<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>& arr, int k) {
        int n=arr.size();
        vector<int>pref(n,0);

        for(int i=0;i<n;i++){
            if(i==0) pref[i]=arr[i];
            else pref[i]=arr[i]+pref[i-1];
        }
        unordered_map<int,int>s;
        int last=0;
        for(int i=n-1;i>=0;i--){
            s[arr[i]+last]=i;
            last=arr[i]+last;
        }

        int ans=1e9;
        auto it=s.find(k);
        if(s.count(k)) ans=min(ans,n-it->second);
        for(int i=0;i<n;i++){
            int x=k-pref[i];
            if(x==0){
                ans=min(ans,i+1);
                break;
            }
            if(x<0) break;
            if(s.find(x)!=s.end()){
                auto it=s.find(x);
                if(it->second>i) ans=min(ans,i+1+n-it->second);
            }
        }
        return ans==1e9?-1:ans;
    }
};
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