To find max continuous subarray sum of size M

Question

I am new to the competitive programming. I am finding trouble in doing the following problem.

The question is that you have given an array or list. And a number M, now you hav to find the continuous subarray of size M having the largest sum.

For example if list is 4,6,10,8,2,1 and M=3 then largest sum window will be 6,8,10 that is sum equal 24 . So answer will be 24

Can anyone help me regarding this question?

Asked By: MagnusEffect

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Source

Answer 1

You can edit the list and remove the largest number successively:

list = [4,6,10,8,2,1]
M = 3
result = 0
# to get unique elements
new_list = set(list)

for i in range(M):
   result += max(new_list)
   # removing the largest element from new_list
   new_list.remove(max(new_list))

print(result)

Answered By: PluggeRo

Answer 2

l=[4,6,10,8,2,1]
ans=0
m=3
for i in range(len(l)+1):
   for j in range(i):
       if len(l[j:i])==m:
          ans=max(ans,sum(l[j:i]))
print(ans)

find the sublist of list and store sum of sublist in variable

Answered By: zeeshan12396

Answer 3

First think of a brute force solution. Here it is simple we can find the answer using two for loops(nested). Outer loop will mark as starting point and inner loop will go to next M elements.
Now comes the hard part i.e., optimization. Since it is a continuous subarray of fixed size you can make window of fixed size(here M) using two pointer(say left and right) and maintain the size of the window and keep moving towards right and keep calculating the sum and updating the ans if required.

    sum = 0
    for i in range(M):
        sum+= arr[i]
 
    ans = sum
    for i in range(M, len(arr)):
        sum += arr[i] - arr[i-M]
        ans = max(res, sum)
 
    return ans

Of course, we have to check for some corner cases like if M is greater than size of array or M=0

Answered By: Vikash

Answer 4

I tried this solution:

l = [4,6,1,8,2,1]
M = 3

largest_sum = -1
output = []
# Getting all possible sequences of M numbers
for i in range(len(l)):
    if (res:=sum(l[i:i+3])) > largest_sum:
        largest_sum = res
        output = l[i:i+3]

print(f"Largest sum is {largest_sum} from array {output}")

Output:

Largest sum is 15 from array [6, 1, 8]

It might not be the most efficient algorithm, but it works. I get all possible sequences o length M and store the largest sum.

Note: it works with @zeeshan12396 case too.

Answered By: Leo Eiji

Answer 5

Here is one-liner solution combining max and sum builtins in a List-Comprehension:

>>> l=[4,6,10,8,2,1]
>>> M=3
>>> max((l[i:i+M] for i in range(len(l)-M+1)), key=sum)
# Continuous sub array with maximum sum
[6, 10, 8]

>>> max((sum(l[i:i+M]) for i in range(len(l)-M+1)))
# Maximum sum for continuous sub array
24

The idea is to iterate from 0 to the maximum index the sub-array can be created from, then take the list slices starting with each iteration number comprising of M items, then get the sub array based on maximum sum of the values in these sub arrays.

Answered By: ThePyGuy

To find max continuous subarray sum of size M

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