<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>

class Solution {

public:

    int longestSubarray(vector<int>& nums) {

        // Sliding Window 

        // int zero=0;

        // int i=0,j=0,res=0;

        // while(j<nums.size()){

        //     if(nums[j]==0)

        //         zero++;

        //     if(zero>1){

        //         while(nums[i++]!=0);

        //         zero--;

        //     }

        //     res=max(res,j-i);                

        //     j++;

        // }

        // return res;

        // DP

        int n=nums.size(),mx=0;

        int left[n];memset(left,0,sizeof(left));

        int right[n];memset(right,0,sizeof(right));

        for(int i=1;i<n;i++){

            if(nums[i-1]==1)

            left[i]=left[i-1]+1;

        }

        for(int i=n-2;i>=0;i--){

            if(nums[i+1]==1)

            right[i]=right[i+1]+1;

        }

        for(int i=0;i<n;i++){

     

            mx=max(mx,left[i]+right[i]);

        }

        return mx;

    }

};
   
        
    

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