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3028. Ant on the Boundary-out

An ant is placed on a width x height grid.

Given the grid dimensions (width and height), the starting point (x, y), and a list of integer steps steps, determine after how many seconds the ant will first be at the boundary.

Clarifying Questions

  1. Can the ant start on the boundary?
    • Yes, the ant can start on the boundary.
  2. What happens if the ant is already on the boundary at the start?
    • It is considered to be on the boundary at time 0.
  3. Are the steps for the ant strictly one per second?
    • Yes, each unit movement happens per second.
  4. What should be returned if the ant never reaches the boundary within the given steps?
    • If the ant never reaches the boundary within the given steps, we should return the number of steps.

Strategy

  1. Initial Check:
    • First, check if the starting point (x, y) is already on the boundary. If so, return 0.
  2. Simulate the Ant’s Movement:
    • We’ll simulate the ant’s movement step by step using BFS until the ant reaches any boundary.
  3. Boundary Conditions:
    • All movements should be checked to make sure they stay within the grid boundary.
  4. Directional Movements:
    • Define possible ant movements: left, right, up, down.
  5. Queue for BFS:
    • Use a queue to keep track of the ant’s position and the time elapsed.
  6. Visit Tracking:
    • Keep track of visited positions to avoid cycles.

Code

from collections import deque

def time_to_boundary(width, height, x, y, steps):
    # Check if already at the boundary
    if x == 0 or x == width - 1 or y == 0 or y == height - 1:
        return 0

    # Possible directions (left, right, up, down)
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
    
    # Initialize the queue with the starting position and time = 0
    queue = deque([(x, y, 0)])
    visited = set((x, y))

    while queue:
        cx, cy, time = queue.popleft()
        
        if time > steps:
            break
        
        for dx, dy in directions:
            nx, ny = cx + dx, cy + dy
            
            if 0 <= nx < width and 0 <= ny < height and (nx, ny) not in visited:
                # Check if the next position is a boundary position
                if nx == 0 or nx == width - 1 or ny == 0 or ny == height - 1:
                    return time + 1
                
                # Mark the new position as visited and add to the queue
                visited.add((nx, ny))
                queue.append((nx, ny, time + 1))
    
    return steps

# Example Usage:
width = 5
height = 5
x = 2
y = 2
steps = 10

print(time_to_boundary(width, height, x, y, steps))  # Expected output should be the time when it hits the boundary based on steps allowed

Time Complexity

This algorithm efficiently simulates the ant’s movement and determines the first time it hits the boundary.

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