After the following code is run, what are the values of n$1$ and n$2?$

Note: we use some dictionary methods like dict.get$()$ and dict.setdefault$()$ in this code snippet because HuskyCT does not allow square brackets in question text. It is better in general to use square brackets for getting and setting items in a dictionary, unless you need the extra functionality provided by these methods.

def fewest$\_$coins$($amt$,$ coins$)$:

return $\_$fewest$\_$coins$($amt$,$ coins, dict$())$

def $\_$fewest$\_$coins$($amt$,$ coins, solved$)$:

$"""$Returns the minimum amount of coins needed to make amt"""

# Base case: we've solved this problem already

if amt in solved: return solved.get$($amt$)$

# Base case: we can make this amount with $1$ coin

elif amt in coins:

solved.setdefault$($amt$,1)$

return solved.get$($amt$)$

solved.setdefault$($amt$,$ float$(\text{'}$inf$\text{'}))$

# Explore every coin from here, find minimum

for coin in coins:

if coin $<$ amt:

n$\_$branch $=1+\_$fewest$\_$coins$($amt$-$coin, coins, solved$)$

if n$\_$branch $<$ solved.get$($amt$)$:

solved.pop$($amt$)$

solved.setdefault$($amt$,$ n$\_$branch$)=$

return solved.get$($amt$)$

L$1=$ list$((1,5,10,25))$

L$2=$ list$((1,5,10,20,25))$

n$1=$ fewest$\_$coins$(40,$ L$1)$

n$2=$ fewest$\_$coins$(40,$ L$2)$