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 {
    public int sumOddLengthSubarrays(int[] arr) {
        int n = arr.length;
        int[] f = new int[n];
        int[] g = new int[n];
        int ans = f[0] = arr[0];
        for (int i = 1; i < n; ++i) {
            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;
    }
}
