2017-10-25 01:56:48 +00:00
|
|
|
"""
|
|
|
|
You have m types of coins available in infinite quantities
|
|
|
|
where the value of each coins is given in the array S=[S0,... Sm-1]
|
|
|
|
Can you determine number of ways of making change for n units using
|
2017-10-25 07:58:46 +00:00
|
|
|
the given types of coins?
|
2017-10-25 01:56:48 +00:00
|
|
|
https://www.hackerrank.com/challenges/coin-change/problem
|
|
|
|
"""
|
2017-11-25 09:23:50 +00:00
|
|
|
from __future__ import print_function
|
2019-02-20 16:54:26 +00:00
|
|
|
|
|
|
|
|
2017-10-25 01:56:48 +00:00
|
|
|
def dp_count(S, m, n):
|
2019-02-20 16:54:26 +00:00
|
|
|
|
|
|
|
# table[i] represents the number of ways to get to amount i
|
2017-10-25 01:56:48 +00:00
|
|
|
table = [0] * (n + 1)
|
|
|
|
|
2019-02-20 16:54:26 +00:00
|
|
|
# There is exactly 1 way to get to zero(You pick no coins).
|
2017-10-25 01:56:48 +00:00
|
|
|
table[0] = 1
|
|
|
|
|
|
|
|
# Pick all coins one by one and update table[] values
|
|
|
|
# after the index greater than or equal to the value of the
|
|
|
|
# picked coin
|
2019-02-20 16:54:26 +00:00
|
|
|
for coin_val in S:
|
|
|
|
for j in range(coin_val, n + 1):
|
|
|
|
table[j] += table[j - coin_val]
|
2017-10-25 01:56:48 +00:00
|
|
|
|
|
|
|
return table[n]
|
|
|
|
|
2019-02-20 16:54:26 +00:00
|
|
|
|
2017-10-25 01:56:48 +00:00
|
|
|
if __name__ == '__main__':
|
2017-11-25 09:23:50 +00:00
|
|
|
print(dp_count([1, 2, 3], 3, 4)) # answer 4
|
|
|
|
print(dp_count([2, 5, 3, 6], 4, 10)) # answer 5
|