<p>You are given an integer array <code>nums</code> and an integer <code>x</code>. In one operation, you can either remove the leftmost or the rightmost element from the array <code>nums</code> and subtract its value from <code>x</code>. Note that this <strong>modifies</strong> the array for future operations.</p>

<p>Return <em>the <strong>minimum number</strong> of operations to reduce </em><code>x</code> <em>to <strong>exactly</strong></em> <code>0</code> <em>if it is possible</em><em>, otherwise, return </em><code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,4,2,3], x = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> The optimal solution is to remove the last two elements to reduce x to zero.
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,6,7,8,9], x = 4
<strong>Output:</strong> -1
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,20,1,1,3], x = 10
<strong>Output:</strong> 5
<strong>Explanation:</strong> The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
</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>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= x &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        vector<int>ans;
        ans=nums;
        int i,j,l=0,r=0,chk=0,t=2*ans.size()+1;
        for(i=0;i<nums.size();i++){ans.push_back(nums[i]);chk+=nums[i];}
        if(chk<x)return -1;
        chk=0;
        // for(i=0;i<ans.size();i++)cout<<ans[i]<<" ";
        // cout<<endl;
        while(r<ans.size()){
            
            if(chk>x){
                chk-=ans[l];
                l++;
            }else if(chk<x){
                chk+=ans[r];
                r++;
            }else{
                r--;
                // cout<<chk<<"  l "<<l<<" r "<<r<<endl;
                if(l<=nums.size()-1 && r>=nums.size()-1 || l<=nums.size() && r>=nums.size()   ){
                    t=min(t,r-l+1);
                }
                r++;
                chk+=ans[r];
                r++;
            }
        }
        if(t==2*nums.size()+1){
            t=-1;
        }
        return t;
    }
};
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