Minimize Grid Inconvenience by Adding One Delivery Center

A binary grid represents delivery coverage: 1 marks a delivery center and 0 marks an ordinary location. The distance between cells (r1, c1) and (r2, c2) is the Chebyshev distance max(abs(r1 - r2), abs(c1 - c2)).

The grid's inconvenience is the largest distance from any ordinary location to its nearest delivery center. You may convert at most one 0 cell into a delivery center. Return the minimum possible inconvenience after this optional conversion.

Function Contract

Implement getMinInconvenience(grid) -> int, where grid is a rectangular integer matrix.

Return the optimal inconvenience value, not the position of the new center.

Constraints and Clarifications

• 1 <= number of rows <= 500.
• 1 <= number of columns <= 500.
• Every entry is 0 or 1.
• Existing delivery centers remain in place.
• Distances depend only on coordinates; there are no obstacles or path restrictions.
• The grid may initially contain no delivery center. In that case, placing one center is allowed and makes the final distances well defined.
• As an explicit edge-case convention, a final grid with no ordinary locations has inconvenience 0.
• Adding no center is permitted when the existing arrangement is already optimal.

Examples

grid = [[0, 0, 0, 1],
 [0, 0, 0, 1]]
Output: 1

Converting (0, 0) to a center leaves every ordinary location within Chebyshev distance 1 of a center. Zero inconvenience is impossible because ordinary locations remain.

grid = [[0, 0, 0, 0],
 [0, 0, 0, 0],
 [0, 0, 0, 0]]
Output: 2

Placing the center at (1, 1) achieves inconvenience 2. No single center can be within distance 1 of both the first and last columns.