Sum of All Odd Length Subarrays
<p>Given an array of positive integers <code>arr</code>, return <em>the sum of all possible <strong>odd-length subarrays</strong> of </em><code>arr</code>.</p>

<p>A <strong>subarray</strong> is a contiguous subsequence of the array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,4,2,5,3]
<strong>Output:</strong> 58
<strong>Explanation: </strong>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</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [10,11,12]
<strong>Output:</strong> 66
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 100</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<p>Could you solve this problem in O(n) time complexity?</p>
Array; Math; Prefix Sum
class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        n = len(arr)
        f = [0] * n
        g = [0] * n
        ans = f[0] = arr[0]
        for i in range(1, n):
            f[i] = g[i - 1] + arr[i] * (i // 2 + 1)
            g[i] = f[i - 1] + arr[i] * ((i + 1) // 2)
            ans += f[i]
        return ans
