How to find the max of continuous subarray of size k

Question

l = [2,1,5,1,3,2]

subarrays are [2,1,5], [1,5,1], [5,1,3], [1,3,2]

output is [5 ,1 ,3] which is 9

My code is

def maxcosu(arr,k):
    maxtotal = 0
    for i in range(len(arr)-k):
        total = 0
        for j in range(i, i+k):
            total = total + arr[j]
        maxtotal = max(maxtotal, total)
    return maxtotal
maxcosu([2,1,5,1,3,2] , 3)

But for [1,1,1,8,8,8] my expected output is 24, but I am getting 17.

I have gone through this link and the answer by zeeshan12396 is wrong for test case

Asked By: abd

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Source

Answer 1

Talking about your solution, I guess you should have passed len(arr)-k+1 instead of len(arr)-k to range builtin.

Here is one-liner solution using List-Comprehension to get continuous sub-array of length k for maximum sum:

>>> max((l[i:i+k] for i in range(len(l)-k+1)), key=sum)
[8, 8, 8]

And to get the maximum sum:

>>> max((sum(l[i:i+k]) for i in range(len(l)-k+1)))
24

Answered By: ThePyGuy

Answer 2

import math

arr =[1,1,1,8,8,8]
k = 3

list = []
iterations = len(arr) + 1 - k


for i in range(iterations):
    list.append(arr[0+i:k+i])
print(list)

suml = []

for l in list:
    suml.append(math.fsum(l))
print(suml)

max_val = max(suml)
print(max_val)

max_idx = suml.index(max_val)
print(max_idx)

correct_arr = list[max_idx]
print(correct_arr)

I think this is what you are looking for?

Answered By: Ryan Tandy

Answer 3

Following your logic, I have got:

def maxcosu(arr,k):
    sub_arr_sum = []
    for i in range(len(arr)-k+1):
        sub_arr_sum.append(sum(arr[i:i+k]))
    return max(sub_arr_sum)

Answered By: Jeremy

How to find the max of continuous subarray of size k

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