can you rewrite this pesudocode to partition in thirds instead of halves?RANDOMIZED$-$SELECT $(A,p,r,i)$

if $p==r$

return $A[p]$

$q=$ RANDOMIZED$-$PARTITION $(A,p,r)$

$k=q-p+1$

if $i==k,//$ the pivot value is the answer

return $A[q]$

elseif $(A,p,q-1,i)(A,q+1,r,i-k)(A,p,r)i=$RANDOM$(p,r)A[r]A[i](A,p,r)iA[p.r]A[pdotsr](A,p,r)x=A[r]i=p-1j=pr-1A[j]xi=i+1A[i]A[j]A[i+1]A[r]i+1i$

return RANDOMIZED$-$SELECT $(A,p,q-1,i)$

else return RANDOMIZED$-$SELECT $(A,q+1,r,i-k)$

RANDOMIZED$-$PARTITION $(A,p,r)$

$i=$RANDOM$(p,r)$

exchange $A[r]$ with $A[i]$

return PARTITION $(A,p,r)$

RANDOMIZED$-$SELECT returns the $ith$ smallest element $of$ the array $A[p.r].$

Name, SID, Date

The key $to$ the algorithm $is$ the PARTITION procedure, which rearranges the subar$-$

ray $A[pdotsr]in$ place.

PARTITION $(A,p,r)$

$x=A[r]$

$i=p-1$

for $j=ptor-1$

$ifA[j]x$

$i=i+1$

exchange $A[i]$ with $A[j]$

exchange $A[i+1]$ with $A[r]$

return $i+1$