<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();
        vector<int> pre(n);
        vector<int> suf(n);
        pre[0] = nums[0];
        suf[n-1] = nums[n-1];
        for(int i = 1;i<n;i++){
            pre[i] = pre[i-1] + nums[i];
            suf[n-i-1] = suf[n-i] + nums[n-i-1];
        }

        int ans = 1e9;

        for(int i = n-1;i>=0;i--){
            int k = x-suf[i];
            auto it = lower_bound(pre.begin(), pre.end(), k);
            if(it<=(pre.begin()+i-1) && (*it) == k){
                cout<<*it<<endl;
                int y = (it-pre.begin())+1;
                ans = min(ans,(n-i+y));
            }
        }

        auto it1 = lower_bound(pre.begin(),pre.end(),x);

        if(it1!=pre.end() && (*it1)==x){
            int y = (it1-pre.begin());
            ans = min(ans,y+1);
        }

        reverse(suf.begin(),suf.end());

        auto it2 = lower_bound(suf.begin(),suf.end(),x);

        if(it2!=suf.end() && (*it2)==x){
            int y = (it2-suf.begin());
            ans = min(ans,y+1);
        }

        if(ans>=1e9){
            return -1;
        }

        return ans;
        
    }
};
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