Medium
Inverse Coin Change — Python
Full explanation · Time O(n^2) · Space O(1)
# Time: O(n^2)
# Space: O(1)
# dp
class Solution(object):
def findCoins(self, numWays):
"""
:type numWays: List[int]
:rtype: List[int]
"""
result = []
for i in xrange(1, len(numWays)+1):
if numWays[i-1] == 1:
result.append(i)
for j in reversed(xrange(i, len(numWays)+1)):
numWays[j-1] -= numWays[(j-i)-1] if (j-i)-1 >= 0 else 1
if numWays[i-1]:
return []
return result
# Time: O(n^2)
# Space: O(n)
# dp
class Solution2(object):
def findCoins(self, numWays):
"""
:type numWays: List[int]
:rtype: List[int]
"""
result = []
dp = [0]*(len(numWays)+1)
dp[0] = 1
for i in xrange(1, len(numWays)+1):
if numWays[i-1]-dp[i] == 1:
result.append(i)
for j in xrange(i, len(numWays)+1):
dp[j] += dp[j-i]
if numWays[i-1]-dp[i]:
return []
return result