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Given the integers R and C representing the number of rows and columns in a matrix, and the integers r0 and c0 representing the coordinates of a starting cell (r0, c0), return the matrix cells in distance order from (r0, c0). The cells are ordered by their Manhattan distance, which is defined as:

[ \text{Manhattan Distance} = r1 - r0 + c1 - c0 ]

Clarifying Questions

  1. Q: Can the starting cell (r0, c0) be outside the boundary of the matrix?
    • A: No, (r0, c0) will always be within the boundary of the matrix.
  2. Q: Is the output expected to be a list of lists where each inner list represents the coordinates of a cell?
    • A: Yes, the output should be a list of coordinate pairs, ordered by their Manhattan distance from (r0, c0).
  3. Q: Are there any constraints on the values of R and C?
    • A: No specific constraints are mentioned, but typical constraints are that R and C are within standard problem limits (0 ≤ R, C ≤ 100).

Strategy

  1. Generate All Cells: Create a list of coordinates for all cells in the matrix.
  2. Calculate Manhattan Distance: Write a function or a lambda expression that calculates the Manhattan distance of any cell (r, c) from the starting cell (r0, c0).
  3. Sort the Cells: Sort the list of coordinates based on their calculated Manhattan distance.
  4. Return the Sorted List: Return the sorted list of coordinates.

Code

def allCellsDistOrder(R, C, r0, c0):
    # Step 1: Generate all coordinates in the matrix
    cells = [(r, c) for r in range(R) for c in range(C)]
    
    # Step 2: Define a function to calculate Manhattan distance
    def manhattanDist(cell):
        r, c = cell
        return abs(r - r0) + abs(c - c0)
    
    # Step 3: Sort the list of cells based on their Manhattan distance
    cells.sort(key=manhattanDist)
    
    # Step 4: Return the sorted list of coordinates
    return cells

# Example usage:
R, C, r0, c0 = 2, 3, 1, 2
print(allCellsDistOrder(R, C, r0, c0))

Time Complexity

The time complexity of this solution is as follows:

  1. Generating all coordinates: O(R * C) because we iterate through each cell in the matrix.
  2. Sorting the cells: O((R * C) log (R * C)) due to the sorting algorithm.

Therefore, the overall time complexity is:

[ \mathbf{O(R \cdot C) + O((R \cdot C) \log (R \cdot C))} = \mathbf{O((R \cdot C) \log (R \cdot C))} ]

This is efficient for the input size constraints typically expected in such problems.

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