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Robot Return to Origin solution and explanation

EasySimulationString6 min read78.0% acceptanceUpdated 9 Apr 2026
Asked atGoldman SachsMicrosoftBloombergGoogleAmazon
Practise Robot Return to Origin
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Robot Return to Origin | Simple Explanation | Leetcode 657 | codestorywithMIKcodestorywithMIK · 3:43 · 3K viewsWatch on YouTube
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The problem

There is a robot starting at the position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.

You are given a string moves that represents the move sequence of the robot where moves[i] represents its ith move. Valid moves are 'R' (right), 'L' (left), 'U' (up), and 'D' (down).

Return true if the robot returns to the origin after it finishes all of its moves, or false otherwise.

Note: The way that the robot is "facing" is irrelevant. 'R' will always make the robot move to the right once, 'L' will always make it move left, etc. Also, assume that the magnitude of the robot's movement is the same for each move.

Example 1:

Input: moves = "UD"
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.

Example 2:

Input: moves = "LL"
Output: false
Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.

Constraints:

  • 1 <= moves.length <= 2 * 104
  • moves only contains the characters 'U', 'D', 'L' and 'R'.

Approach overview

Problem Overview: You receive a string of moves where each character represents a step taken by a robot on a 2D grid: U, D, L, and R. The robot starts at coordinate (0,0). The task is to determine whether the sequence of moves brings the robot back to the origin after executing all steps.

Approach 1: Balancing Moves (O(n) time, O(1) space)

This approach simulates the robot’s movement directly. Initialize two integers x and y to represent the current coordinates. Iterate through the move string once. For each character, update the coordinates: U increments y, D decrements y, R increments x, and L decrements x. After processing all characters, check whether both coordinates returned to zero.

The key insight is that every movement changes exactly one axis by one unit. Tracking coordinates directly mirrors how the robot moves in a real grid. This solution works well for problems involving step‑by‑step movement simulation and is a common pattern in simulation problems. The algorithm performs a single pass over the input string, so the time complexity is O(n), and it uses only two integer variables, resulting in O(1) space.

Approach 2: Net Balance Using Counts (O(n) time, O(1) space)

This approach relies on the observation that opposite moves cancel each other. A robot returns to the origin only if the number of U moves equals the number of D moves and the number of L moves equals the number of R moves. Instead of updating coordinates step by step, iterate through the string and maintain four counters or use built‑in character counts.

After counting, compare the totals: count('U') == count('D') and count('L') == count('R'). If both conditions hold, the robot ends at the origin. This approach focuses on the net displacement rather than the path taken. Since it processes the string once and stores only a few counters, the complexity remains O(n) time and O(1) space.

This counting technique is common in string processing tasks where the relative frequency of characters determines the result. It removes coordinate tracking and makes the logic concise.

Recommended for interviews: Interviewers typically expect the coordinate simulation approach because it clearly models the robot’s movement and demonstrates understanding of grid traversal. The counting approach is equally optimal and slightly shorter, but explaining the cancellation idea shows deeper reasoning about the problem. Both solutions run in O(n) time with constant space, which is optimal for this problem.

Approach 1: Balancing Moves

This approach involves counting the number of moves in each direction and checking if the horizontal movements ('L' and 'R') balance each other and the vertical movements ('U' and 'D') balance each other to return to the origin.

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        x = y = 0
        for move in moves:
            if move == 'U': y += 1
            elif move == 'D': y -= 1
            elif move == 'L': x -= 1
            elif move == 'R': x += 1
        return x == 0 and y == 0
class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        
        for (char move : moves.toCharArray()) {
            if (move == 'U') y++;
            else if (move == 'D') y--;
            else if (move == 'L') x--;
            else if (move == 'R') x++;
        }
        
        return x == 0 && y == 0;
    }
}
#include <string>
bool judgeCircle(std::string moves) {
    int x = 0, y = 0;
    
    for (char move : moves) {
        if (move == 'U') y++;
        else if (move == 'D') y--;
        else if (move == 'L') x--;
        else if (move == 'R') x++;
    }
    
    return x == 0 && y == 0;
}
var judgeCircle = function(moves) {
    let x = 0, y = 0;
    
    for (let move of moves) {
        if (move === 'U') y++;
        else if (move === 'D') y--;
        else if (move === 'L') x--;
        else if (move === 'R') x++;
    }
    
    return x === 0 && y === 0;
};
#include <stdbool.h>

bool judgeCircle(char* moves) {
    int x = 0, y = 0;
    
    for (int i = 0; moves[i] != '\0'; i++) {
        char move = moves[i];
        if (move == 'U') y++;
        else if (move == 'D') y--;
        else if (move == 'L') x--;
        else if (move == 'R') x++;
    }
    
    return x == 0 && y == 0;
}
public class Solution {
    public bool JudgeCircle(string moves) {
        int x = 0, y = 0;
        
        foreach (char move in moves) {
            switch (move) {
                case 'U': y++; break;
                case 'D': y--; break;
                case 'L': x--; break;
                case 'R': x++; break;
            }
        }
        
        return x == 0 && y == 0;
    }
}

The Python function uses two variables x and y to keep track of the robot's position. For each character in the input moves, it adjusts the position accordingly. The function returns true if the position is back to the origin after processing all moves.

Complexity

Time Complexity: O(n), where n is the number of characters in moves.
Space Complexity: O(1), as the space requirement does not increase with input size.

Try this approach in the editor

Approach 2: Net Balance Using Counts

Instead of tracking coordinates, this approach counts the occurrences of each move character to determine if the robot returns to the origin. The idea is that the number of 'U' moves should equal 'D' moves and 'L' moves should equal 'R'.

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        U = moves.count('U')
        D = moves.count('D')
        L = moves.count('L')
        R = moves.count('R')
        
        return U == D and L == R
class Solution {
    public boolean judgeCircle(String moves) {
        int U = 0, D = 0, L = 0, R = 0;
        
        for (char move : moves.toCharArray()) {
            switch (move) {
                case 'U': U++; break;
                case 'D': D++; break;
                case 'L': L++; break;
                case 'R': R++; break;
            }
        }
        
        return U == D && L == R;
    }
}
#include <string>

bool judgeCircle(std::string moves) {
    int U = 0, D = 0, L = 0, R = 0;
    
    for (char move : moves) {
        switch (move) {
            case 'U': U++; break;
            case 'D': D++; break;
            case 'L': L++; break;
            case 'R': R++; break;
        }
    }
    
    return U == D && L == R;
}
var judgeCircle = function(moves) {
    let U = 0, D = 0, L = 0, R = 0;
    
    for (let move of moves) {
        switch (move) {
            case 'U': U++; break;
            case 'D': D++; break;
            case 'L': L++; break;
            case 'R': R++; break;
        }
    }
    
    return U === D && L === R;
};
#include <stdbool.h>
#include <string.h>

bool judgeCircle(char* moves) {
    int U = 0, D = 0, L = 0, R = 0;
    
    for (int i = 0; moves[i] != '\0'; i++) {
        switch (moves[i]) {
            case 'U': U++; break;
            case 'D': D++; break;
            case 'L': L++; break;
            case 'R': R++; break;
        }
    }
    
    return U == D && L == R;
}
public class Solution {
    public bool JudgeCircle(string moves) {
        int U = 0, D = 0, L = 0, R = 0;
        
        foreach (char move in moves) {
            switch (move) {
                case 'U': U++; break;
                case 'D': D++; break;
                case 'L': L++; break;
                case 'R': R++; break;
            }
        }
        
        return U == D && L == R;
    }
}

Using Python string's count method, this solution aggregates move counts directly. The robot is back to the origin if the count of 'U' equals 'D' and the count of 'L' equals 'R'. This method simplifies counting through built-in functions.

Complexity

Time Complexity: O(n), since count traverses the string each time.
Space Complexity: O(1), accounts for fixed space of integer counters.

Try this approach in the editor

Approach 3: Maintain Coordinates

We can maintain a coordinate (x, y) to represent the robot's movement in the horizontal and vertical directions.

Traverse the string moves and update the coordinate (x, y) based on the current character:

  • If the current character is 'U', then y increases by 1;
  • If the current character is `'D', then y decreases by 1;
  • If the current character is `'L', then x decreases by 1;
  • If the current character is `'R', then x increases by 1.

Finally, check if both x and y are 0.

The time complexity is O(n), where n is the length of the string moves. The space complexity is O(1)$.

function judgeCircle(moves: string): boolean {
    let [x, y] = [0, 0];
    for (const c of moves) {
        if (c === 'U') {
            y++;
        } else if (c === 'D') {
            y--;
        } else if (c === 'L') {
            x--;
        } else {
            x++;
        }
    }
    return x === 0 && y === 0;
}
func judgeCircle(moves string) bool {
	x, y := 0, 0
	for _, c := range moves {
		switch c {
		case 'U':
			y++
		case 'D':
			y--
		case 'L':
			x--
		case 'R':
			x++
		}
	}
	return x == 0 && y == 0
}
Try this approach in the editor

Complexity comparison

ApproachComplexity
Balancing Moves

Time Complexity: O(n), where n is the number of characters in moves.
Space Complexity: O(1), as the space requirement does not increase with input size.

Net Balance Using Counts

Time Complexity: O(n), since count traverses the string each time.
Space Complexity: O(1), accounts for fixed space of integer counters.

Maintain Coordinates–

Complexity analysis

ApproachTimeSpaceWhen to Use
Balancing Moves (Coordinate Simulation)O(n)O(1)Best for interview explanations and grid movement problems where you simulate position changes.
Net Balance Using CountsO(n)O(1)Useful when opposite operations cancel out and only final displacement matters.

Frequently asked questions

Is Robot Return to Origin easy or hard?

Robot Return to Origin is classified as an Easy problem on LeetCode with a high acceptance rate. The challenge mainly tests basic string traversal and simple simulation logic rather than advanced algorithms.

Robot Return to Origin Python/Java solution

In Python or Java, iterate through the string and update coordinates for each move. For example, increment y for 'U', decrement y for 'D', increment x for 'R', and decrement x for 'L'. After the loop, return true if both coordinates equal zero. The implementation runs in O(n) time and O(1) space.

How to solve Robot Return to Origin in O(n)?

Iterate through the move string once and track the robot’s position using two integers for x and y. Increment or decrement the coordinates based on the move character. After processing all moves, return true if x == 0 and y == 0; otherwise return false.

What is the best approach for Robot Return to Origin?

The best approach is simulating the robot's coordinates while iterating through the move string. Update x and y positions for each move (U, D, L, R) and check if both return to zero. This solution runs in O(n) time with O(1) space and clearly models the robot’s movement on a grid.

Is Robot Return to Origin asked at Google/Amazon/Meta?

Robot Return to Origin is a common easy-level interview question used by companies like Amazon and Google to test basic string processing and simulation skills. It checks whether candidates can model movement logic and reason about net displacement efficiently.

What data structure is used in Robot Return to Origin?

The problem typically uses simple variables or counters rather than complex data structures. Most solutions rely on integer variables for coordinates or character frequency counts while iterating through the string.

What is the time complexity of Robot Return to Origin?

The optimal time complexity is O(n), where n is the length of the move string. The algorithm scans the string once and performs constant-time updates for each character. Space complexity is O(1) since only a few counters or coordinate variables are required.

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