There is no single answer to this question since it depends on the size of the array and the number of combinations needed. However, here are some possible ways to efficiently sum the n-length combinations of an array:
def sum_combinations(arr, n):
if n == 0:
return [[]]
else:
result = []
for i in range(len(arr)):
subarr = arr[i+1:]
for combination in sum_combinations(subarr, n-1):
result.append([arr[i]] + combination)
return result
arr = [1, 2, 3, 4]
n = 2
combinations = sum_combinations(arr, n)
sums = [sum(combination) for combination in combinations]
print(sums)
This code generates all possible n-length combinations of the array arr
using recursion and then calculates their sums using a list comprehension. The time complexity of this approach is O(n * 2^n) since there are 2^n possible combinations and each combination has n elements that need to be summed.
def sum_combinations(arr, n):
table = [[0]*(n+1) for i in range(len(arr)+1)]
for i in range(1, len(arr)+1):
table[i][1] = arr[i-1]
for j in range(2, n+1):
if i > j:
table[i][j] = table[i-1][j] + table[i-1][j-1] + arr[i-1]
return table[len(arr)][n]
arr = [1, 2, 3, 4]
n = 2
sums = sum_combinations(arr, n)
print(sums)
This code generates the sum of all possible n-length combinations of the array arr
using dynamic programming. The time complexity of this approach is O(n * len(arr) * len(arr)) since we need to fill up an n x len(arr) table. However, it's much faster than the recursive approach for large values of n and len(arr).
sum_combinations(arr, n) = sum(arr) choose n
where "choose" denotes the binomial coefficient. Here's an example Python code for this approach:
import math
arr = [1, 2, 3, 4]
n = 2
sums = math.comb(sum(arr), n)
print(sums)
This code calculates the sum of all possible n-length combinations of the array arr
using the mathematical formula. The time complexity of this approach is O(1), but it only works for non-negative integers.
Asked: 2023-06-12 01:16:50 +0000
Seen: 13 times
Last updated: Jun 12 '23