LeetCode #3661 — HARD

Maximum Walls Destroyed by Robots

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

The Problem

Problem Statement

There is an endless straight line populated with some robots and walls. You are given integer arrays robots, distance, and walls:

robots[i] is the position of the i^th robot.
distance[i] is the maximum distance the i^th robot's bullet can travel.
walls[j] is the position of the j^th wall.

Every robot has one bullet that can either fire to the left or the right at most distance[i] meters.

A bullet destroys every wall in its path that lies within its range. Robots are fixed obstacles: if a bullet hits another robot before reaching a wall, it immediately stops at that robot and cannot continue.

Return the maximum number of unique walls that can be destroyed by the robots.

Notes:

A wall and a robot may share the same position; the wall can be destroyed by the robot at that position.
Robots are not destroyed by bullets.

Example 1:

Input: robots = [4], distance = [3], walls = [1,10]

Output: 1

Explanation:

robots[0] = 4 fires left with distance[0] = 3, covering [1, 4] and destroys walls[0] = 1.
Thus, the answer is 1.

Example 2:

Input: robots = [10,2], distance = [5,1], walls = [5,2,7]

Output: 3

Explanation:

robots[0] = 10 fires left with distance[0] = 5, covering [5, 10] and destroys walls[0] = 5 and walls[2] = 7.
robots[1] = 2 fires left with distance[1] = 1, covering [1, 2] and destroys walls[1] = 2.
Thus, the answer is 3.

Example 3:

Input: robots = [1,2], distance = [100,1], walls = [10]

Output: 0

Explanation:

In this example, only robots[0] can reach the wall, but its shot to the right is blocked by robots[1]; thus the answer is 0.

Constraints:

1 <= robots.length == distance.length <= 10⁵
1 <= walls.length <= 10⁵
1 <= robots[i], walls[j] <= 10⁹
1 <= distance[i] <= 10⁵
All values in robots are unique
All values in walls are unique

Step 01

Brute Force Baseline

Problem summary: There is an endless straight line populated with some robots and walls. You are given integer arrays robots, distance, and walls: robots[i] is the position of the ith robot. distance[i] is the maximum distance the ith robot's bullet can travel. walls[j] is the position of the jth wall. Every robot has one bullet that can either fire to the left or the right at most distance[i] meters. A bullet destroys every wall in its path that lies within its range. Robots are fixed obstacles: if a bullet hits another robot before reaching a wall, it immediately stops at that robot and cannot continue. Return the maximum number of unique walls that can be destroyed by the robots. Notes: A wall and a robot may share the same position; the wall can be destroyed by the robot at that position. Robots are not destroyed by bullets.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[4]
[3]
[1,10]

Example 2

[10,2]
[5,1]
[5,2,7]

Example 3

[1,2]
[100,1]
[10]

Step 02

Core Insight

What unlocks the optimal approach

Sort both the robots and walls arrays. This will help in efficiently processing positions and performing range queries.
Each robot can shoot either left or right. However, if a robot fires and another robot is in its path, the bullet stops. You need to use the positions of neighboring robots to limit the shooting range.
Use binary search (lower_bound and upper_bound) to count how many walls fall within a certain range.
You can use dynamic programming to keep track of the maximum number of walls destroyed so far, depending on the direction the previous robot shot.

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

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 #3661: Maximum Walls Destroyed by Robots
class Solution {
    private Integer[][] f;
    private int[][] arr;
    private int[] walls;
    private int n;

    public int maxWalls(int[] robots, int[] distance, int[] walls) {
        n = robots.length;
        arr = new int[n][2];
        for (int i = 0; i < n; i++) {
            arr[i][0] = robots[i];
            arr[i][1] = distance[i];
        }
        Arrays.sort(arr, Comparator.comparingInt(a -> a[0]));
        Arrays.sort(walls);
        this.walls = walls;
        f = new Integer[n][2];
        return dfs(n - 1, 1);
    }

    private int dfs(int i, int j) {
        if (i < 0) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }

        int left = arr[i][0] - arr[i][1];
        if (i > 0) {
            left = Math.max(left, arr[i - 1][0] + 1);
        }
        int l = lowerBound(walls, left);
        int r = lowerBound(walls, arr[i][0] + 1);
        int ans = dfs(i - 1, 0) + (r - l);

        int right = arr[i][0] + arr[i][1];
        if (i + 1 < n) {
            if (j == 0) {
                right = Math.min(right, arr[i + 1][0] - arr[i + 1][1] - 1);
            } else {
                right = Math.min(right, arr[i + 1][0] - 1);
            }
        }
        l = lowerBound(walls, arr[i][0]);
        r = lowerBound(walls, right + 1);
        ans = Math.max(ans, dfs(i - 1, 1) + (r - l));
        return f[i][j] = ans;
    }

    private int lowerBound(int[] arr, int target) {
        int idx = Arrays.binarySearch(arr, target);
        if (idx < 0) {
            return -idx - 1;
        }
        return idx;
    }
}

// Accepted solution for LeetCode #3661: Maximum Walls Destroyed by Robots
func maxWalls(robots []int, distance []int, walls []int) int {
	type pair struct {
		x, d int
	}
	n := len(robots)
	arr := make([]pair, n)
	for i := 0; i < n; i++ {
		arr[i] = pair{robots[i], distance[i]}
	}
	sort.Slice(arr, func(i, j int) bool {
		return arr[i].x < arr[j].x
	})
	sort.Ints(walls)

	f := make(map[[2]int]int)

	var dfs func(int, int) int
	dfs = func(i, j int) int {
		if i < 0 {
			return 0
		}
		key := [2]int{i, j}
		if v, ok := f[key]; ok {
			return v
		}

		left := arr[i].x - arr[i].d
		if i > 0 {
			left = max(left, arr[i-1].x+1)
		}
		l := sort.SearchInts(walls, left)
		r := sort.SearchInts(walls, arr[i].x+1)
		ans := dfs(i-1, 0) + (r - l)

		right := arr[i].x + arr[i].d
		if i+1 < n {
			if j == 0 {
				right = min(right, arr[i+1].x-arr[i+1].d-1)
			} else {
				right = min(right, arr[i+1].x-1)
			}
		}
		l = sort.SearchInts(walls, arr[i].x)
		r = sort.SearchInts(walls, right+1)
		ans = max(ans, dfs(i-1, 1)+(r-l))

		f[key] = ans
		return ans
	}

	return dfs(n-1, 1)
}

# Accepted solution for LeetCode #3661: Maximum Walls Destroyed by Robots
class Solution:
    def maxWalls(self, robots: List[int], distance: List[int], walls: List[int]) -> int:
        n = len(robots)
        arr = sorted(zip(robots, distance), key=lambda x: x[0])
        walls.sort()

        @cache
        def dfs(i: int, j: int) -> int:
            if i < 0:
                return 0
            left = arr[i][0] - arr[i][1]
            if i > 0:
                left = max(left, arr[i - 1][0] + 1)
            l = bisect_left(walls, left)
            r = bisect_left(walls, arr[i][0] + 1)
            ans = dfs(i - 1, 0) + r - l
            right = arr[i][0] + arr[i][1]
            if i + 1 < n:
                if j == 0:
                    right = min(right, arr[i + 1][0] - arr[i + 1][1] - 1)
                else:
                    right = min(right, arr[i + 1][0] - 1)
            l = bisect_left(walls, arr[i][0])
            r = bisect_left(walls, right + 1)
            ans = max(ans, dfs(i - 1, 1) + r - l)
            return ans

        return dfs(n - 1, 1)

// Accepted solution for LeetCode #3661: Maximum Walls Destroyed by Robots
// 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 #3661: Maximum Walls Destroyed by Robots
// class Solution {
//     private Integer[][] f;
//     private int[][] arr;
//     private int[] walls;
//     private int n;
// 
//     public int maxWalls(int[] robots, int[] distance, int[] walls) {
//         n = robots.length;
//         arr = new int[n][2];
//         for (int i = 0; i < n; i++) {
//             arr[i][0] = robots[i];
//             arr[i][1] = distance[i];
//         }
//         Arrays.sort(arr, Comparator.comparingInt(a -> a[0]));
//         Arrays.sort(walls);
//         this.walls = walls;
//         f = new Integer[n][2];
//         return dfs(n - 1, 1);
//     }
// 
//     private int dfs(int i, int j) {
//         if (i < 0) {
//             return 0;
//         }
//         if (f[i][j] != null) {
//             return f[i][j];
//         }
// 
//         int left = arr[i][0] - arr[i][1];
//         if (i > 0) {
//             left = Math.max(left, arr[i - 1][0] + 1);
//         }
//         int l = lowerBound(walls, left);
//         int r = lowerBound(walls, arr[i][0] + 1);
//         int ans = dfs(i - 1, 0) + (r - l);
// 
//         int right = arr[i][0] + arr[i][1];
//         if (i + 1 < n) {
//             if (j == 0) {
//                 right = Math.min(right, arr[i + 1][0] - arr[i + 1][1] - 1);
//             } else {
//                 right = Math.min(right, arr[i + 1][0] - 1);
//             }
//         }
//         l = lowerBound(walls, arr[i][0]);
//         r = lowerBound(walls, right + 1);
//         ans = Math.max(ans, dfs(i - 1, 1) + (r - l));
//         return f[i][j] = ans;
//     }
// 
//     private int lowerBound(int[] arr, int target) {
//         int idx = Arrays.binarySearch(arr, target);
//         if (idx < 0) {
//             return -idx - 1;
//         }
//         return idx;
//     }
// }

// Accepted solution for LeetCode #3661: Maximum Walls Destroyed by Robots
function maxWalls(robots: number[], distance: number[], walls: number[]): number {
    type Pair = [number, number];
    const n = robots.length;
    const arr: Pair[] = robots.map((r, i) => [r, distance[i]]);

    _.sortBy(arr, p => p[0]).forEach((p, i) => (arr[i] = p));
    walls.sort((a, b) => a - b);
    const f: number[][] = Array.from({ length: n }, () => Array(2).fill(-1));

    function dfs(i: number, j: number): number {
        if (i < 0) {
            return 0;
        }
        if (f[i][j] !== -1) {
            return f[i][j];
        }

        let left = arr[i][0] - arr[i][1];
        if (i > 0) left = Math.max(left, arr[i - 1][0] + 1);
        let l = _.sortedIndex(walls, left);
        let r = _.sortedIndex(walls, arr[i][0] + 1);
        let ans = dfs(i - 1, 0) + (r - l);

        let right = arr[i][0] + arr[i][1];
        if (i + 1 < n) {
            if (j === 0) {
                right = Math.min(right, arr[i + 1][0] - arr[i + 1][1] - 1);
            } else {
                right = Math.min(right, arr[i + 1][0] - 1);
            }
        }
        l = _.sortedIndex(walls, arr[i][0]);
        r = _.sortedIndex(walls, right + 1);
        ans = Math.max(ans, dfs(i - 1, 1) + (r - l));

        f[i][j] = ans;
        return ans;
    }

    return dfs(n - 1, 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

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.