Medium
Maximum and Minimum Sums of at Most Size K Subsequences — Python
Full explanation · Time O(nlogn) · Space O(n)
# Time: O(nlogn)
# Space: O(n)
# sort, combinatorics, two pointers, sliding window
MOD = 10**9+7
FACT, INV, INV_FACT = [[1]*2 for _ in xrange(3)]
def nCr(n, k):
if n < k:
return 0
while len(INV) <= n: # lazy initialization
FACT.append(FACT[-1]*len(INV) % MOD)
INV.append(INV[MOD%len(INV)]*(MOD-MOD//len(INV)) % MOD) # https://cp-algorithms.com/algebra/module-inverse.html
INV_FACT.append(INV_FACT[-1]*INV[-1] % MOD)
return (FACT[n]*INV_FACT[n-k] % MOD) * INV_FACT[k] % MOD
class Solution(object):
def minMaxSums(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
MOD = 10**9+7
nums.sort()
result = 0
cnt = 1
for i in xrange(len(nums)):
result = (result+(nums[i]+nums[~i])*cnt)%MOD
cnt = (cnt*2-nCr(i, k-1)) % MOD
return result
# Time: O(nlogn + n * k)
# Space: O(n)
# sort, combinatorics
class Solution2(object):
def minMaxSums(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
MOD = 10**9+7
nums.sort()
result = 0
cnt = 1
for i in xrange(len(nums)):
cnt = reduce(lambda accu, x: (accu+x)%MOD, (nCr(i, j) for j in xrange(min(i, k-1)+1)), 0)
result = (result+(nums[i]+nums[~i])*cnt)%MOD
return result