LeetCode 657: Robot Return to Origin Solution in Python – A Step-by-Step Guide

Imagine you’re guiding a robot on a 2D grid, starting at the origin (0, 0), and it follows a string of moves like "UDLR"—up, down, left, right. Your task is to figure out if it ends up back where it started after all its steps. For "UD", it does; for "UDL", it doesn’t. That’s the essence of LeetCode 657: Robot Return to Origin, an easy-level problem that’s all about tracking movement to check a simple condition. Using Python, we’ll explore two solutions: the Best Solution, a coordinate tracking approach for clarity and efficiency, and an Alternative Solution, a count-based balancing method that’s equally simple but different in style. With detailed examples, beginner-friendly breakdowns—especially for the coordinate method—and clear code, this guide will help you steer that robot, whether you’re new to coding or leveling up. Let’s start moving and see where we land!

What Is LeetCode 657: Robot Return to Origin?

In LeetCode 657: Robot Return to Origin, you’re given a string moves where each character is one of:

'U' (up): Move y + 1.
'D' (down): Move y - 1.
'L' (left): Move x - 1.
'R' (right): Move x + 1.

Starting at the origin (0, 0) on a 2D plane, your task is to determine if the robot returns to (0, 0) after executing all moves, returning True if it does, False otherwise. For example, with moves = "UD", the robot goes up (0, 1) then down (0, 0), returning True. This problem tests your ability to track positional changes in a straightforward way.

Problem Statement

Input:

moves: String of 'U', 'D', 'L', 'R' characters.

Output: A boolean—True if robot returns to (0, 0), False otherwise.
Rules:

Start at (0, 0).
'U': y + 1, 'D': y - 1, 'L': x - 1, 'R': x + 1.
Check if final position is (0, 0).

Constraints

1 ≤ moves.length ≤ 2 * 10⁴.
moves contains only 'U', 'D', 'L', 'R'.

Examples

Input: moves = "UD"

Output: True ((0, 0) → (0, 1) → (0, 0))

Input: moves = "LL"

Output: False ((0, 0) → (-1, 0) → (-2, 0))

Input: moves = "URDL"

Output: True ((0, 0) → (0, 1) → (1, 1) → (1, 0) → (0, 0))

These examples set the moves—let’s solve it!

Understanding the Problem: Tracking the Robot

To solve LeetCode 657: Robot Return to Origin in Python, we need to track a robot’s position on a 2D plane starting at (0, 0), applying a sequence of moves, and checking if it returns to the origin. A brute-force approach—simulating each move explicitly—is actually fine here due to the problem’s simplicity, but we can optimize how we process it. Since moves affect x and y independently, we can use coordinates or counts. We’ll explore:

Best Solution (Coordinate Tracking): O(n) time, O(1) space—fast and intuitive.
Alternative Solution (Count-Based Balancing): O(n) time, O(1) space—simple and elegant.

Let’s start with the coordinate solution, breaking it down for beginners.

Best Solution: Coordinate Tracking

Why This Is the Best Solution

The coordinate tracking approach is the top choice for LeetCode 657 because it’s efficient—O(n) time with O(1) space—and directly mirrors the robot’s movement by updating x and y coordinates for each move, making it crystal clear for beginners. It fits constraints (n ≤ 2 * 10⁴) perfectly and requires minimal code. It’s like walking the robot step-by-step and checking its final spot!

How It Works

Think of this as a robot walker:

Step 1: Initialize Coordinates:

Start at (x, y) = (0, 0).

Step 2: Process Moves:

For each move:

'U': y += 1.
'D': y -= 1.
'L': x -= 1.
'R': x += 1.

Step 3: Check Origin:

Return True if (x, y) = (0, 0), False otherwise.

It’s like a position tracker—move and check!

Step-by-Step Example

Example: moves = "URDL"

Step 1: Init: x = 0, y = 0.
Step 2: Process:

'U': (0, 0) → (0, 1), y += 1.
'R': (0, 1) → (1, 1), x += 1.
'D': (1, 1) → (1, 0), y -= 1.
'L': (1, 0) → (0, 0), x -= 1.

Step 3: Final (x, y) = (0, 0), return True.
Output: True.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        # Step 1: Initialize coordinates
        x = 0
        y = 0

        # Step 2: Process each move
        for move in moves:
            if move == 'U':
                y += 1
            elif move == 'D':
                y -= 1
            elif move == 'L':
                x -= 1
            elif move == 'R':
                x += 1

        # Step 3: Check if back at origin
        return x == 0 and y == 0

Lines 4-5: Init: Start at (0, 0).
Lines 8-16: Process:

Update x or y based on move direction.

Line 19: Check: Return True if (x, y) = (0, 0).
Time Complexity: O(n)—single pass through moves.
Space Complexity: O(1)—two variables.

This is like a robot guide—track and confirm!

Alternative Solution: Count-Based Balancing

Why an Alternative Approach?

The count-based balancing approach also runs in O(n) time with O(1) space but uses a different lens—counting moves in each direction and checking if they balance out. It’s equally simple, avoiding explicit coordinates, but might feel less visual. It’s like tallying steps to see if they cancel!

How It Works

Picture this as a move counter:

Step 1: Initialize counters:

up_down: Net vertical moves (U - D).
left_right: Net horizontal moves (L - R).

Step 2: Count moves:

'U': up_down += 1.
'D': up_down -= 1.
'L': left_right += 1.
'R': left_right -= 1.

Step 3: Check balance:

Return True if both counters are 0, False otherwise.

It’s like a balance checker—tally and test!

Step-by-Step Example

Example: moves = "URDL"

Step 1: Init: up_down = 0, left_right = 0.
Step 2: Count:

'U': up_down += 1 → 1.
'R': left_right -= 1 → -1.
'D': up_down -= 1 → 0.
'L': left_right += 1 → 0.

Step 3: up_down = 0, left_right = 0, return True.
Output: True.

Code for Count-Based Approach

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        # Step 1: Initialize counters
        up_down = 0      # U - D
        left_right = 0   # L - R

        # Step 2: Count moves
        for move in moves:
            if move == 'U':
                up_down += 1
            elif move == 'D':
                up_down -= 1
            elif move == 'L':
                left_right += 1
            elif move == 'R':
                left_right -= 1

        # Step 3: Check if balanced
        return up_down == 0 and left_right == 0

Lines 4-5: Init: Counters for net moves.
Lines 8-16: Count:

Adjust counters based on move direction.

Line 19: Check: Return True if both zero.
Time Complexity: O(n)—single pass.
Space Complexity: O(1)—two variables.

It’s a move balancer—count and verify!

Comparing the Two Solutions

Coordinate Tracking (Best):

Pros: O(n) time, O(1) space, visually intuitive.
Cons: Slightly more operations per move.

Count-Based (Alternative):

Pros: O(n) time, O(1) space, minimalistic.
Cons: Less spatial representation.

Coordinate wins for clarity.

Additional Examples and Edge Cases

Input: moves = ""

Output: True (No moves, stays at origin).

Input: moves = "UDUD"

Output: True (Up-down pairs cancel).

Both handle these well.

Complexity Breakdown

Coordinate: Time O(n), Space O(1).
Count-Based: Time O(n), Space O(1).

Both are optimal, coordinate clearer.

Key Takeaways

Coordinate: Position tracking—smart!
Count-Based: Move balancing—clear!
Robot: Moving is fun.
Python Tip: Loops rock—see Python Basics.

Final Thoughts: Guide That Robot

LeetCode 657: Robot Return to Origin in Python is a fun movement challenge. Coordinate tracking offers a visual, straightforward solution, while count-based balancing provides an elegant alternative. Want more? Try LeetCode 66: Plus One or LeetCode 463: Island Perimeter. Ready to move? Head to Solve LeetCode 657 on LeetCode and guide that robot back home today!