Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.

A subarray is a contiguous subsequence of the array.

Return the sum of all odd-length subarrays of arr.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66


Constraints:

• 1 <= arr.length <= 100
• 1 <= arr[i] <= 1000

## Solution 0: Brute Force

Enumerate all odd length subarrys: O(n^2), each take O(n) to compute the sum.

Total time complexity: O(n^3)
Space complexity: O(1)

## Solution 1: Running Prefix Sum

Reduce the time complexity to O(n^2)

## Solution 2: Math

Count how many times arr[i] can be in of an odd length subarray
we chose the start, which can be 0, 1, 2, … i, i + 1 choices
we chose the end, which can be i, i + 1, … n – 1, n – i choices
Among those 1/2 are odd length.
So there will be upper((i + 1) * (n – i) / 2) odd length subarrays contain arr[i]

ans = sum(((i + 1) * (n – i) + 1) / 2 * arr[i] for in range(n))

Time complexity: O(n)
Space complexity: O(1)

## C++

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