<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>

class Solution {
public:
    int sumOddLengthSubarrays(vector<int>& arr) {
        vector<int> presum(arr.size());
        presum[0] = arr[0];
        int ans = 0;

        for (int i = 1; i < arr.size(); i++) {
            presum[i] = presum[i - 1] + arr[i];
        }

        for (int i = 0; i < arr.size(); i++) {
            int x = i;
            if (x % 2 == 0) ans += presum[i];
            x--;
            for (int j = x; j>= 0; j = j - 2) {
                ans += presum[i] - presum[j];
            }
        }
        return ans;
    }
};

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