algoadvance

You are given R rows and C columns representing a matrix. Initially, you start from the cell (rStart, cStart). Your task is to return a list of coordinates representing the cells in the order you visit them in a spiral order.

Example:

Input: R = 1, C = 4, rStart = 0, cStart = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Constraints:

• 1 <= R <= 100
• 1 <= C <= 100
• 0 <= rStart < R
• 0 <= cStart < C

Clarifying Questions

1. Bounds Check: Do we need to handle scenarios where the traversal may go out of the bounds of the matrix?
   • Yes, we should ensure that the spiral order proceeds only within the boundaries of the given matrix.
2. Output Order: Should the coordinates be outputted in the exact order they’re visited in the spiral?
   • Yes, the coordinates should be in the exact order they’re visited.
3. Traversal Details: Should we start by traversing right and then follow the traditional spiral (right, down, left, up)?
   • Yes, typically we start by moving right and follow the order right, down, left, and up repetitively.

Strategy

1. Initialization:
   • Define direction vectors for right, down, left, and up movements.
   • Start from the initial position (rStart, cStart).
2. Spiral Traversal:
   • Use a step increment approach: move a certain number of steps in the current direction, add all valid coordinates to the result, then change direction.
   • After every two directions (one horizontal, one vertical), increase the number of steps to be taken by 1.
3. Bounds Check:
   • Only add coordinates that are within the matrix boundaries to the result list.
4. Termination:
   • Stop once the result list contains R * C coordinates.

Code

def spiralMatrixIII(R, C, rStart, cStart):
    # Direction vectors for moving right, down, left, and up
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    
    result = []
    x, y = rStart, cStart
    steps = 1

    result.append([x, y])
    if R * C == 1:  # Edge case where the matrix is 1x1
        return result
    
    while len(result) < R * C:
        for direction in directions:
            for _ in range(steps):
                x += direction[0]
                y += direction[1]
                if 0 <= x < R and 0 <= y < C:
                    result.append([x, y])
            if direction == directions[1] or direction == directions[3]:
                steps += 1
                
    return result

Time Complexity

• Time Complexity: O(R * C)
  • Because every cell in the matrix is visited exactly once.
• Space Complexity: O(R * C)
  • The space complexity is directly related to storing the result, which is the list of coordinates.

This ensures an optimal traversal while guaranteeing that the spiral order is maintained and compliance with matrix boundaries is enforced.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?