LeetCode #2768 — MEDIUM

Number of Black Blocks

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

The Problem

Problem Statement

You are given two integers m and n representing the dimensions of a 0-indexed m x n grid.

You are also given a 0-indexed 2D integer matrix coordinates, where coordinates[i] = [x, y] indicates that the cell with coordinates [x, y] is colored black. All cells in the grid that do not appear in coordinates are white.

A block is defined as a 2 x 2 submatrix of the grid. More formally, a block with cell [x, y] as its top-left corner where 0 <= x < m - 1 and 0 <= y < n - 1 contains the coordinates [x, y], [x + 1, y], [x, y + 1], and [x + 1, y + 1].

Return a 0-indexed integer array arr of size 5 such that arr[i] is the number of blocks that contains exactly i black cells.

Example 1:

Input: m = 3, n = 3, coordinates = [[0,0]]
Output: [3,1,0,0,0]
Explanation: The grid looks like this:

There is only 1 block with one black cell, and it is the block starting with cell [0,0].
The other 3 blocks start with cells [0,1], [1,0] and [1,1]. They all have zero black cells. 
Thus, we return [3,1,0,0,0].

Example 2:

Input: m = 3, n = 3, coordinates = [[0,0],[1,1],[0,2]]
Output: [0,2,2,0,0]
Explanation: The grid looks like this:

There are 2 blocks with two black cells (the ones starting with cell coordinates [0,0] and [0,1]).
The other 2 blocks have starting cell coordinates of [1,0] and [1,1]. They both have 1 black cell.
Therefore, we return [0,2,2,0,0].

Constraints:

2 <= m <= 10⁵
2 <= n <= 10⁵
0 <= coordinates.length <= 10⁴
coordinates[i].length == 2
0 <= coordinates[i][0] < m
0 <= coordinates[i][1] < n
It is guaranteed that coordinates contains pairwise distinct coordinates.

Step 01

Brute Force Baseline

Problem summary: You are given two integers m and n representing the dimensions of a 0-indexed m x n grid. You are also given a 0-indexed 2D integer matrix coordinates, where coordinates[i] = [x, y] indicates that the cell with coordinates [x, y] is colored black. All cells in the grid that do not appear in coordinates are white. A block is defined as a 2 x 2 submatrix of the grid. More formally, a block with cell [x, y] as its top-left corner where 0 <= x < m - 1 and 0 <= y < n - 1 contains the coordinates [x, y], [x + 1, y], [x, y + 1], and [x + 1, y + 1]. Return a 0-indexed integer array arr of size 5 such that arr[i] is the number of blocks that contains exactly i black cells.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

3
3
[[0,0]]

Example 2

3
3
[[0,0],[1,1],[0,2]]

Step 02

Core Insight

What unlocks the optimal approach

The number of blocks is too much but the number of black cells is less than that.
It means the number of blocks with at least one black cell is O(|coordinates|). let’s just hold them.
Iterate through the coordinates and update the block counts accordingly. For each coordinate, determine which block(s) it belongs to and increment the count of black cells for those block(s).
After processing all the coordinates, count the number of blocks with different numbers of black cells. You can use another data structure to keep track of the counts of blocks with 0 black cells, 1 black cell, and so on.

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 #2768: Number of Black Blocks
class Solution {
    public long[] countBlackBlocks(int m, int n, int[][] coordinates) {
        Map<Long, Integer> cnt = new HashMap<>(coordinates.length);
        int[] dirs = {0, 0, -1, -1, 0};
        for (var e : coordinates) {
            int x = e[0], y = e[1];
            for (int k = 0; k < 4; ++k) {
                int i = x + dirs[k], j = y + dirs[k + 1];
                if (i >= 0 && i < m - 1 && j >= 0 && j < n - 1) {
                    cnt.merge(1L * i * n + j, 1, Integer::sum);
                }
            }
        }
        long[] ans = new long[5];
        ans[0] = (m - 1L) * (n - 1);
        for (int x : cnt.values()) {
            ++ans[x];
            --ans[0];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2768: Number of Black Blocks
func countBlackBlocks(m int, n int, coordinates [][]int) []int64 {
	cnt := map[int64]int{}
	dirs := [5]int{0, 0, -1, -1, 0}
	for _, e := range coordinates {
		x, y := e[0], e[1]
		for k := 0; k < 4; k++ {
			i, j := x+dirs[k], y+dirs[k+1]
			if i >= 0 && i < m-1 && j >= 0 && j < n-1 {
				cnt[int64(i*n+j)]++
			}
		}
	}
	ans := make([]int64, 5)
	ans[0] = int64((m - 1) * (n - 1))
	for _, x := range cnt {
		ans[x]++
		ans[0]--
	}
	return ans
}

# Accepted solution for LeetCode #2768: Number of Black Blocks
class Solution:
    def countBlackBlocks(
        self, m: int, n: int, coordinates: List[List[int]]
    ) -> List[int]:
        cnt = Counter()
        for x, y in coordinates:
            for a, b in pairwise((0, 0, -1, -1, 0)):
                i, j = x + a, y + b
                if 0 <= i < m - 1 and 0 <= j < n - 1:
                    cnt[(i, j)] += 1
        ans = [0] * 5
        for x in cnt.values():
            ans[x] += 1
        ans[0] = (m - 1) * (n - 1) - len(cnt.values())
        return ans

// Accepted solution for LeetCode #2768: Number of Black Blocks
// 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 #2768: Number of Black Blocks
// class Solution {
//     public long[] countBlackBlocks(int m, int n, int[][] coordinates) {
//         Map<Long, Integer> cnt = new HashMap<>(coordinates.length);
//         int[] dirs = {0, 0, -1, -1, 0};
//         for (var e : coordinates) {
//             int x = e[0], y = e[1];
//             for (int k = 0; k < 4; ++k) {
//                 int i = x + dirs[k], j = y + dirs[k + 1];
//                 if (i >= 0 && i < m - 1 && j >= 0 && j < n - 1) {
//                     cnt.merge(1L * i * n + j, 1, Integer::sum);
//                 }
//             }
//         }
//         long[] ans = new long[5];
//         ans[0] = (m - 1L) * (n - 1);
//         for (int x : cnt.values()) {
//             ++ans[x];
//             --ans[0];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2768: Number of Black Blocks
function countBlackBlocks(m: number, n: number, coordinates: number[][]): number[] {
    const cnt: Map<number, number> = new Map();
    const dirs: number[] = [0, 0, -1, -1, 0];
    for (const [x, y] of coordinates) {
        for (let k = 0; k < 4; ++k) {
            const [i, j] = [x + dirs[k], y + dirs[k + 1]];
            if (i >= 0 && i < m - 1 && j >= 0 && j < n - 1) {
                const key = i * n + j;
                cnt.set(key, (cnt.get(key) || 0) + 1);
            }
        }
    }
    const ans: number[] = Array(5).fill(0);
    ans[0] = (m - 1) * (n - 1);
    for (const [_, x] of cnt) {
        ++ans[x];
        --ans[0];
    }
    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(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.