Vocabulary - Block Size

The block size $B$ determines your time complexity. For most problems, $B = \lfloor\sqrt{n}\rfloor$ is optimal. You'll have $\lceil n/B \rceil$ blocks.

The last block might be smaller than $B$ if $n$ isn't perfectly divisible. Given index $i$, its block number is $\lfloor i/B \rfloor$.

Elements $i$ and $j$ are in the same block if $\lfloor i/B \rfloor = \lfloor j/B \rfloor$. Block $k$ covers indices from $k \cdot B$ to $\min((k+1) \cdot B - 1, n-1)$.

This indexing math is the foundation. Every sqrt decomposition algorithm uses it.