Snake in Matrix

Problem

There is a snake in an n x n matrix grid and can move in four possible directions. Each cell in the grid is identified by the position: grid[i][j] = (i * n) + j.

The snake starts at cell 0 and follows a sequence of commands.

You are given an integer n representing the size of the grid and an array of strings commands where each command[i] is either "UP", "RIGHT", "DOWN", and "LEFT". It's guaranteed that the snake will remain within the grid boundaries throughout its movement.

Return the position of the final cell where the snake ends up after executing commands.

Examples

Example 1

Input: n = 2, commands = ["RIGHT","DOWN"]

Output: 3

Explanation:

0 | 1  
---|---  
2 | 3  
0 | 1  
---|---  
2 | 3  
0 | 1  
---|---  
2 | 3  
  

Example 2

Input: n = 3, commands = ["DOWN","RIGHT","UP"]

Output: 1

Explanation:

0 | 1 | 2  
---|---|---  
3 | 4 | 5  
6 | 7 | 8  
0 | 1 | 2  
---|---|---  
3 | 4 | 5  
6 | 7 | 8  
0 | 1 | 2  
---|---|---  
3 | 4 | 5  
6 | 7 | 8  
0 | 1 | 2  
---|---|---  
3 | 4 | 5  
6 | 7 | 8  
  

Constraints

• 2 <= n <= 10
• 1 <= commands.length <= 100
• commands consists only of "UP", "RIGHT", "DOWN", and "LEFT".
• The input is generated such the snake will not move outside of the boundaries.

Solution

Method 1 – Simulate Movement with Direction Mapping

Intuition

The snake's movement can be simulated by tracking its current position (row, col) and updating it according to each command. Since the grid is small and the commands are guaranteed to keep the snake within bounds, we can use a simple direction-to-delta mapping.

Approach

1. Start at position (0, 0).
2. For each command, update the row or column based on the direction:
   • "UP": row -= 1
   • "DOWN": row += 1
   • "LEFT": col -= 1
   • "RIGHT": col += 1
3. After all commands, return the final position as (row * n + col).

Code

C++

class Solution {
public:
    int finalCell(int n, vector<string>& cmds) {
        int r = 0, c = 0;
        for (auto& s : cmds) {
            if (s == "UP") r--;
            else if (s == "DOWN") r++;
            else if (s == "LEFT") c--;
            else if (s == "RIGHT") c++;
        }
        return r * n + c;
    }
};

Go

func finalCell(n int, cmds []string) int {
    r, c := 0, 0
    for _, s := range cmds {
        switch s {
        case "UP": r--
        case "DOWN": r++
        case "LEFT": c--
        case "RIGHT": c++
        }
    }
    return r*n + c
}

Java

class Solution {
    public int finalCell(int n, String[] cmds) {
        int r = 0, c = 0;
        for (String s : cmds) {
            if (s.equals("UP")) r--;
            else if (s.equals("DOWN")) r++;
            else if (s.equals("LEFT")) c--;
            else if (s.equals("RIGHT")) c++;
        }
        return r * n + c;
    }
}

Kotlin

class Solution {
    fun finalCell(n: Int, cmds: Array<String>): Int {
        var r = 0; var c = 0
        for (s in cmds) {
            when (s) {
                "UP" -> r--
                "DOWN" -> r++
                "LEFT" -> c--
                "RIGHT" -> c++
            }
        }
        return r * n + c
    }
}

Python

class Solution:
    def finalCell(self, n: int, cmds: list[str]) -> int:
        r, c = 0, 0
        for s in cmds:
            if s == "UP": r -= 1
            elif s == "DOWN": r += 1
            elif s == "LEFT": c -= 1
            elif s == "RIGHT": c += 1
        return r * n + c

Rust

impl Solution {
    pub fn final_cell(n: i32, cmds: Vec<String>) -> i32 {
        let (mut r, mut c) = (0, 0);
        for s in cmds.iter() {
            match s.as_str() {
                "UP" => r -= 1,
                "DOWN" => r += 1,
                "LEFT" => c -= 1,
                "RIGHT" => c += 1,
                _ => {}
            }
        }
        r * n + c
    }
}

TypeScript

class Solution {
    finalCell(n: number, cmds: string[]): number {
        let r = 0, c = 0;
        for (const s of cmds) {
            if (s === "UP") r--;
            else if (s === "DOWN") r++;
            else if (s === "LEFT") c--;
            else if (s === "RIGHT") c++;
        }
        return r * n + c;
    }
}

Complexity

• ⏰ Time complexity: O(m), where m is the number of commands. Each command is processed once.
• 🧺 Space complexity: O(1), only a constant number of variables are used.