<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(),sum=0,mini=INT_MAX,i;
        vector<int> prfx(n);
        vector<int> sufx(n);
        for(i=0;i<n;i++){
            sum+=nums[i];prfx[i]=sum;
        }
        sum=0;
        for(i=n-1;i>=0;i--){
            sum+=nums[i];sufx[i]=sum;
        }
        for(i=0;i<n;i++){
            if(prfx[i]==x){
                mini=min(mini,i+1);
            }else if(prfx[i]>x){
                break;
            }
            int high=i+1,low=n-1,target=x-prfx[i],ind=-1;
            while(high<=low){
                int mid=(low+high)/2;
                if(target==sufx[mid]){
                    ind=mid;break;
                }else if(target<sufx[mid]){
                    high=mid+1;
                }else{
                    low=mid-1;
                }
            }
            if(ind!=-1){
                mini=min(mini,i+1+n-ind);
            }
        }
        for(i=n-1;i>=0;i--){
            if(sufx[i]==x){
                mini=min(mini,n-i);
            }else if(sufx[i]>x){
                break;
            } 
            int low=0,high=i-1,target=x-sufx[i],ind=-1;
            while(low<=high){
                int mid=(low+high)/2;
                if(target==prfx[mid]){
                    ind=i;break;
                }else if(target>prfx[mid]){
                    low=mid+1;
                }else{
                    high=mid-1;
                }
            }
            if(ind!=-1){
                mini=min(mini,ind+1+n-i);
            }
        }
        if(mini==INT_MAX){
            return -1;
        }
        return mini;
    }
};
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