[0].v < v {
			p := heap.Pop(&q).(tuple)
			i, j := p.i, p.j
			cnt++
			for h := 0; h < 4; h++ {
				x, y := i+dirs[h], j+dirs[h+1]
				if x >= 0 && x < m && y >= 0 && y < n && !vis[x*n+y] {
					vis[x*n+y] = true
					heap.Push(&q, tuple{grid[x][y], x, y})
				}
			}
		}
		ans[k] = cnt
	}
	return ans
}

type pair struct{ v, i int }

type tuple struct{ v, i, j int }
type hp []tuple

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].v < h[j].v }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(tuple)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
class Solution:
    def maxPoints(self, grid: List[List[int]], queries: List[int]) -> List[int]:
        m, n = len(grid), len(grid[0])
        qs = sorted((v, i) for i, v in enumerate(queries))
        ans = [0] * len(qs)
        q = [(grid[0][0], 0, 0)]
        cnt = 0
        vis = [[False] * n for _ in range(m)]
        vis[0][0] = True
        for v, k in qs:
            while q and q[0][0] < v:
                _, i, j = heappop(q)
                cnt += 1
                for a, b in pairwise((-1, 0, 1, 0, -1)):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and not vis[x][y]:
                        heappush(q, (grid[x][y], x, y))
                        vis[x][y] = True
            ans[k] = cnt
        return ans

// Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
// class Solution {
//     public int[] maxPoints(int[][] grid, int[] queries) {
//         int k = queries.length;
//         int[][] qs = new int[k][2];
//         for (int i = 0; i < k; ++i) {
//             qs[i] = new int[] {queries[i], i};
//         }
//         Arrays.sort(qs, (a, b) -> a[0] - b[0]);
//         int[] ans = new int[k];
//         int m = grid.length, n = grid[0].length;
//         boolean[][] vis = new boolean[m][n];
//         vis[0][0] = true;
//         PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
//         q.offer(new int[] {grid[0][0], 0, 0});
//         int[] dirs = new int[] {-1, 0, 1, 0, -1};
//         int cnt = 0;
//         for (var e : qs) {
//             int v = e[0];
//             k = e[1];
//             while (!q.isEmpty() && q.peek()[0] < v) {
//                 var p = q.poll();
//                 ++cnt;
//                 for (int h = 0; h < 4; ++h) {
//                     int x = p[1] + dirs[h], y = p[2] + dirs[h + 1];
//                     if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
//                         vis[x][y] = true;
//                         q.offer(new int[] {grid[x][y], x, y});
//                     }
//                 }
//             }
//             ans[k] = cnt;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2503: Maximum Number of Points From Grid Queries
// class Solution {
//     public int[] maxPoints(int[][] grid, int[] queries) {
//         int k = queries.length;
//         int[][] qs = new int[k][2];
//         for (int i = 0; i < k; ++i) {
//             qs[i] = new int[] {queries[i], i};
//         }
//         Arrays.sort(qs, (a, b) -> a[0] - b[0]);
//         int[] ans = new int[k];
//         int m = grid.length, n = grid[0].length;
//         boolean[][] vis = new boolean[m][n];
//         vis[0][0] = true;
//         PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
//         q.offer(new int[] {grid[0][0], 0, 0});
//         int[] dirs = new int[] {-1, 0, 1, 0, -1};
//         int cnt = 0;
//         for (var e : qs) {
//             int v = e[0];
//             k = e[1];
//             while (!q.isEmpty() && q.peek()[0] < v) {
//                 var p = q.poll();
//                 ++cnt;
//                 for (int h = 0; h < 4; ++h) {
//                     int x = p[1] + dirs[h], y = p[2] + dirs[h + 1];
//                     if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
//                         vis[x][y] = true;
//                         q.offer(new int[] {grid[x][y], x, y});
//                     }
//                 }
//             }
//             ans[k] = cnt;
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(k × log k + m × n log(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.