Given a target amount n and a list (array) of distinct coin values, what’s the fewest coins needed to make the change amount.
For example:
If n = 10 and coins = [1,5,10]. Then there are 4 possible ways to make change:
With 1 coin being the minimum amount.
Implement your solution below:
def rec_coin(target,coins):
min_coins = target
if target in coins:
return 1
else:
for value in [ c for c in coins if c<= target]:
num_coins = rec_coin(target-value, coins) + 1
min_coins = min(num_coins, min_coins)
return min_coins
pass
rec_coin(10,[1,5])
2
rec_coin(63,[1,5,10,25])
6
"""
RUN THIS CELL TO TEST YOUR FUNCTION.
NOTE: NON-DYNAMIC FUNCTIONS WILL TAKE A LONG TIME TO TEST.
"""
from nose.tools import assert_equal
class TestCoins(object):
def check(self,solution):
coins = [1,5,10,25]
assert_equal(solution(45,coins),3)
assert_equal(solution(23,coins),5)
assert_equal(solution(74,coins),8)
print ('Passed all tests.')
# Run Test
test = TestCoins()
test.check(rec_coin)
Recommended: Introduction to Recursion
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