LeetCode #3809 — MEDIUM

Best Reachable Tower

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

The Problem

Problem Statement

You are given a 2D integer array towers, where towers[i] = [x_i, y_i, q_i] represents the coordinates (x_i, y_i) and quality factor q_i of the i^th tower.

You are also given an integer array center = [cx, cy] representing your location, and an integer radius.

A tower is reachable if its Manhattan distance from center is less than or equal to radius.

Among all reachable towers:

Return the coordinates of the tower with the maximum quality factor.
If there is a tie, return the tower with the lexicographically smallest coordinate. If no tower is reachable, return [-1, -1].

The Manhattan Distance between two cells (x_i, y_i) and (x_j, y_j) is |x_i - x_j| + |y_i - y_j|.

A coordinate [x_i, y_i] is lexicographically smaller than [x_j, y_j] if x_i < x_j, or x_i == x_j and y_i < y_j.

|x| denotes the absolute value of x.

Example 1:

Input: towers = [[1,2,5], [2,1,7], [3,1,9]], center = [1,1], radius = 2

Output: [3,1]

Explanation:

Tower [1, 2, 5]: Manhattan distance = |1 - 1| + |2 - 1| = 1, reachable.
Tower [2, 1, 7]: Manhattan distance = |2 - 1| + |1 - 1| = 1, reachable.
Tower [3, 1, 9]: Manhattan distance = |3 - 1| + |1 - 1| = 2, reachable.

All towers are reachable. The maximum quality factor is 9, which corresponds to tower [3, 1].

Example 2:

Input: towers = [[1,3,4], [2,2,4], [4,4,7]], center = [0,0], radius = 5

Output: [1,3]

Explanation:

Tower [1, 3, 4]: Manhattan distance = |1 - 0| + |3 - 0| = 4, reachable.
Tower [2, 2, 4]: Manhattan distance = |2 - 0| + |2 - 0| = 4, reachable.
Tower [4, 4, 7]: Manhattan distance = |4 - 0| + |4 - 0| = 8, not reachable.

Among the reachable towers, the maximum quality factor is 4. Both [1, 3] and [2, 2] have the same quality, so the lexicographically smaller coordinate is [1, 3].

Example 3:

Input: towers = [[5,6,8], [0,3,5]], center = [1,2], radius = 1

Output: [-1,-1]

Explanation:

Tower [5, 6, 8]: Manhattan distance = |5 - 1| + |6 - 2| = 8, not reachable.
Tower [0, 3, 5]: Manhattan distance = |0 - 1| + |3 - 2| = 2, not reachable.

No tower is reachable within the given radius, so [-1, -1] is returned.

Constraints:

1 <= towers.length <= 10⁵
towers[i] = [x_i, y_i, q_i]
center = [cx, cy]
0 <= x_i, y_i, q_i, cx, cy <= 10⁵
0 <= radius <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array towers, where towers[i] = [xi, yi, qi] represents the coordinates (xi, yi) and quality factor qi of the ith tower. You are also given an integer array center = [cx, cy] representing your location, and an integer radius. A tower is reachable if its Manhattan distance from center is less than or equal to radius. Among all reachable towers: Return the coordinates of the tower with the maximum quality factor. If there is a tie, return the tower with the lexicographically smallest coordinate. If no tower is reachable, return [-1, -1]. The Manhattan Distance between two cells (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|. A coordinate [xi, yi] is lexicographically smaller than [xj, yj] if xi < xj, or xi == xj and yi < yj. |x| denotes the absolute value of x.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2,5],[2,1,7],[3,1,9]]
[1,1]
2

Example 2

[[1,3,4],[2,2,4],[4,4,7]]
[0,0]
5

Example 3

[[5,6,8],[0,3,5]]
[1,2]
1

Step 02

Core Insight

What unlocks the optimal approach

Iterate through all towers and compute the Manhattan distance <code>abs(xi - cx) + abs(yi - cy)</code>.
Consider only towers with distance less than or equal to <code>radius</code>.
Track the reachable tower with the highest <code>qi</code>; break ties using lexicographical order on <code>[xi, yi]</code>.
If no tower is reachable, return <code>[-1, -1]</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 #3809: Best Reachable Tower
class Solution {
    public int[] bestTower(int[][] towers, int[] center, int radius) {
        int cx = center[0], cy = center[1];
        int idx = -1;
        for (int i = 0; i < towers.length; i++) {
            int x = towers[i][0], y = towers[i][1], q = towers[i][2];
            int dist = Math.abs(x - cx) + Math.abs(y - cy);
            if (dist > radius) {
                continue;
            }
            if (idx == -1 || towers[idx][2] < q
                || (towers[idx][2] == q
                    && (x < towers[idx][0] || (x == towers[idx][0] && y < towers[idx][1])))) {
                idx = i;
            }
        }
        return idx == -1 ? new int[] {-1, -1} : new int[] {towers[idx][0], towers[idx][1]};
    }
}

// Accepted solution for LeetCode #3809: Best Reachable Tower
func bestTower(towers [][]int, center []int, radius int) []int {
	cx, cy := center[0], center[1]
	idx := -1
	for i, a := range towers {
		x, y, q := a[0], a[1], a[2]
		dist := abs(x-cx) + abs(y-cy)
		if dist > radius {
			continue
		}
		if idx == -1 ||
			towers[idx][2] < q ||
			(towers[idx][2] == q &&
				(x < towers[idx][0] ||
					(x == towers[idx][0] && y < towers[idx][1]))) {
			idx = i
		}
	}
	if idx == -1 {
		return []int{-1, -1}
	}
	return []int{towers[idx][0], towers[idx][1]}
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3809: Best Reachable Tower
class Solution:
    def bestTower(
        self, towers: List[List[int]], center: List[int], radius: int
    ) -> List[int]:
        cx, cy = center
        idx = -1
        for i, (x, y, q) in enumerate(towers):
            dist = abs(x - cx) + abs(y - cy)
            if dist > radius:
                continue
            if (
                idx == -1
                or towers[idx][2] < q
                or (towers[idx][2] == q and towers[i][:2] < towers[idx][:2])
            ):
                idx = i
        return [-1, -1] if idx == -1 else towers[idx][:2]

// Accepted solution for LeetCode #3809: Best Reachable Tower
// 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 #3809: Best Reachable Tower
// class Solution {
//     public int[] bestTower(int[][] towers, int[] center, int radius) {
//         int cx = center[0], cy = center[1];
//         int idx = -1;
//         for (int i = 0; i < towers.length; i++) {
//             int x = towers[i][0], y = towers[i][1], q = towers[i][2];
//             int dist = Math.abs(x - cx) + Math.abs(y - cy);
//             if (dist > radius) {
//                 continue;
//             }
//             if (idx == -1 || towers[idx][2] < q
//                 || (towers[idx][2] == q
//                     && (x < towers[idx][0] || (x == towers[idx][0] && y < towers[idx][1])))) {
//                 idx = i;
//             }
//         }
//         return idx == -1 ? new int[] {-1, -1} : new int[] {towers[idx][0], towers[idx][1]};
//     }
// }

// Accepted solution for LeetCode #3809: Best Reachable Tower
function bestTower(towers: number[][], center: number[], radius: number): number[] {
    const [cx, cy] = center;
    let idx = -1;
    for (let i = 0; i < towers.length; i++) {
        const [x, y, q] = towers[i];
        const dist = Math.abs(x - cx) + Math.abs(y - cy);
        if (dist > radius) {
            continue;
        }
        if (
            idx === -1 ||
            towers[idx][2] < q ||
            (towers[idx][2] === q &&
                (x < towers[idx][0] || (x === towers[idx][0] && y < towers[idx][1])))
        ) {
            idx = i;
        }
    }
    return idx === -1 ? [-1, -1] : [towers[idx][0], towers[idx][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(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.