问题
A list is defined as follows: [1, 2, 3]
and the sub-lists of this are:
[1], [2], [3],
[1,2]
[1,3]
[2,3]
[1,2,3]
Given K for example 3 the task is to find the largest length of sublist with sum of elements is less than equal to k.
I am aware of itertools
in python but it will result in segmentation fault for larger lists. Is there any other efficient algorithm to achieve this? Any help would be appreciated.
My code is as allows:
from itertools import combinations
def maxLength(a, k):
#print a,k
l= []
i = len(a)
while(i>=0):
lst= list(combinations(sorted(a),i))
for j in lst:
#rint list(j)
lst = list(j)
#print sum(lst)
sum1=0
sum1 = sum(lst)
if sum1<=k:
return len(lst)
i=i-1
回答1:
As far as I can see (since you treat sub array as any items of the initial array) you can use greedy algorithm with O(N*log(N))
complexity (you have to sort the array):
1. Assign entire array to the sub array
2. If sum(sub array) <= k then stop and return sub array
3. Remove maximim item from the sub array
4. goto 2
Example
[1, 2, 3, 5, 10, 25]
k = 12
Solution
sub array = [1, 2, 3, 5, 10, 25], sum = 46 > 12, remove 25
sub array = [1, 2, 3, 5, 10], sum = 21 > 12, remove 10
sub array = [1, 2, 3, 5], sum = 11 <= 12, stop and return
As an alternative you can start with an empty sub array and add up items from minimum to maximum while sum is less or equal then k
:
sub array = [], sum = 0 <= 12, add 1
sub array = [1], sum = 1 <= 12, add 2
sub array = [1, 2], sum = 3 <= 12, add 3
sub array = [1, 2, 3], sum = 6 <= 12, add 5
sub array = [1, 2, 3, 5], sum = 11 <= 12, add 10
sub array = [1, 2, 3, 5, 10], sum = 21 > 12, stop,
return prior one: [1, 2, 3, 5]
回答2:
You can use the dynamic programming solution that @Apy linked to. Here's a Python example:
def largest_subset(items, k):
res = 0
# We can form subset with value 0 from empty set,
# items[0], items[0...1], items[0...2]
arr = [[True] * (len(items) + 1)]
for i in range(1, k + 1):
# Subset with value i can't be formed from empty set
cur = [False] * (len(items) + 1)
for j, val in enumerate(items, 1):
# cur[j] is True if we can form a set with value of i from
# items[0...j-1]
# There are two possibilities
# - Set can be formed already without even considering item[j-1]
# - There is a subset with value i - val formed from items[0...j-2]
cur[j] = cur[j-1] or ((i >= val) and arr[i-val][j-1])
if cur[-1]:
# If subset with value of i can be formed store
# it as current result
res = i
arr.append(cur)
return res
ITEMS = [5, 4, 1]
for i in range(sum(ITEMS) + 1):
print('{} -> {}'.format(i, largest_subset(ITEMS, i)))
Output:
0 -> 0
1 -> 1
2 -> 1
3 -> 1
4 -> 4
5 -> 5
6 -> 6
7 -> 6
8 -> 6
9 -> 9
10 -> 10
In above arr[i][j]
is True
if set with value of i
can be chosen from items[0...j-1]
. Naturally arr[0]
contains only True
values since empty set can be chosen. Similarly for all the successive rows the first cell is False
since there can't be empty set with non-zero value.
For rest of the cells there are two options:
- If there already is a subset with value of
i
even without consideringitem[j-1]
the value isTrue
- If there is a subset with value of
i - items[j - 1]
then we can add item to it and have a subset with value ofi
.
回答3:
Look, for generating the power-set it takes O(2^n) time. It's pretty bad. You can instead use the dynamic programming approach.
Check in here for the algorithm. http://www.geeksforgeeks.org/dynamic-programming-subset-sum-problem/
And yes, https://www.youtube.com/watch?v=s6FhG--P7z0 (Tushar explains everything well) :D
来源:https://stackoverflow.com/questions/41284951/largest-subset-whose-sum-is-less-than-equal-to-a-given-sum