<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) {
        
        
        int n=nums.size()-1;
        if(x<min(nums[0],nums[n]))return -1;
        
        vector<int>leftr(n+1),rightr(n+1);
        leftr[0]=nums[0],rightr[0]=nums[n];
        for(int i=1;i<=n;i++)leftr[i]=leftr[i-1]+nums[i];
        for(int i=n-1;i>=0;i--)rightr[n-i]=rightr[n-i-1]+nums[i];
        int ans=INT_MAX;
        int a=lower_bound(leftr.begin(),leftr.end(),x)-leftr.begin();
        int b=lower_bound(rightr.begin(),rightr.end(),x)-rightr.begin();
        if(a!=n+1)if(leftr[a]==x)ans=min(ans,a+1);
        if(b!=n+1)if(rightr[b]==x)ans=min(ans,b+1);
        
        int j=0;
        if(rightr[n]<x)return -1;
        while(j<=n)
        {
           
            if(x<leftr[j])break;
            else
            {
                int c=lower_bound(rightr.begin(),rightr.end(),x-leftr[j])-rightr.begin();
                if(c!=n+1)if(leftr[j]+rightr[c]==x)ans=min(c+1+j+1,ans);
            }
            j++;
        }
        return (ans==INT_MAX)?-1:ans;
    }
};
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