LEFT$($i$)$

return $2$i

RIGHT$($i$)$

return $2$i $+1$

MaxHeapify$($A$,$ i$)$

l $=$ LEFT$($i$)$;

r $=$ RIGHT$($i$)$;

if l $<=$ A$.$heap$-$size and A$[$l$]>$ A$[$i$]$

largest $=$ l;

else

largest $=$ i;

if l $<=$ A$.$heap$-$size and A$[$r$]>$ A$[$largest$]$

largest $=$ r;

if largest $!=$ i

exchange A$[$i$]$ with A$[$largest$]$

MaxHeapify$($A$,$ largest$)$

BuildMaxHeap$($A$)$

A$.$heap$-$size $=$ A$.$length

for i $=$ floor$($A$.$length$/2)$ downto $1$

MaxHeapify$($A$,$ i$)$

If we modify the for loop in BuildMaxHeap$($A$)$ to increase from $1$ to floor$($A$.$length$/2)$ rather than decrease from floor$($A$.$length$/2)$ to $1,$ would the algorithm work?

Group of answer choices

True

False