LeetCode #3531 — MEDIUM

Count Covered Buildings

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

The Problem

Problem Statement

You are given a positive integer n, representing an n x n city. You are also given a 2D grid buildings, where buildings[i] = [x, y] denotes a unique building located at coordinates [x, y].

A building is covered if there is at least one building in all four directions: left, right, above, and below.

Return the number of covered buildings.

Example 1:

Input: n = 3, buildings = [[1,2],[2,2],[3,2],[2,1],[2,3]]

Output: 1

Explanation:

Only building [2,2] is covered as it has at least one building:
- above ([1,2])
- below ([3,2])
- left ([2,1])
- right ([2,3])
Thus, the count of covered buildings is 1.

Example 2:

Input: n = 3, buildings = [[1,1],[1,2],[2,1],[2,2]]

Output: 0

Explanation:

No building has at least one building in all four directions.

Example 3:

Input: n = 5, buildings = [[1,3],[3,2],[3,3],[3,5],[5,3]]

Output: 1

Explanation:

Only building [3,3] is covered as it has at least one building:
- above ([1,3])
- below ([5,3])
- left ([3,2])
- right ([3,5])
Thus, the count of covered buildings is 1.

Constraints:

2 <= n <= 10⁵
1 <= buildings.length <= 10⁵
buildings[i] = [x, y]
1 <= x, y <= n
All coordinates of buildings are unique.

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer n, representing an n x n city. You are also given a 2D grid buildings, where buildings[i] = [x, y] denotes a unique building located at coordinates [x, y]. A building is covered if there is at least one building in all four directions: left, right, above, and below. Return the number of covered buildings.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Group buildings with the same x or y value together, and sort each group.
In each sorted list, the buildings that are not at the first or last positions are covered in that direction.

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 #3531: Count Covered Buildings
class Solution {
    public int countCoveredBuildings(int n, int[][] buildings) {
        Map<Integer, List<Integer>> g1 = new HashMap<>();
        Map<Integer, List<Integer>> g2 = new HashMap<>();

        for (int[] building : buildings) {
            int x = building[0], y = building[1];
            g1.computeIfAbsent(x, k -> new ArrayList<>()).add(y);
            g2.computeIfAbsent(y, k -> new ArrayList<>()).add(x);
        }

        for (var e : g1.entrySet()) {
            Collections.sort(e.getValue());
        }
        for (var e : g2.entrySet()) {
            Collections.sort(e.getValue());
        }

        int ans = 0;

        for (int[] building : buildings) {
            int x = building[0], y = building[1];
            List<Integer> l1 = g1.get(x);
            List<Integer> l2 = g2.get(y);

            if (l2.get(0) < x && x < l2.get(l2.size() - 1) && l1.get(0) < y
                && y < l1.get(l1.size() - 1)) {
                ans++;
            }
        }

        return ans;
    }
}

// Accepted solution for LeetCode #3531: Count Covered Buildings
func countCoveredBuildings(n int, buildings [][]int) (ans int) {
	g1 := make(map[int][]int)
	g2 := make(map[int][]int)

	for _, building := range buildings {
		x, y := building[0], building[1]
		g1[x] = append(g1[x], y)
		g2[y] = append(g2[y], x)
	}

	for _, list := range g1 {
		sort.Ints(list)
	}
	for _, list := range g2 {
		sort.Ints(list)
	}

	for _, building := range buildings {
		x, y := building[0], building[1]
		l1 := g1[x]
		l2 := g2[y]

		if l2[0] < x && x < l2[len(l2)-1] && l1[0] < y && y < l1[len(l1)-1] {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3531: Count Covered Buildings
class Solution:
    def countCoveredBuildings(self, n: int, buildings: List[List[int]]) -> int:
        g1 = defaultdict(list)
        g2 = defaultdict(list)
        for x, y in buildings:
            g1[x].append(y)
            g2[y].append(x)
        for x in g1:
            g1[x].sort()
        for y in g2:
            g2[y].sort()
        ans = 0
        for x, y in buildings:
            l1 = g1[x]
            l2 = g2[y]
            if l2[0] < x < l2[-1] and l1[0] < y < l1[-1]:
                ans += 1
        return ans

// Accepted solution for LeetCode #3531: Count Covered Buildings
use std::collections::HashMap;

impl Solution {
    pub fn count_covered_buildings(_n: i32, buildings: Vec<Vec<i32>>) -> i32 {
        let mut g1: HashMap<i32, Vec<i32>> = HashMap::new();
        let mut g2: HashMap<i32, Vec<i32>> = HashMap::new();

        for b in &buildings {
            let x = b[0];
            let y = b[1];
            g1.entry(x).or_insert(Vec::new()).push(y);
            g2.entry(y).or_insert(Vec::new()).push(x);
        }

        for v in g1.values_mut() {
            v.sort();
        }
        for v in g2.values_mut() {
            v.sort();
        }

        let mut ans: i32 = 0;

        for b in &buildings {
            let x = b[0];
            let y = b[1];

            let l1 = g1.get(&x).unwrap();
            let l2 = g2.get(&y).unwrap();

            if l2[0] < x && x < l2[l2.len() - 1] && l1[0] < y && y < l1[l1.len() - 1] {
                ans += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3531: Count Covered Buildings
function countCoveredBuildings(n: number, buildings: number[][]): number {
    const g1: Map<number, number[]> = new Map();
    const g2: Map<number, number[]> = new Map();

    for (const [x, y] of buildings) {
        if (!g1.has(x)) g1.set(x, []);
        g1.get(x)?.push(y);

        if (!g2.has(y)) g2.set(y, []);
        g2.get(y)?.push(x);
    }

    for (const list of g1.values()) {
        list.sort((a, b) => a - b);
    }
    for (const list of g2.values()) {
        list.sort((a, b) => a - b);
    }

    let ans = 0;

    for (const [x, y] of buildings) {
        const l1 = g1.get(x)!;
        const l2 = g2.get(y)!;

        if (l2[0] < x && x < l2[l2.length - 1] && l1[0] < y && y < l1[l1.length - 1]) {
            ans++;
        }
    }

    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.