Recursion: Make Changes With Fewest Coins

I'm studing the data structure and algorithm in python. Here is the classic problem of recursion involving making change with the fewest coins. Here are the codes. What I do not un

Solution 1:

minCoins = change: minCoins is initialized with the value of change which is the maximum value that recMC can return as the minimum value of a coin is 1, assuming integer values of coins.

if change in coinValueList: return 1: base case 1 - if some coin has a value of that of change we just need to grab this 1 coin, thus returning 1

for i in [c for c in coinValueList if c <= change]:: The function then loops through all possible values of 1 single coin to

numCoins = 1 + recMC(coinValueList,change-i): deduct the coin value i from change, add 1 coin to the number of coins needed which is recursively calculated for the leftover change (change-i). This works inductively from the 2 base cases

if numCoins < minCoins: minCoins = numCoins inside this loop effectively assigns the smallest number of coins possible to minCoins

If change still had the initialized value of minCoins (implying no value c in coinValueList satisfies c <= change), this means that the number of coins needed is also the value of change, i.e. change 1-unit coins. This is base case 2 that is based on the pre-condition that the coin with value 1 is always available.