I'm confused about why this solution exceeds the two-second limit. The grid size N is at most 2000, and there can be up to 100,000 updates. I estimated the setup as O(N²) and each update as O(1), so I expected roughly O(N² + U), which seemed manageable. The code checks groups of four symmetric cells, then processes each update by toggling one cell and recalculating the required number of changes. Is my complexity analysis wrong, or is Python overhead causing the timeout?
3 Answers
There is another major performance problem: calling print() once for every update. That can produce 100,000 separate output operations, which is unnecessarily slow. Store the answers in a list of strings and write them all at once with sys.stdout.write(), or use an output buffer.
The broad complexity estimate is close, but the constants and repeated work matter a lot here. With N = 2000, the initialization loop still visits about one million groups. For every group you create a list, append four values, and call count() several times, so each iteration performs multiple scans and Python-level operations. That is much slower than treating it as one abstract O(1) operation. The update phase is closer to O(U), but the initialization and output overhead can still exceed the limit.
Big-O notation only describes how the running time scales; it does not guarantee that a particular input size will finish within a time limit. The setup has roughly N² work, and for N = 2000 that is about a million iterations before accounting for all the operations inside each iteration. Avoid rebuilding temporary lists and repeatedly calling count(). Count the four cell values directly or use a small lookup table, and make sure the initial pass and all output are optimized.
Also, the often-quoted hundreds of millions of operations per second is generally for optimized C++, not regular Python. CPython may only manage around 10 million simple operations per second, and list creation, indexing, counting, and nested loops are not free.