LeetCode #2069 — MEDIUM

Walking Robot Simulation II

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

The Problem

Problem Statement

A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the top-right cell at (width - 1, height - 1). The grid is aligned with the four cardinal directions ("North", "East", "South", and "West"). A robot is initially at cell (0, 0) facing direction "East".

The robot can be instructed to move for a specific number of steps. For each step, it does the following.

Attempts to move forward one cell in the direction it is facing.
If the cell the robot is moving to is out of bounds, the robot instead turns 90 degrees counterclockwise and retries the step.

After the robot finishes moving the number of steps required, it stops and awaits the next instruction.

Implement the Robot class:

Robot(int width, int height) Initializes the width x height grid with the robot at (0, 0) facing "East".
void step(int num) Instructs the robot to move forward num steps.
int[] getPos() Returns the current cell the robot is at, as an array of length 2, [x, y].
String getDir() Returns the current direction of the robot, "North", "East", "South", or "West".

Example 1:

Input
["Robot", "step", "step", "getPos", "getDir", "step", "step", "step", "getPos", "getDir"]
[[6, 3], [2], [2], [], [], [2], [1], [4], [], []]
Output
[null, null, null, [4, 0], "East", null, null, null, [1, 2], "West"]

Explanation
Robot robot = new Robot(6, 3); // Initialize the grid and the robot at (0, 0) facing East.
robot.step(2);  // It moves two steps East to (2, 0), and faces East.
robot.step(2);  // It moves two steps East to (4, 0), and faces East.
robot.getPos(); // return [4, 0]
robot.getDir(); // return "East"
robot.step(2);  // It moves one step East to (5, 0), and faces East.
                // Moving the next step East would be out of bounds, so it turns and faces North.
                // Then, it moves one step North to (5, 1), and faces North.
robot.step(1);  // It moves one step North to (5, 2), and faces North (not West).
robot.step(4);  // Moving the next step North would be out of bounds, so it turns and faces West.
                // Then, it moves four steps West to (1, 2), and faces West.
robot.getPos(); // return [1, 2]
robot.getDir(); // return "West"

Constraints:

2 <= width, height <= 100
1 <= num <= 10⁵
At most 10⁴ calls in total will be made to step, getPos, and getDir.

Step 01

Brute Force Baseline

Problem summary: A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the top-right cell at (width - 1, height - 1). The grid is aligned with the four cardinal directions ("North", "East", "South", and "West"). A robot is initially at cell (0, 0) facing direction "East". The robot can be instructed to move for a specific number of steps. For each step, it does the following. Attempts to move forward one cell in the direction it is facing. If the cell the robot is moving to is out of bounds, the robot instead turns 90 degrees counterclockwise and retries the step. After the robot finishes moving the number of steps required, it stops and awaits the next instruction. Implement the Robot class: Robot(int width, int height) Initializes the width x height grid with the robot at (0, 0) facing "East". void step(int num) Instructs the robot to move forward num steps. int[] getPos()

Baseline thinking

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

Pattern signal: Design

Example 1

["Robot","step","step","getPos","getDir","step","step","step","getPos","getDir"]
[[6,3],[2],[2],[],[],[2],[1],[4],[],[]]

Core Insight

What unlocks the optimal approach

The robot only moves along the perimeter of the grid. Can you think if modulus can help you quickly compute which cell it stops at?
After the robot moves one time, whenever the robot stops at some cell, it will always face a specific direction. i.e., The direction it faces is determined by the cell it stops at.
Can you precompute what direction it faces when it stops at each cell along the perimeter, and reuse the results?

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 #2069: Walking Robot Simulation II
class Robot {
  public Robot(int width, int height) {
    pos.add(new Pair<>(new int[] {0, 0}, "South"));
    for (int i = 1; i < width; ++i)
      pos.add(new Pair<>(new int[] {i, 0}, "East"));
    for (int j = 1; j < height; ++j)
      pos.add(new Pair<>(new int[] {width - 1, j}, "North"));
    for (int i = width - 2; i >= 0; --i)
      pos.add(new Pair<>(new int[] {i, height - 1}, "West"));
    for (int j = height - 2; j > 0; --j)
      pos.add(new Pair<>(new int[] {0, j}, "South"));
  }

  public void step(int num) {
    isOrigin = false;
    i = (i + num) % pos.size();
  }

  public int[] getPos() {
    return pos.get(i).getKey();
  }

  public String getDir() {
    return isOrigin ? "East" : pos.get(i).getValue();
  }

  private boolean isOrigin = true;
  private int i = 0;
  private List<Pair<int[], String>> pos = new ArrayList<>();
}

// Accepted solution for LeetCode #2069: Walking Robot Simulation II
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #2069: Walking Robot Simulation II
// class Robot {
//   public Robot(int width, int height) {
//     pos.add(new Pair<>(new int[] {0, 0}, "South"));
//     for (int i = 1; i < width; ++i)
//       pos.add(new Pair<>(new int[] {i, 0}, "East"));
//     for (int j = 1; j < height; ++j)
//       pos.add(new Pair<>(new int[] {width - 1, j}, "North"));
//     for (int i = width - 2; i >= 0; --i)
//       pos.add(new Pair<>(new int[] {i, height - 1}, "West"));
//     for (int j = height - 2; j > 0; --j)
//       pos.add(new Pair<>(new int[] {0, j}, "South"));
//   }
// 
//   public void step(int num) {
//     isOrigin = false;
//     i = (i + num) % pos.size();
//   }
// 
//   public int[] getPos() {
//     return pos.get(i).getKey();
//   }
// 
//   public String getDir() {
//     return isOrigin ? "East" : pos.get(i).getValue();
//   }
// 
//   private boolean isOrigin = true;
//   private int i = 0;
//   private List<Pair<int[], String>> pos = new ArrayList<>();
// }

# Accepted solution for LeetCode #2069: Walking Robot Simulation II
class Robot:
  def __init__(self, width: int, height: int):
    self.isOrigin = True
    self.i = 0
    self.pos = ([((0, 0), 'South')] +
                [((i, 0), 'East') for i in range(1, width)] +
                [((width - 1, j), 'North') for j in range(1, height)] +
                [((i, height - 1), 'West') for i in range(width - 2, -1, -1)] +
                [((0, j), 'South') for j in range(height - 2, 0, -1)])

  def step(self, num: int) -> None:
    self.isOrigin = False
    self.i = (self.i + num) % len(self.pos)

  def getPos(self) -> list[int]:
    return self.pos[self.i][0]

  def getDir(self) -> str:
    return 'East' if self.isOrigin else self.pos[self.i][1]

// Accepted solution for LeetCode #2069: Walking Robot Simulation II
/**
 * [2069] Walking Robot Simulation II
 *
 * A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the top-right cell at (width - 1, height - 1). The grid is aligned with the four cardinal directions ("North", "East", "South", and "West"). A robot is initially at cell (0, 0) facing direction "East".
 * The robot can be instructed to move for a specific number of steps. For each step, it does the following.
 * <ol>
 * 	Attempts to move forward one cell in the direction it is facing.
 * 	If the cell the robot is moving to is out of bounds, the robot instead turns 90 degrees counterclockwise and retries the step.
 * </ol>
 * After the robot finishes moving the number of steps required, it stops and awaits the next instruction.
 * Implement the Robot class:
 *
 * 	Robot(int width, int height) Initializes the width x height grid with the robot at (0, 0) facing "East".
 * 	void step(int num) Instructs the robot to move forward num steps.
 * 	int[] getPos() Returns the current cell the robot is at, as an array of length 2, [x, y].
 * 	String getDir() Returns the current direction of the robot, "North", "East", "South", or "West".
 *
 *  
 * Example 1:
 * <img alt="example-1" src="https://assets.leetcode.com/uploads/2021/10/09/example-1.png" style="width: 498px; height: 268px;" />
 * Input
 * ["Robot", "step", "step", "getPos", "getDir", "step", "step", "step", "getPos", "getDir"]
 * [[6, 3], [2], [2], [], [], [2], [1], [4], [], []]
 * Output
 * [null, null, null, [4, 0], "East", null, null, null, [1, 2], "West"]
 * Explanation
 * Robot robot = new Robot(6, 3); // Initialize the grid and the robot at (0, 0) facing East.
 * robot.step(2);  // It moves two steps East to (2, 0), and faces East.
 * robot.step(2);  // It moves two steps East to (4, 0), and faces East.
 * robot.getPos(); // return [4, 0]
 * robot.getDir(); // return "East"
 * robot.step(2);  // It moves one step East to (5, 0), and faces East.
 *                 // Moving the next step East would be out of bounds, so it turns and faces North.
 *                 // Then, it moves one step North to (5, 1), and faces North.
 * robot.step(1);  // It moves one step North to (5, 2), and faces North (not West).
 * robot.step(4);  // Moving the next step North would be out of bounds, so it turns and faces West.
 *                 // Then, it moves four steps West to (1, 2), and faces West.
 * robot.getPos(); // return [1, 2]
 * robot.getDir(); // return "West"
 *
 *  
 * Constraints:
 *
 * 	2 <= width, height <= 100
 * 	1 <= num <= 10^5
 * 	At most 10^4 calls in total will be made to step, getPos, and getDir.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/walking-robot-simulation-ii/
// discuss: https://leetcode.com/problems/walking-robot-simulation-ii/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

enum Direction {
    North,
    East,
    South,
    West,
}

struct Robot {
    max_x: i32,
    max_y: i32,
    current_route_num: i32,
    position_cache: (i32, i32, Direction),
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl Robot {
    fn new(width: i32, height: i32) -> Self {
        Self {
            max_x: width - 1,
            max_y: height - 1,
            current_route_num: -1,
            position_cache: (0, 0, Direction::East),
        }
    }

    fn step(&mut self, num: i32) {
        if self.current_route_num < 0 {
            self.current_route_num = 0;
        }

        let perim = (self.max_x + 1) * 2 + (self.max_y - 1) * 2;

        self.current_route_num += num;
        self.current_route_num %= perim;
        self.set_position_data();
    }

    fn get_pos(&self) -> Vec<i32> {
        vec![self.position_cache.0.clone(), self.position_cache.1.clone()]
    }

    fn get_dir(&self) -> String {
        match self.position_cache.2 {
            Direction::North => "North",
            Direction::East => "East",
            Direction::South => "South",
            Direction::West => "West",
        }
        .to_string()
    }

    fn set_position_data(&mut self) {
        self.position_cache = if self.current_route_num < 0 {
            (0, 0, Direction::East)
        } else if self.current_route_num == 0 {
            (self.current_route_num, 0, Direction::South)
        } else if self.current_route_num <= self.max_x {
            (self.current_route_num, 0, Direction::East)
        } else if self.current_route_num <= self.max_x + self.max_y {
            (
                self.max_x,
                self.current_route_num - self.max_x,
                Direction::North,
            )
        } else if self.current_route_num <= 2 * self.max_x + self.max_y {
            (
                self.max_x - (self.current_route_num - self.max_x - self.max_y),
                self.max_y,
                Direction::West,
            )
        } else {
            (
                0,
                self.max_y - (self.current_route_num - 2 * self.max_x - self.max_y),
                Direction::South,
            )
        }
    }
}

/**
 * Your Robot object will be instantiated and called as such:
 * let obj = Robot::new(width, height);
 * obj.step(num);
 * let ret_2: Vec<i32> = obj.get_pos();
 * let ret_3: String = obj.get_dir();
 */

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2069_example_1() {
        let mut robot = Robot::new(6, 3); // Initialize the grid and the robot at (0, 0) facing East.
        robot.step(2); // It moves two steps East to (2, 0), and faces East.
        robot.step(2); // It moves two steps East to (4, 0), and faces East.
        assert_eq!(robot.get_pos(), vec![4, 0]); // return [4, 0]
        assert_eq!(robot.get_dir(), "East".to_string()); // return "East"
        robot.step(2); // It moves one step East to (5, 0), and faces East.
        // Moving the next step East would be out of bounds, so it turns and faces North.
        // Then, it moves one step North to (5, 1), and faces North.
        robot.step(1); // It moves one step North to (5, 2), and faces North (not West).
        robot.step(4); // Moving the next step North would be out of bounds, so it turns and faces West.
        // Then, it moves four steps West to (1, 2), and faces West.
        assert_eq!(robot.get_pos(), vec![1, 2]); // return [1, 2]
        assert_eq!(robot.get_dir(), "West".to_string()); // return "West"
    }
}

// Accepted solution for LeetCode #2069: Walking Robot Simulation II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2069: Walking Robot Simulation II
// class Robot {
//   public Robot(int width, int height) {
//     pos.add(new Pair<>(new int[] {0, 0}, "South"));
//     for (int i = 1; i < width; ++i)
//       pos.add(new Pair<>(new int[] {i, 0}, "East"));
//     for (int j = 1; j < height; ++j)
//       pos.add(new Pair<>(new int[] {width - 1, j}, "North"));
//     for (int i = width - 2; i >= 0; --i)
//       pos.add(new Pair<>(new int[] {i, height - 1}, "West"));
//     for (int j = height - 2; j > 0; --j)
//       pos.add(new Pair<>(new int[] {0, j}, "South"));
//   }
// 
//   public void step(int num) {
//     isOrigin = false;
//     i = (i + num) % pos.size();
//   }
// 
//   public int[] getPos() {
//     return pos.get(i).getKey();
//   }
// 
//   public String getDir() {
//     return isOrigin ? "East" : pos.get(i).getValue();
//   }
// 
//   private boolean isOrigin = true;
//   private int i = 0;
//   private List<Pair<int[], String>> pos = new ArrayList<>();
// }

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(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

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.