Longest Subarray of 1's After Deleting One Element
<p>Given a binary array <code>nums</code>, you should delete one element from it.</p>

<p>Return <em>the size of the longest non-empty subarray containing only </em><code>1</code><em>&#39;s in the resulting array</em>. Return <code>0</code> if there is no such subarray.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,0,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1&#39;s.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,1,1,0,1,1,0,1]
<strong>Output:</strong> 5
<strong>Explanation:</strong> After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1&#39;s is [1,1,1,1,1].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You must delete one element.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Dynamic Programming; Sliding Window
class Solution {
public:
    int longestSubarray(vector<int>& nums) {
        int n = nums.size();
        vector<int> left(n + 1);
        vector<int> right(n + 1);
        for (int i = 1; i <= n; ++i) {
            if (nums[i - 1]) {
                left[i] = left[i - 1] + 1;
            }
        }
        for (int i = n - 1; ~i; --i) {
            if (nums[i]) {
                right[i] = right[i + 1] + 1;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = max(ans, left[i] + right[i + 1]);
        }
        return ans;
    }
};