LeetCode #3648 — MEDIUM

Minimum Sensors to Cover Grid

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given n × m grid and an integer k.

A sensor placed on cell (r, c) covers all cells whose Chebyshev distance from (r, c) is at most k.

The Chebyshev distance between two cells (r₁, c₁) and (r₂, c₂) is max(|r₁ − r₂|,|c₁ − c₂|).

Your task is to return the minimum number of sensors required to cover every cell of the grid.

Example 1:

Input: n = 5, m = 5, k = 1

Output: 4

Explanation:

Placing sensors at positions (0, 3), (1, 0), (3, 3), and (4, 1) ensures every cell in the grid is covered. Thus, the answer is 4.

Example 2:

Input: n = 2, m = 2, k = 2

Output: 1

Explanation:

With k = 2, a single sensor can cover the entire 2 * 2 grid regardless of its position. Thus, the answer is 1.

Constraints:

1 <= n <= 10³
1 <= m <= 10³
0 <= k <= 10³

Step 01

Brute Force Baseline

Problem summary: You are given n × m grid and an integer k. A sensor placed on cell (r, c) covers all cells whose Chebyshev distance from (r, c) is at most k. The Chebyshev distance between two cells (r1, c1) and (r2, c2) is max(|r1 − r2|,|c1 − c2|). Your task is to return the minimum number of sensors required to cover every cell of the grid.

Baseline thinking

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

Pattern signal: Math

Example 1

5
5
1

Example 2

2
2
2

Step 02

Core Insight

What unlocks the optimal approach

Let <code>s = 2*k + 1</code>, the side length each sensor covers.
The minimum number is <code>ceil(n / s) * ceil(m / s)</code>.

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

Upper-end input sizes

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 #3648: Minimum Sensors to Cover Grid
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minSensors(n, m, k int) int {
// 	size := k*2 + 1
// 	return ((n-1)/size + 1) * ((m-1)/size + 1)
// }

// Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
package main

// https://space.bilibili.com/206214
func minSensors(n, m, k int) int {
	size := k*2 + 1
	return ((n-1)/size + 1) * ((m-1)/size + 1)
}

# Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
# Time:  O(1)
# Space: O(1)

# math
class Solution(object):
    def minSensors(self, n, m, k):
        """
        :type n: int
        :type m: int
        :type k: int
        :rtype: int
        """
        def ceil_divide(a, b):
            return (a+b-1)//b
    
        return ceil_divide(n, 2*k+1)*ceil_divide(m, 2*k+1)

// Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
// Rust example auto-generated from go 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 (go):
// // Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minSensors(n, m, k int) int {
// 	size := k*2 + 1
// 	return ((n-1)/size + 1) * ((m-1)/size + 1)
// }

// Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3648: Minimum Sensors to Cover Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minSensors(n, m, k int) int {
// 	size := k*2 + 1
// 	return ((n-1)/size + 1) * ((m-1)/size + 1)
// }

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.