LeetCode #1041 — MEDIUM

Robot Bounded In Circle

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

Solve on LeetCode

The Problem

Problem Statement

On an infinite plane, a robot initially stands at (0, 0) and faces north. Note that:

The north direction is the positive direction of the y-axis.
The south direction is the negative direction of the y-axis.
The east direction is the positive direction of the x-axis.
The west direction is the negative direction of the x-axis.

The robot can receive one of three instructions:

"G": go straight 1 unit.
"L": turn 90 degrees to the left (i.e., anti-clockwise direction).
"R": turn 90 degrees to the right (i.e., clockwise direction).

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Example 1:

Input: instructions = "GGLLGG"
Output: true
Explanation: The robot is initially at (0, 0) facing the north direction.
"G": move one step. Position: (0, 1). Direction: North.
"G": move one step. Position: (0, 2). Direction: North.
"L": turn 90 degrees anti-clockwise. Position: (0, 2). Direction: West.
"L": turn 90 degrees anti-clockwise. Position: (0, 2). Direction: South.
"G": move one step. Position: (0, 1). Direction: South.
"G": move one step. Position: (0, 0). Direction: South.
Repeating the instructions, the robot goes into the cycle: (0, 0) --> (0, 1) --> (0, 2) --> (0, 1) --> (0, 0).
Based on that, we return true.

Example 2:

Input: instructions = "GG"
Output: false
Explanation: The robot is initially at (0, 0) facing the north direction.
"G": move one step. Position: (0, 1). Direction: North.
"G": move one step. Position: (0, 2). Direction: North.
Repeating the instructions, keeps advancing in the north direction and does not go into cycles.
Based on that, we return false.

Example 3:

Input: instructions = "GL"
Output: true
Explanation: The robot is initially at (0, 0) facing the north direction.
"G": move one step. Position: (0, 1). Direction: North.
"L": turn 90 degrees anti-clockwise. Position: (0, 1). Direction: West.
"G": move one step. Position: (-1, 1). Direction: West.
"L": turn 90 degrees anti-clockwise. Position: (-1, 1). Direction: South.
"G": move one step. Position: (-1, 0). Direction: South.
"L": turn 90 degrees anti-clockwise. Position: (-1, 0). Direction: East.
"G": move one step. Position: (0, 0). Direction: East.
"L": turn 90 degrees anti-clockwise. Position: (0, 0). Direction: North.
Repeating the instructions, the robot goes into the cycle: (0, 0) --> (0, 1) --> (-1, 1) --> (-1, 0) --> (0, 0).
Based on that, we return true.

Constraints:

1 <= instructions.length <= 100
instructions[i] is 'G', 'L' or, 'R'.

Step 01

Brute Force Baseline

Problem summary: On an infinite plane, a robot initially stands at (0, 0) and faces north. Note that: The north direction is the positive direction of the y-axis. The south direction is the negative direction of the y-axis. The east direction is the positive direction of the x-axis. The west direction is the negative direction of the x-axis. The robot can receive one of three instructions: "G": go straight 1 unit. "L": turn 90 degrees to the left (i.e., anti-clockwise direction). "R": turn 90 degrees to the right (i.e., clockwise direction). The robot performs the instructions given in order, and repeats them forever. Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Baseline thinking

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

Pattern signal: Math

Example 1

"GGLLGG"

Example 2

"GG"

Example 3

"GL"

Step 02

Core Insight

What unlocks the optimal approach

Calculate the final vector of how the robot travels after executing all instructions once - it consists of a change in position plus a change in direction.
The robot stays in the circle if and only if (looking at the final vector) it changes direction (ie. doesn't stay pointing north), or it moves 0.

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 #1041: Robot Bounded In Circle
class Solution {
    public boolean isRobotBounded(String instructions) {
        int k = 0;
        int[] dist = new int[4];
        for (int i = 0; i < instructions.length(); ++i) {
            char c = instructions.charAt(i);
            if (c == 'L') {
                k = (k + 1) % 4;
            } else if (c == 'R') {
                k = (k + 3) % 4;
            } else {
                ++dist[k];
            }
        }
        return (dist[0] == dist[2] && dist[1] == dist[3]) || (k != 0);
    }
}

// Accepted solution for LeetCode #1041: Robot Bounded In Circle
func isRobotBounded(instructions string) bool {
	dist := [4]int{}
	k := 0
	for _, c := range instructions {
		if c == 'L' {
			k = (k + 1) % 4
		} else if c == 'R' {
			k = (k + 3) % 4
		} else {
			dist[k]++
		}
	}
	return (dist[0] == dist[2] && dist[1] == dist[3]) || k != 0
}

# Accepted solution for LeetCode #1041: Robot Bounded In Circle
class Solution:
    def isRobotBounded(self, instructions: str) -> bool:
        k = 0
        dist = [0] * 4
        for c in instructions:
            if c == 'L':
                k = (k + 1) % 4
            elif c == 'R':
                k = (k + 3) % 4
            else:
                dist[k] += 1
        return (dist[0] == dist[2] and dist[1] == dist[3]) or k != 0

// Accepted solution for LeetCode #1041: Robot Bounded In Circle
struct Solution;

impl Solution {
    fn is_robot_bounded(instructions: String) -> bool {
        let mut x = 0;
        let mut y = 0;
        let mut i = 0;
        let d = [[1, 0], [0, 1], [-1, 0], [0, -1]];
        for c in instructions.chars() {
            match c {
                'G' => {
                    x += d[i][0];
                    y += d[i][1];
                }
                'L' => {
                    i = (i + 1) % 3;
                }
                'R' => {
                    i = (i + 3) % 3;
                }
                _ => (),
            }
        }
        x == 0 && y == 0 || i != 0
    }
}

#[test]
fn test() {
    assert_eq!(Solution::is_robot_bounded("GGLLGG".to_string()), true);
    assert_eq!(Solution::is_robot_bounded("GG".to_string()), false);
    assert_eq!(Solution::is_robot_bounded("GL".to_string()), true);
}

// Accepted solution for LeetCode #1041: Robot Bounded In Circle
function isRobotBounded(instructions: string): boolean {
    const dist: number[] = new Array(4).fill(0);
    let k = 0;
    for (const c of instructions) {
        if (c === 'L') {
            k = (k + 1) % 4;
        } else if (c === 'R') {
            k = (k + 3) % 4;
        } else {
            ++dist[k];
        }
    }
    return (dist[0] === dist[2] && dist[1] === dist[3]) || k !== 0;
}

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

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.