 Solution

For the flattened Boolean array, the abstraction function selects the true cells. A suitable representation invariant requires the occupied cells to form a order ideal: an occupied $(x,y)$ implies occupancy of every $(i,j)$ with $i\le x,j\le y$. Deleting a northeast rectangle preserves that property.

For column heights, the invariant is $N\ge h(0)\ge h(1)\ge \cdots \ge h(N-1)\ge 0$, and the abstraction function is $B=\{(x,y):0\le x<N,0\le y<h(x)\}$. A legal move replaces every $h(i)$ for $i\ge x$ by $\min (h(i),y)$, preserving the invariant. This is an order-ideal board representation.

For a fair comparison, allow either representation a scalar cached token count: Count and Legal then take $O(1)$ in both. Without that cache, counting needs $\Theta (N^2)$ Boolean inspections versus $\Theta (N)$ height additions. A move takes $O(N^2)$ in the worst case in the flattened representation and $O(N)$ with heights. Displaying the full square takes $\Theta (N^2)$ character positions in either. Heights use $\Theta (N)$ array entries rather than $\Theta (N^2)$, so they are superior for the requested operations; their initialization also takes only $\Theta (N)$ time.