<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> nums2 ;
        for (int n: nums) {
            nums2.push_back(n);
        }
        for (int n: nums) {
            nums2.push_back(n);
        }
        int n = nums2.size();

        int l = 0;
        int r = 0;
        int sum = 0;
        int ans = INT_MAX;
        while (r < n) {
             sum += nums2[r];
            if (r-l+1 > n/2) {
                sum -= nums2[l];
                l++;
                 if (sum == x) {
                  ans = min(ans,  r-l+1);
                }
                r++;
                continue;
            }

            if (sum == x) {
                ans = min(ans,  r-l+1);
                r++;
            } else if (sum > x && (r == n/2 -1 || (r > n/2 -1 && l < n/2))) {
                while (sum > x && (r == n/2 -1 || (r > n/2 -1 && l < n/2))) {
                    sum -= nums2[l];
                    l++;
                }
                if (sum == x) {
                  ans = min(ans,  r-l+1);
                }
                r++;
            } else {
                r++;
            }
        }
        return ans == INT_MAX ? -1: ans;
        
    }

    //1,1,4,2,3 1 1 4 2 3

};
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