Maximum Sum Sublist

Finding the subarray with the Maximum sum in the given ArrayList

Here is a more concise solution

private List<Integer> solution(List<Integer> nums) {
    int biggestSumSoFar = Integer.MIN_VALUE;
    List<Integer> biggestSubListSoFar = new ArrayList<>();
    for (int left = 0; left < nums.size(); ++left) {
        for (int right = left + 1; right < nums.size(); ++right) {
            List<Integer> currentSubList = subListSum(nums, left, right);
            int currentSum = sum(currentSubList);
            if (currentSum > biggestSumSoFar) {
                biggestSumSoFar = currentSum;
                biggestSubListSoFar = currentSubList;
            }
        }
    }
    return biggestSubListSoFar;
}

private List<Integer> subListSum(final List<Integer> nums, final int left, final int right)
{
    final List<Integer> sublist = new ArrayList<>();
    for (int i = left; i < right; i++)
    {
        sublist.add(nums.get(i));
    }
    return sublist;
}

private int sum(List<Integer> arr) {
    int sum = 0;
    for(int a : arr){
        sum += a;
    }
    return sum;
}

Find maximum sum of sublist in list of positive integers under O(n^2) of specified length Python 3.5

1) Find sum current of first w elements, assign it to best.

2) Starting from i = w: current = current + lst[i]-lst[i-w], best = max(best, current).

3) Done.

Maximum sum of sublist with a specific length

Let's clarify to make sure we're on the same page.

Inputs: 1) a list li of digits; 2) n

Output: the slice from li of length n that has maximal sum.

li = [4,2,1,7,1,3,8,4,7,8,1]

n = 2

slices = (li[x:x+n] for x in range(len(li)-n+1))

max(map(sum,slices))
Out[113]: 15

To find max continuous subarray sum of size M

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)

Index of the highest sum() of sublists

Since you explicitly asked for a one-liner, here it is:

>>> list(map(sum, my_list)).index(max(map(sum, my_list)))
4

This line creates a list of sums using the map function and finds a value in said list using index. The value that is searched for is the max value of the list of sums mentioned earlier.

Note that this computes the sums of my_list's elements twice, which isn't necessary. Solution without the one-line requirement:

>>> sum_list = list(map(sum, my_list))
>>> sum_list.index(max(sum_list))
4

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