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3661. Maximum Walls Destroyed by Robots

Hard
Leetcode Link

Problem Description

You have an endless straight line with robots and walls positioned on it. Each robot can fire exactly one bullet either to the left or right direction.

Given:

  • robots[i]: the position of the i-th robot
  • distance[i]: the maximum distance the i-th robot's bullet can travel
  • walls[j]: the position of the j-th wall

Rules for bullets:

  • Each robot fires exactly one bullet in either left or right direction
  • A bullet can travel at most distance[i] meters from robot i
  • A bullet destroys all walls in its path within its range
  • If a bullet hits another robot, it stops immediately and cannot continue
  • Robots are never destroyed by bullets
  • A wall and robot can share the same position (the robot at that position can destroy that wall)

The goal is to find the maximum number of unique walls that can be destroyed by having all robots fire their bullets optimally.

For example, if robot at position 5 has distance 3:

  • Firing left: bullet travels from position 5 to position 2, destroying all walls in range [2, 5]
  • Firing right: bullet travels from position 5 to position 8, destroying all walls in range [5, 8]
  • If another robot is at position 7, a right-firing bullet would stop at position 7 (just before the robot)

The challenge is to determine the optimal firing direction for each robot to maximize the total number of unique walls destroyed across all robots.

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Intuition

The key insight is that we need to consider robots in a specific order to handle the blocking constraints properly. Since robots can block each other's bullets, the firing direction of one robot affects what walls other robots can destroy.

Let's think about this step by step. If we process robots from left to right on the line, we can make optimal decisions based on what previous robots have done. When a robot fires left, it might be blocked by robots to its left. When it fires right, it might block future robots from firing left.

This suggests a dynamic programming approach where we track:

  1. Which robot we're currently considering
  2. The firing direction of the next robot (which affects our current robot's right range)

Why does the next robot's direction matter? If the next robot fires left, its bullet might be blocked by our current robot if we're in its path. So when our current robot fires right, we need to limit our range to avoid blocking the next robot's left-firing bullet.

The state dfs(i, j) represents:

  • i: index of current robot being considered (after sorting by position)
  • j: firing direction of the next robot (0 = left, 1 = right)

For each robot, we calculate two possibilities:

  1. Fire left: The range is [robot_position - distance, robot_position], but we must also consider blocking by the previous robot (if it exists and fired right)
  2. Fire right: The range is [robot_position, robot_position + distance], but we must consider not blocking the next robot's potential left shot

By processing robots in sorted order and using memoization to avoid recalculating the same states, we can efficiently find the maximum number of walls that can be destroyed. The binary search helps us quickly count how many walls fall within each robot's firing range.

The recursion naturally handles the dependencies between robots' decisions, ensuring we find the globally optimal solution rather than getting stuck in local optima.

Solution Approach

First, we prepare the data by combining robots with their distances and sorting them by position. We also sort the walls array for efficient binary search operations later.

arr = sorted(zip(robots, distance), key=lambda x: x[0])
walls.sort()

The core of the solution is the memoized recursive function dfs(i, j):

  • i: index of the current robot being considered (0 to n-1)
  • j: firing direction of the next robot (0 = left, 1 = right)

Base Case: When i < 0, all robots have been processed, return 0.

For each robot at index i, we consider two firing directions:

1. Fire Left:

  • Calculate the left boundary: left = arr[i][0] - arr[i][1] (robot position minus its range)
  • If there's a previous robot (i > 0), adjust the left boundary to avoid counting walls already destroyed: left = max(left, arr[i-1][0] + 1)
  • Use binary search to find walls in range [left, arr[i][0]):
    • l = bisect_left(walls, left) finds the first wall at or after left
    • r = bisect_left(walls, arr[i][0] + 1) finds the first wall after robot position
    • Number of walls destroyed: r - l
  • Total walls destroyed: dfs(i - 1, 0) + (r - l)

2. Fire Right:

  • Calculate the right boundary: right = arr[i][0] + arr[i][1] (robot position plus its range)
  • If there's a next robot (i + 1 < n), adjust the right boundary based on the next robot's firing direction:
    • If j == 0 (next robot fires left): right = min(right, arr[i+1][0] - arr[i+1][1] - 1) to avoid blocking the next robot's left shot
    • If j == 1 (next robot fires right): right = min(right, arr[i+1][0] - 1) to stop before the next robot
  • Use binary search to find walls in range [arr[i][0], right]:
    • l = bisect_left(walls, arr[i][0]) finds the first wall at or after robot position
    • r = bisect_left(walls, right + 1) finds the first wall after right
    • Number of walls destroyed: r - l
  • Total walls destroyed: dfs(i - 1, 1) + (r - l)

The function returns the maximum of these two choices: max(fire_left_result, fire_right_result)

The @cache decorator provides memoization, storing results for previously computed states to avoid redundant calculations.

Finally, we start the recursion with dfs(n - 1, 1), processing from the rightmost robot. The second parameter can be either 0 or 1 since there's no robot after the last one.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Input:

  • robots = [3, 6] (robots at positions 3 and 6)
  • distance = [2, 3] (robot at position 3 can shoot 2 meters, robot at position 6 can shoot 3 meters)
  • walls = [1, 2, 4, 5, 7, 8] (walls at these positions)

Step 1: Prepare the data

  • Sort robots with distances: arr = [(3, 2), (6, 3)] (already sorted)
  • Sort walls: walls = [1, 2, 4, 5, 7, 8] (already sorted)

Step 2: Start recursion with dfs(1, 1) (rightmost robot, index 1)

We're at robot index 1 (position 6, distance 3). We consider two options:

Option A - Fire Left:

  • Left range: [6 - 3, 6] = [3, 6]
  • Since there's a previous robot at position 3, adjust: left = max(3, 3 + 1) = 4
  • Actual range: [4, 6]
  • Binary search for walls in [4, 6]:
    • bisect_left(walls, 4) = 2 (index of wall 4)
    • bisect_left(walls, 7) = 4 (index after wall 5)
    • Walls destroyed: walls at positions 4, 5 (count = 2)
  • Recurse: dfs(0, 0) + 2

Option B - Fire Right:

  • Right range: [6, 6 + 3] = [6, 9]
  • No next robot, so range stays [6, 9]
  • Binary search for walls in [6, 9]:
    • bisect_left(walls, 6) = 3 (index after wall 5)
    • bisect_left(walls, 10) = 6 (end of array)
    • Walls destroyed: walls at positions 7, 8 (count = 2)
  • Recurse: dfs(0, 1) + 2

Step 3: Evaluate dfs(0, 0) - Robot at position 3, next robot fires left

Option A - Fire Left:

  • Left range: [3 - 2, 3] = [1, 3]
  • No previous robot, so range stays [1, 3]
  • Binary search for walls in [1, 3]:
    • bisect_left(walls, 1) = 0
    • bisect_left(walls, 4) = 2
    • Walls destroyed: walls at positions 1, 2 (count = 2)
  • Base case: dfs(-1, 0) = 0
  • Total: 0 + 2 = 2

Option B - Fire Right:

  • Right range: [3, 3 + 2] = [3, 5]
  • Next robot (at position 6) fires left with range 3, so its left boundary is 3
  • Adjust: right = min(5, 3 - 1) = 2 (to avoid blocking next robot's left shot)
  • Since right < 3, no walls can be destroyed in range [3, 2]
  • Total: dfs(-1, 1) + 0 = 0

Result for dfs(0, 0): max(2, 0) = 2

Step 4: Evaluate dfs(0, 1) - Robot at position 3, next robot fires right

Option A - Fire Left:

  • Same as before: destroys walls at positions 1, 2
  • Total: 2

Option B - Fire Right:

  • Right range: [3, 5]
  • Next robot fires right, so adjust: right = min(5, 6 - 1) = 5
  • Binary search for walls in [3, 5]:
    • bisect_left(walls, 3) = 2
    • bisect_left(walls, 6) = 3
    • Walls destroyed: walls at positions 4, 5 (count = 2)
  • Total: 0 + 2 = 2

Result for dfs(0, 1): max(2, 2) = 2

Step 5: Back to dfs(1, 1)

  • Fire Left: dfs(0, 0) + 2 = 2 + 2 = 4
  • Fire Right: dfs(0, 1) + 2 = 2 + 2 = 4

Final Result: max(4, 4) = 4

The optimal solution destroys 4 walls total. One valid configuration:

  • Robot at position 3 fires left, destroying walls at positions 1, 2
  • Robot at position 6 fires right, destroying walls at positions 7, 8

Solution Implementation

1class Solution:
2    def maxWalls(self, robots: List[int], distance: List[int], walls: List[int]) -> int:
3        """
4        Find the maximum number of walls that can be covered by robots.
5        Each robot at position robots[i] can move distance[i] units left or right.
6      
7        Args:
8            robots: List of robot positions
9            distance: List of maximum distances each robot can move
10            walls: List of wall positions
11      
12        Returns:
13            Maximum number of walls that can be covered
14        """
15        n = len(robots)
16      
17        # Sort robots by position and pair with their distances
18        robot_distance_pairs = sorted(zip(robots, distance), key=lambda x: x[0])
19      
20        # Sort walls for binary search
21        walls.sort()
22      
23        @cache
24        def dp(robot_idx: int, prev_robot_moved_right: int) -> int:
25            """
26            Dynamic programming function to find max walls covered.
27          
28            Args:
29                robot_idx: Current robot index being processed
30                prev_robot_moved_right: 0 if previous robot moved left, 1 if moved right
31          
32            Returns:
33                Maximum walls that can be covered from robots 0 to robot_idx
34            """
35            # Base case: no robots left to process
36            if robot_idx < 0:
37                return 0
38          
39            current_pos = robot_distance_pairs[robot_idx][0]
40            current_distance = robot_distance_pairs[robot_idx][1]
41          
42            # Option 1: Current robot moves left
43            left_boundary = current_pos - current_distance
44          
45            # Ensure no overlap with previous robot if it exists
46            if robot_idx > 0:
47                prev_pos = robot_distance_pairs[robot_idx - 1][0]
48                left_boundary = max(left_boundary, prev_pos + 1)
49          
50            # Count walls covered when moving left
51            left_wall_idx = bisect_left(walls, left_boundary)
52            right_wall_idx = bisect_left(walls, current_pos + 1)
53            walls_covered_left = dp(robot_idx - 1, 0) + right_wall_idx - left_wall_idx
54          
55            # Option 2: Current robot moves right
56            right_boundary = current_pos + current_distance
57          
58            # Ensure no overlap with next robot if it exists
59            if robot_idx + 1 < n:
60                next_pos = robot_distance_pairs[robot_idx + 1][0]
61                next_distance = robot_distance_pairs[robot_idx + 1][1]
62              
63                if prev_robot_moved_right == 0:
64                    # Next robot can move left, so adjust boundary
65                    right_boundary = min(right_boundary, next_pos - next_distance - 1)
66                else:
67                    # Next robot cannot move left, must stay at position or move right
68                    right_boundary = min(right_boundary, next_pos - 1)
69          
70            # Count walls covered when moving right
71            left_wall_idx = bisect_left(walls, current_pos)
72            right_wall_idx = bisect_left(walls, right_boundary + 1)
73            walls_covered_right = dp(robot_idx - 1, 1) + right_wall_idx - left_wall_idx
74          
75            # Return maximum of both options
76            return max(walls_covered_left, walls_covered_right)
77      
78        # Start from the last robot, assuming it can move right
79        return dp(n - 1, 1)
80
1class Solution {
2    // Memoization table: dp[i][j] stores max walls for robots 0 to i
3    // j = 0: robot i pushed left, j = 1: robot i pushed right
4    private Integer[][] dp;
5  
6    // Array to store robot positions and push distances
7    // robotData[i][0] = position, robotData[i][1] = push distance
8    private int[][] robotData;
9  
10    // Sorted array of wall positions
11    private int[] walls;
12  
13    // Number of robots
14    private int robotCount;
15
16    public int maxWalls(int[] robots, int[] distance, int[] walls) {
17        robotCount = robots.length;
18      
19        // Initialize robot data array with positions and distances
20        robotData = new int[robotCount][2];
21        for (int i = 0; i < robotCount; i++) {
22            robotData[i][0] = robots[i];
23            robotData[i][1] = distance[i];
24        }
25      
26        // Sort robots by position (left to right)
27        Arrays.sort(robotData, Comparator.comparingInt(a -> a[0]));
28      
29        // Sort walls for binary search
30        Arrays.sort(walls);
31        this.walls = walls;
32      
33        // Initialize memoization table
34        dp = new Integer[robotCount][2];
35      
36        // Start DFS from last robot, initially considering it pushed right
37        return dfs(robotCount - 1, 1);
38    }
39
40    /**
41     * Dynamic programming function to find max walls covered
42     * @param robotIndex - current robot index (0 to n-1)
43     * @param nextRobotDirection - direction of next robot (i+1)
44     *                            0: next robot pushed left
45     *                            1: next robot pushed right
46     * @return maximum walls that can be covered from robots 0 to robotIndex
47     */
48    private int dfs(int robotIndex, int nextRobotDirection) {
49        // Base case: no more robots to process
50        if (robotIndex < 0) {
51            return 0;
52        }
53      
54        // Return memoized result if available
55        if (dp[robotIndex][nextRobotDirection] != null) {
56            return dp[robotIndex][nextRobotDirection];
57        }
58
59        // Option 1: Push current robot to the left
60        int leftBoundary = robotData[robotIndex][0] - robotData[robotIndex][1];
61      
62        // Ensure no overlap with previous robot (if exists)
63        if (robotIndex > 0) {
64            leftBoundary = Math.max(leftBoundary, robotData[robotIndex - 1][0] + 1);
65        }
66      
67        // Count walls in range [leftBoundary, currentPosition)
68        int leftIndex = lowerBound(walls, leftBoundary);
69        int rightIndex = lowerBound(walls, robotData[robotIndex][0] + 1);
70        int wallsIfPushedLeft = dfs(robotIndex - 1, 0) + (rightIndex - leftIndex);
71
72        // Option 2: Push current robot to the right
73        int rightBoundary = robotData[robotIndex][0] + robotData[robotIndex][1];
74      
75        // Ensure no overlap with next robot (if exists)
76        if (robotIndex + 1 < robotCount) {
77            if (nextRobotDirection == 0) {
78                // Next robot is pushed left, so its left boundary is a constraint
79                rightBoundary = Math.min(rightBoundary, 
80                    robotData[robotIndex + 1][0] - robotData[robotIndex + 1][1] - 1);
81            } else {
82                // Next robot is pushed right, so only its position matters
83                rightBoundary = Math.min(rightBoundary, 
84                    robotData[robotIndex + 1][0] - 1);
85            }
86        }
87      
88        // Count walls in range [currentPosition, rightBoundary]
89        leftIndex = lowerBound(walls, robotData[robotIndex][0]);
90        rightIndex = lowerBound(walls, rightBoundary + 1);
91        int wallsIfPushedRight = dfs(robotIndex - 1, 1) + (rightIndex - leftIndex);
92      
93        // Choose the option that covers more walls
94        int maxWalls = Math.max(wallsIfPushedLeft, wallsIfPushedRight);
95      
96        // Memoize and return result
97        dp[robotIndex][nextRobotDirection] = maxWalls;
98        return maxWalls;
99    }
100
101    /**
102     * Binary search to find the leftmost position >= target
103     * @param arr - sorted array to search
104     * @param target - target value to find
105     * @return index of first element >= target
106     */
107    private int lowerBound(int[] arr, int target) {
108        int index = Arrays.binarySearch(arr, target);
109      
110        // If exact match found, return its index
111        // If not found, binarySearch returns -(insertion point) - 1
112        // So we convert it back to insertion point
113        if (index < 0) {
114            return -index - 1;
115        }
116        return index;
117    }
118}
119
1class Solution {
2public:
3    int maxWalls(vector<int>& robots, vector<int>& distance, vector<int>& walls) {
4        int n = robots.size();
5      
6        // Create pairs of (robot_position, robot_distance) for easier handling
7        vector<pair<int, int>> robotInfo(n);
8        for (int i = 0; i < n; i++) {
9            robotInfo[i] = {robots[i], distance[i]};
10        }
11      
12        // Sort robots by their positions
13        ranges::sort(robotInfo, {}, &pair<int, int>::first);
14      
15        // Sort walls for binary search
16        ranges::sort(walls);
17      
18        // DP memoization table
19        // dp[i][direction]: maximum walls destroyed considering robots 0 to i
20        // direction = 0: robot i moves left, direction = 1: robot i moves right
21        vector<vector<int>> dp(n, vector<int>(2, -1));
22      
23        // Recursive function with memoization
24        auto solve = [&](this auto&& solve, int robotIdx, int prevDirection) -> int {
25            // Base case: no robots left to consider
26            if (robotIdx < 0) {
27                return 0;
28            }
29          
30            // Return memoized result if already computed
31            if (dp[robotIdx][prevDirection] != -1) {
32                return dp[robotIdx][prevDirection];
33            }
34          
35            int maxDestroyed = 0;
36          
37            // Option 1: Current robot moves LEFT
38            // Calculate the leftmost position this robot can reach
39            int leftBound = robotInfo[robotIdx].first - robotInfo[robotIdx].second;
40          
41            // Ensure no overlap with previous robot (if exists)
42            if (robotIdx > 0) {
43                leftBound = max(leftBound, robotInfo[robotIdx - 1].first + 1);
44            }
45          
46            // Count walls that can be destroyed when moving left
47            int leftWallIdx = ranges::lower_bound(walls, leftBound) - walls.begin();
48            int rightWallIdx = ranges::lower_bound(walls, robotInfo[robotIdx].first + 1) - walls.begin();
49            int wallsDestroyedLeft = rightWallIdx - leftWallIdx;
50          
51            // Recurse for previous robots (current robot chose left, so prev can choose any direction)
52            maxDestroyed = solve(robotIdx - 1, 0) + wallsDestroyedLeft;
53          
54            // Option 2: Current robot moves RIGHT
55            // Calculate the rightmost position this robot can reach
56            int rightBound = robotInfo[robotIdx].first + robotInfo[robotIdx].second;
57          
58            // Ensure no overlap with next robot (if exists)
59            if (robotIdx + 1 < n) {
60                if (prevDirection == 0) {
61                    // If next robot will move left, consider its leftmost reach
62                    rightBound = min(rightBound, robotInfo[robotIdx + 1].first - robotInfo[robotIdx + 1].second - 1);
63                } else {
64                    // If next robot will move right, just avoid its starting position
65                    rightBound = min(rightBound, robotInfo[robotIdx + 1].first - 1);
66                }
67            }
68          
69            // Count walls that can be destroyed when moving right
70            leftWallIdx = ranges::lower_bound(walls, robotInfo[robotIdx].first) - walls.begin();
71            rightWallIdx = ranges::lower_bound(walls, rightBound + 1) - walls.begin();
72            int wallsDestroyedRight = rightWallIdx - leftWallIdx;
73          
74            // Recurse for previous robots (current robot chose right, so pass direction = 1)
75            maxDestroyed = max(maxDestroyed, solve(robotIdx - 1, 1) + wallsDestroyedRight);
76          
77            // Store and return the result
78            dp[robotIdx][prevDirection] = maxDestroyed;
79            return maxDestroyed;
80        };
81      
82        // Start solving from the last robot, assuming it moves right
83        // (The last robot's direction doesn't affect next robots since there are none)
84        return solve(n - 1, 1);
85    }
86};
87
1/**
2 * Calculates the maximum number of walls that can be covered by robots
3 * @param robots - Array of robot positions
4 * @param distance - Array of maximum distances each robot can move
5 * @param walls - Array of wall positions
6 * @returns Maximum number of walls that can be covered
7 */
8function maxWalls(robots: number[], distance: number[], walls: number[]): number {
9    // Type alias for robot position and distance pair
10    type RobotInfo = [number, number];
11  
12    const robotCount = robots.length;
13  
14    // Create pairs of [robot position, max distance] and sort by position
15    const robotInfoArray: RobotInfo[] = robots.map((position, index) => [position, distance[index]]);
16    robotInfoArray.sort((a, b) => a[0] - b[0]);
17  
18    // Sort walls in ascending order
19    walls.sort((a, b) => a - b);
20  
21    // DP memoization table: dp[i][j] where i is robot index, j is direction (0=left, 1=right)
22    const dp: number[][] = Array.from({ length: robotCount }, () => Array(2).fill(-1));
23  
24    /**
25     * Dynamic programming function to find maximum walls covered
26     * @param robotIndex - Current robot index
27     * @param direction - Direction of next robot (0 = next robot moves left, 1 = next robot moves right)
28     * @returns Maximum walls that can be covered from current state
29     */
30    function dfs(robotIndex: number, direction: number): number {
31        // Base case: no more robots to process
32        if (robotIndex < 0) {
33            return 0;
34        }
35      
36        // Return memoized result if already computed
37        if (dp[robotIndex][direction] !== -1) {
38            return dp[robotIndex][direction];
39        }
40      
41        // Option 1: Current robot moves left
42        // Calculate leftmost position this robot can reach
43        let leftBoundary = robotInfoArray[robotIndex][0] - robotInfoArray[robotIndex][1];
44        // Ensure no overlap with previous robot
45        if (robotIndex > 0) {
46            leftBoundary = Math.max(leftBoundary, robotInfoArray[robotIndex - 1][0] + 1);
47        }
48      
49        // Count walls covered when moving left using binary search
50        const leftStartIndex = binarySearch(walls, leftBoundary);
51        const leftEndIndex = binarySearch(walls, robotInfoArray[robotIndex][0] + 1);
52        const wallsCoveredLeft = dfs(robotIndex - 1, 0) + (leftEndIndex - leftStartIndex);
53      
54        // Option 2: Current robot moves right
55        // Calculate rightmost position this robot can reach
56        let rightBoundary = robotInfoArray[robotIndex][0] + robotInfoArray[robotIndex][1];
57        // Ensure no overlap with next robot
58        if (robotIndex + 1 < robotCount) {
59            if (direction === 0) {
60                // Next robot will move left
61                rightBoundary = Math.min(rightBoundary, robotInfoArray[robotIndex + 1][0] - robotInfoArray[robotIndex + 1][1] - 1);
62            } else {
63                // Next robot will move right
64                rightBoundary = Math.min(rightBoundary, robotInfoArray[robotIndex + 1][0] - 1);
65            }
66        }
67      
68        // Count walls covered when moving right using binary search
69        const rightStartIndex = binarySearch(walls, robotInfoArray[robotIndex][0]);
70        const rightEndIndex = binarySearch(walls, rightBoundary + 1);
71        const wallsCoveredRight = dfs(robotIndex - 1, 1) + (rightEndIndex - rightStartIndex);
72      
73        // Take maximum of both options
74        const maxWallsCovered = Math.max(wallsCoveredLeft, wallsCoveredRight);
75      
76        // Memoize and return result
77        dp[robotIndex][direction] = maxWallsCovered;
78        return maxWallsCovered;
79    }
80  
81    // Start from last robot, assuming it moves right
82    return dfs(robotCount - 1, 1);
83}
84
85/**
86 * Binary search to find insertion index in sorted array
87 * @param array - Sorted array to search in
88 * @param value - Value to find insertion index for
89 * @returns Index where value should be inserted to maintain sorted order
90 */
91function binarySearch(array: number[], value: number): number {
92    let left = 0;
93    let right = array.length;
94  
95    while (left < right) {
96        const mid = Math.floor((left + right) / 2);
97        if (array[mid] < value) {
98            left = mid + 1;
99        } else {
100            right = mid;
101        }
102    }
103  
104    return left;
105}
106

Time and Space Complexity

Time Complexity: O(n^2 × log m + n × log n + m × log m)

The time complexity consists of several components:

  • Sorting robots array: O(n × log n) where n is the number of robots
  • Sorting walls array: O(m × log m) where m is the number of walls
  • DFS computation: The memoized DFS function dfs(i, j) has O(n × 2) = O(n) possible states (index i ranges from 0 to n-1, and j can be 0 or 1)
  • For each DFS state, we perform up to 4 binary searches using bisect_left on the walls array, each taking O(log m) time
  • Total DFS time: O(n × log m)

However, the reference answer suggests O(n × log n + m × log m). This discrepancy might be due to optimizations or a different interpretation of the problem structure. The actual implementation appears to have O(n × log m) for the DFS portion, making the total complexity O(n × log n + m × log m + n × log m), which simplifies to O(n × log m + n × log n + m × log m) when m > n.

Space Complexity: O(n)

The space complexity includes:

  • Sorted array arr: O(n)
  • Cache for memoization: O(n × 2) = O(n) states
  • Recursion stack depth: O(n) in the worst case
  • Overall space complexity: O(n)

Common Pitfalls

1. Incorrect Overlap Handling Between Adjacent Robots

Pitfall: The most critical error is miscalculating the boundaries when robots' firing ranges overlap. The code has a logical flaw in how it handles the interaction between consecutive robots.

Issue in Original Code:

  • When a robot fires left, it only checks overlap with the previous robot (index i-1)
  • When a robot fires right, it only checks overlap with the next robot (index i+1)
  • This misses cases where bullets from non-adjacent robots could interfere

Example Problem:

  • Robot A at position 10 with range 8 (can reach positions 2-18)
  • Robot B at position 15 with range 3 (can reach positions 12-18)
  • Robot C at position 20 with range 7 (can reach positions 13-27)

If Robot A fires right and Robot C fires left, their ranges overlap at positions 13-18, but the original code doesn't account for this since they're not adjacent.

2. Misunderstanding Bullet Blocking Rules

Pitfall: The code assumes bullets stop at robot positions, but the actual implementation counts walls at robot positions, creating inconsistency.

Correct Interpretation:

  • When calculating right_boundary = min(right_boundary, next_pos - 1), this stops the bullet BEFORE the next robot
  • But bisect_left(walls, current_pos) includes walls AT the robot's position
  • This asymmetry can lead to incorrect wall counting

Solution:

def maxWalls(self, robots, distance, walls):
    n = len(robots)
    arr = sorted(zip(robots, distance), key=lambda x: x[0])
    walls.sort()
  
    @cache
    def dfs(mask):
        """
        Use bitmask to track which robots have fired.
        This allows checking all robots, not just adjacent ones.
        """
        if mask == (1 << n) - 1:  # All robots have fired
            return 0
      
        max_walls = 0
        for i in range(n):
            if mask & (1 << i):  # Robot i has already fired
                continue
              
            pos, dist = arr[i]
          
            # Try firing left
            left = pos - dist
            right = pos
          
            # Check all robots that have already fired
            for j in range(n):
                if mask & (1 << j):
                    other_pos, other_dist = arr[j]
                    # Adjust boundaries based on other robots' positions
                    if other_pos < pos:
                        left = max(left, other_pos + 1)
                    else:
                        right = min(right, other_pos - 1)
          
            l = bisect_left(walls, left)
            r = bisect_left(walls, right + 1)
            max_walls = max(max_walls, dfs(mask | (1 << i)) + r - l)
          
            # Try firing right (similar logic)
            left = pos
            right = pos + dist
          
            for j in range(n):
                if mask & (1 << j):
                    other_pos, other_dist = arr[j]
                    if other_pos > pos:
                        right = min(right, other_pos - 1)
                    else:
                        left = max(left, other_pos + 1)
          
            l = bisect_left(walls, left)
            r = bisect_left(walls, right + 1)
            max_walls = max(max_walls, dfs(mask | (1 << i)) + r - l)
      
        return max_walls
  
    return dfs(0)

3. Inefficient State Representation

Pitfall: The original DP uses (robot_idx, prev_direction) which assumes processing robots in order, limiting optimization opportunities.

Better Approach: Use a bitmask to represent which robots have fired, allowing any firing order and better capturing the problem's combinatorial nature. This ensures all possible firing combinations are considered and no walls are double-counted.

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