Minimum Moves to Reach Target with Rotations

Minimum Moves to Reach Target with Rotations - Problem

Imagine controlling a snake robot that spans exactly 2 cells and needs to navigate through an obstacle course! Your snake starts at the top-left corner of an n×n grid, initially positioned horizontally at cells (0, 0) and (0, 1).

The grid contains:

Empty cells (represented by 0) - safe to move through
Blocked cells (represented by 1) - obstacles that block movement

Your goal is to reach the bottom-right corner at positions (n-1, n-2) and (n-1, n-1) in the minimum number of moves.

The snake can perform these moves:

Move Right: Both cells move one position right (if no obstacles)
Move Down: Both cells move one position down (if no obstacles)
Rotate Clockwise: When horizontal, rotate to vertical if both cells below are empty
Rotate Counterclockwise: When vertical, rotate to horizontal if both cells to the right are empty

Return the minimum moves needed, or -1 if impossible to reach the target.

Input & Output

example_1.py — Simple 3x3 Grid

$ Input: grid = [[0,0,0,0,0,1],[1,1,0,0,1,0],[0,0,0,0,1,1],[0,0,1,0,1,0],[0,1,1,0,0,0],[0,1,1,0,0,0]]

› Output: 11

💡 Note: The snake needs to navigate around obstacles using a combination of moves and rotations. The optimal path requires 11 moves to reach the target position.

example_2.py — Simple Path

$ Input: grid = [[0,0,1,1,1,1],[0,0,0,0,1,1],[1,1,0,0,0,1],[1,1,1,0,0,0],[0,1,1,0,0,0],[0,1,1,0,0,0]]

› Output: -1

💡 Note: There is no valid path for the snake to reach the target due to the obstacle layout blocking all possible routes.

example_3.py — Minimum Grid

$ Input: grid = [[0,0,0],[0,1,0],[0,0,0]]

› Output: 4

💡 Note: In a 3x3 grid with one obstacle, the snake can reach target (2,1)-(2,2) in 4 moves: right, down, right, down.

Constraints

2 ≤ grid.length ≤ 100
2 ≤ grid[i].length ≤ 100
grid[i][j] is 0 or 1
It is guaranteed that the snake and target area are empty
The snake always starts horizontally at (0,0) and (0,1)
Target is always at (n-1, n-2) and (n-1, n-1) horizontally

Visualization

Tap to expand

Understanding the Visualization

Initial State

Snake starts horizontally at top-left: (0,0)-(0,1)

Explore Moves

BFS explores right, down, and rotate moves simultaneously

Track States

Each unique (position, orientation) combination is a state

Find Shortest

First time we reach target state gives minimum moves

Key Takeaway

🎯 Key Insight: By treating each (position, orientation) as a unique state and using BFS, we guarantee finding the shortest path through the state space in O(n²) time.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses BFS (Breadth-First Search) to explore all possible snake states level by level. Each state represents the snake's position and orientation. BFS guarantees finding the shortest path with O(n²) time complexity, as there are at most 2n² possible states (each cell in 2 orientations).

Common Approaches

Approach	Time	Space	Notes
✓ BFS with State Tracking (Optimal)	O(n²)	O(n²)	Use BFS to find shortest path by exploring states level by level
Recursive Backtracking (DFS)	O(4^(n²))	O(n²)	Try all possible move sequences using depth-first search

BFS with State Tracking (Optimal) — Algorithm Steps

Initialize BFS queue with starting state: horizontal at (0,0)-(0,1)
For each state, generate all valid next moves
Add new states to queue if not visited before
Continue until target state is reached or queue is empty
Return the number of moves when target is found

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with snake at (0,0)-(0,1) horizontally, moves = 0

Level 1

Explore all states reachable in 1 move: right, down, rotate

Level 2

From each Level 1 state, explore all reachable states in 1 more move

Continue

Keep exploring until target state is reached or no more states

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

#define MAX_STATES 10000

typedef struct {
    int pos[2][2];
    int moves;
} State;

typedef struct {
    State states[MAX_STATES];
    int front, rear;
} Queue;

bool isValid(int r, int c, int n, int** grid) {
    return r >= 0 && r < n && c >= 0 && c < n && grid[r][c] == 0;
}

void enqueue(Queue* q, State s) {
    q->states[q->rear++] = s;
}

State dequeue(Queue* q) {
    return q->states[q->front++];
}

bool isEmpty(Queue* q) {
    return q->front == q->rear;
}

int stateHash(int pos[2][2]) {
    return pos[0][0] * 1000 + pos[0][1] * 100 + pos[1][0] * 10 + pos[1][1];
}

int minimumMoves(int** grid, int gridSize, int* gridColSize) {
    int n = gridSize;
    Queue queue = {.front = 0, .rear = 0};
    bool visited[100000] = {false};
    
    State start = {{{0, 0}, {0, 1}}, 0};
    enqueue(&queue, start);
    visited[stateHash(start.pos)] = true;
    
    while (!isEmpty(&queue)) {
        State current = dequeue(&queue);
        
        // Check if reached target
        if (current.pos[0][0] == n-1 && current.pos[0][1] == n-2 && 
            current.pos[1][0] == n-1 && current.pos[1][1] == n-1) {
            return current.moves;
        }
        
        int r1 = current.pos[0][0], c1 = current.pos[0][1];
        int r2 = current.pos[1][0], c2 = current.pos[1][1];
        
        // Try all possible moves
        State nextStates[3];
        int validCount = 0;
        
        // Move right
        if (isValid(r1, c1 + 1, n, grid) && isValid(r2, c2 + 1, n, grid)) {
            nextStates[validCount].pos[0][0] = r1;
            nextStates[validCount].pos[0][1] = c1 + 1;
            nextStates[validCount].pos[1][0] = r2;
            nextStates[validCount].pos[1][1] = c2 + 1;
            nextStates[validCount].moves = current.moves + 1;
            validCount++;
        }
        
        // Move down
        if (isValid(r1 + 1, c1, n, grid) && isValid(r2 + 1, c2, n, grid)) {
            nextStates[validCount].pos[0][0] = r1 + 1;
            nextStates[validCount].pos[0][1] = c1;
            nextStates[validCount].pos[1][0] = r2 + 1;
            nextStates[validCount].pos[1][1] = c2;
            nextStates[validCount].moves = current.moves + 1;
            validCount++;
        }
        
        // Rotate
        if (r1 == r2) { // Horizontal
            if (isValid(r1 + 1, c1, n, grid) && isValid(r1 + 1, c2, n, grid)) {
                nextStates[validCount].pos[0][0] = r1;
                nextStates[validCount].pos[0][1] = c1;
                nextStates[validCount].pos[1][0] = r1 + 1;
                nextStates[validCount].pos[1][1] = c1;
                nextStates[validCount].moves = current.moves + 1;
                validCount++;
            }
        } else { // Vertical
            if (isValid(r1, c1 + 1, n, grid) && isValid(r2, c1 + 1, n, grid)) {
                nextStates[validCount].pos[0][0] = r1;
                nextStates[validCount].pos[0][1] = c1;
                nextStates[validCount].pos[1][0] = r1;
                nextStates[validCount].pos[1][1] = c1 + 1;
                nextStates[validCount].moves = current.moves + 1;
                validCount++;
            }
        }
        
        for (int i = 0; i < validCount; i++) {
            int hash = stateHash(nextStates[i].pos);
            if (!visited[hash]) {
                visited[hash] = true;
                enqueue(&queue, nextStates[i]);
            }
        }
    }
    
    return -1;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Each of the n² cells can be visited in at most 2 orientations, so 2n² total states

⚠ Quadratic Growth

Space Complexity

O(n²)

Queue and visited set can hold up to 2n² states

⚠ Quadratic Space

21.0K Views

Medium Frequency

~25 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

BFS with State Tracking (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler