Hard
Number of Great Partitions — Python
Full explanation · Time O(n * k) · Space O(k)
# Time: O(n * k)
# Space: O(k)
# knapsack dp
class Solution(object):
def countPartitions(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
MOD = 10**9+7
if sum(nums) < 2*k:
return 0
dp = [0]*k
dp[0] = 1
for x in nums:
for i in reversed(xrange(k-x)):
dp[i+x] = (dp[i+x]+dp[i])%MOD
return (pow(2, len(nums), MOD)-2*reduce(lambda total, x: (total+x)%MOD, dp, 0))%MOD