\KwIn{A matrix $A$ of size $n \times m$}
    \KwOut{A column index $i$ representing the location of the tallest building.}
    \tcc{Initialize location variable.}
    $tallestLocation \gets 0$

    $tallestHeight \gets 0$

    \For{$i \in \{0,\ldots,m-1\}$}{
        \tcc{Use binary search to get the height of the column and compare it to the current tallest height}
        \tcc{Now choose an index p to partition around}
	    $p \gets  \lfloor n/2 \rfloor$

	    \While{$p \leq n-1$ \text{and} $p \geq 0$} {
	        \If {$A[i, p] = 0$} {
	            \tcc{Height found}
	            \If {the number above = 1 OR $p = n-1$ OR $p=0$} {
	                \If {$p \geq tallestHeight$} {
	                    $tallestHeight \gets j$

	                    $tallestLocation \gets i$
	                }
	                break out of while loop
	            }
	            \Else {
	                \tcc{Set new partition point}
	                $p \gets midpoint of $
	            }
	        }
	    }
    }
    \caption{\textsc{findTallest1} returns the location of the tallest building.}\label{alg:alg2}