LeetCode #1001 — HARD

Grid Illumination

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

Solve on LeetCode
The Problem

Problem Statement

There is a 2D grid of size n x n where each cell of this grid has a lamp that is initially turned off.

You are given a 2D array of lamp positions lamps, where lamps[i] = [rowi, coli] indicates that the lamp at grid[rowi][coli] is turned on. Even if the same lamp is listed more than once, it is turned on.

When a lamp is turned on, it illuminates its cell and all other cells in the same row, column, or diagonal.

You are also given another 2D array queries, where queries[j] = [rowj, colj]. For the jth query, determine whether grid[rowj][colj] is illuminated or not. After answering the jth query, turn off the lamp at grid[rowj][colj] and its 8 adjacent lamps if they exist. A lamp is adjacent if its cell shares either a side or corner with grid[rowj][colj].

Return an array of integers ans, where ans[j] should be 1 if the cell in the jth query was illuminated, or 0 if the lamp was not.

Example 1:

Input: n = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,0]]
Output: [1,0]
Explanation: We have the initial grid with all lamps turned off. In the above picture we see the grid after turning on the lamp at grid[0][0] then turning on the lamp at grid[4][4].
The 0th query asks if the lamp at grid[1][1] is illuminated or not (the blue square). It is illuminated, so set ans[0] = 1. Then, we turn off all lamps in the red square.

The 1st query asks if the lamp at grid[1][0] is illuminated or not (the blue square). It is not illuminated, so set ans[1] = 0. Then, we turn off all lamps in the red rectangle.

Example 2:

Input: n = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,1]]
Output: [1,1]

Example 3:

Input: n = 5, lamps = [[0,0],[0,4]], queries = [[0,4],[0,1],[1,4]]
Output: [1,1,0]

Constraints:

  • 1 <= n <= 109
  • 0 <= lamps.length <= 20000
  • 0 <= queries.length <= 20000
  • lamps[i].length == 2
  • 0 <= rowi, coli < n
  • queries[j].length == 2
  • 0 <= rowj, colj < n

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is a 2D grid of size n x n where each cell of this grid has a lamp that is initially turned off. You are given a 2D array of lamp positions lamps, where lamps[i] = [rowi, coli] indicates that the lamp at grid[rowi][coli] is turned on. Even if the same lamp is listed more than once, it is turned on. When a lamp is turned on, it illuminates its cell and all other cells in the same row, column, or diagonal. You are also given another 2D array queries, where queries[j] = [rowj, colj]. For the jth query, determine whether grid[rowj][colj] is illuminated or not. After answering the jth query, turn off the lamp at grid[rowj][colj] and its 8 adjacent lamps if they exist. A lamp is adjacent if its cell shares either a side or corner with grid[rowj][colj]. Return an array of integers ans, where ans[j] should be 1 if the cell in the jth query was illuminated, or 0 if the lamp was not.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

5
[[0,0],[4,4]]
[[1,1],[1,0]]

Example 2

5
[[0,0],[4,4]]
[[1,1],[1,1]]

Example 3

5
[[0,0],[0,4]]
[[0,4],[0,1],[1,4]]

Related Problems

  • N-Queens (n-queens)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #1001: Grid Illumination
class Solution {
    private int n;
    public int[] gridIllumination(int n, int[][] lamps, int[][] queries) {
        this.n = n;
        Set<Long> s = new HashSet<>();
        Map<Integer, Integer> row = new HashMap<>();
        Map<Integer, Integer> col = new HashMap<>();
        Map<Integer, Integer> diag1 = new HashMap<>();
        Map<Integer, Integer> diag2 = new HashMap<>();
        for (var lamp : lamps) {
            int i = lamp[0], j = lamp[1];
            if (s.add(f(i, j))) {
                merge(row, i, 1);
                merge(col, j, 1);
                merge(diag1, i - j, 1);
                merge(diag2, i + j, 1);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int k = 0; k < m; ++k) {
            int i = queries[k][0], j = queries[k][1];
            if (exist(row, i) || exist(col, j) || exist(diag1, i - j) || exist(diag2, i + j)) {
                ans[k] = 1;
            }
            for (int x = i - 1; x <= i + 1; ++x) {
                for (int y = j - 1; y <= j + 1; ++y) {
                    if (x < 0 || x >= n || y < 0 || y >= n || !s.contains(f(x, y))) {
                        continue;
                    }
                    s.remove(f(x, y));
                    merge(row, x, -1);
                    merge(col, y, -1);
                    merge(diag1, x - y, -1);
                    merge(diag2, x + y, -1);
                }
            }
        }
        return ans;
    }

    private void merge(Map<Integer, Integer> cnt, int x, int d) {
        if (cnt.merge(x, d, Integer::sum) == 0) {
            cnt.remove(x);
        }
    }

    private boolean exist(Map<Integer, Integer> cnt, int x) {
        return cnt.getOrDefault(x, 0) > 0;
    }

    private long f(long i, long j) {
        return i * n + j;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.