I have an N×N grid containing '.' and '#'. Cells that are rotations or reflections of one another form groups of up to four positions, and I need to calculate the minimum number of changes required to make every group contain the same character. After that, there are U updates, where each update toggles one cell and I must print the new minimum after each toggle.
My initial calculation loops through the four-cell symmetry groups, and each update examines only the affected group. The constraints allow as many as 100,000 updates, but my Python solution still exceeds the time limit. Is there a more efficient way to organize the calculation or update the affected groups?
2 Answers
Your update loop is already close to constant time because it only inspects four cells. The expensive work is the initial grid copy and the O(N²) symmetry scan, which is normally unavoidable if the whole grid must be read. Avoid unnecessary copies and repeated list construction where possible, and use fast input if the input size is large. Also make sure the problem's actual constraints permit an N×N grid—if N itself is very large, storing and reading that many cells will dominate the runtime regardless of the number of updates.
The key idea is to treat each set of symmetric cells as an independent group. For a group containing some dots and hashes, its contribution to the answer is min(number_of_dots, number_of_hashes). Compute that contribution once for every group during the initial O(N²) pass.
When a cell is toggled, find its symmetric group, subtract that group's old contribution, update the character, recalculate the contribution for just those four cells, and add the new contribution. Since each update touches only a constant-sized group, every update is O(1), making the total complexity O(N² + U), rather than rebuilding the entire answer after every query.
You do not need to keep a separate count of all groups needing one or two changes. A single running total is enough: remove the old min(dot_count, hash_count), toggle the cell, then add the new value.