Follow the loop$-$based algorithm time complexity analysis guideline and discuss with your team members about time complexity for the following pseudocodes. Note that in pseudocode, we ignore the data type declaration.

Find the matching runtime complexity for each psedudocode:

A$)$

findMin$($x$,$ y$)\{$

if $($x $<$ y$)$

return x

else

return y

$\}$

B$)$

LinearSearch$($numbers$,$ N$,$ key$)\{$

for $($i $=0$; i $<$ N; $++$i$)$

if $($numbers$[$i$]==$ key$)$

return i

return $-1//$ not found

$\}$

C$)$

SelectionSort$($number$,$ N$)\{$

for $($i $=0$; i $<$ N; $++$i$)\{$

indexSmallest $=$ i

for $($j $=$ i $+1$; j $<$ N; $++$j$)$

if $($numbers$[$j$]<$ numbers$[$indexSmallest$])$

indexSmallest $=$ j

temp $=$ numbers$[$i$]$

numbers$[$i$]=$ numbers$[$indexSmallest$]$

numbers$[$indexSmallest$]=$ temp

$\}$

$\}$

D$)$

BinarySearch$($sortedNumbers$,$ N$,$ key$)\{$

mid $=0$

low $=0$

high $=0$

high $=$ N $-1$

while $($high $>=$ low$)\{$

mid $=($high $+$ low$)/2$

if $($sortedNumbers $[$mid$]<$ key$)$

low $=$ mid $+1$

else if $($sortedNumbers $[$mid$]>$ key$)$

high $=$ mid $-1$

else

return mid

$\}$

return $-1//$ not found

$\}$

E$)$

for $($int i $=0$; i $<$ n; i$++)$

for $($int j $=1$; j $<$ n; j $=$ j $*2)$

System.out.println$("$hello$")$;

F$)$

for $($int i $=0$; i $<$ n; i $=$ i $+2)$

for $($int j $=1$; j $<$ n; j $=$ j $*3)$

System.out.println$("$hello$")$;

G$)$

for $($int i $=1$; i $<$ n; i $=$ i $*2)$

for $($int j $=1$; j $<$ n; j $=$ j $*3)$

System.out.println$("$hello$")$;

Group of answer choices

A

O$(1)$

B

O$($N$)$

C

O$($N $*$ N$)$

D

O$($ N $*$ log N$)$

E

O$($ log N$)$

F

O$($N$)$

G

O$($N$)$