<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) {
        unordered_map<int, int> left;
        int res = INT_MAX;
        for (auto l = 0, sum = 0; l <= nums.size() && sum <= x; ++l) {
            left[sum] = l;
            if (l < nums.size())
                sum += nums[l];
        }
        for (int r = nums.size() - 1, sum = 0; r >= 0 && sum <= x; --r) {
            auto it = left.find(x - sum);
            if (it != end(left) && r + 1 >= it->second) {
                res = min(res, (int)nums.size() - r - 1 + it->second);
            }
            sum += nums[r];
        }
        return res == INT_MAX ? -1 : res;
    }
};
// class Solution {
//    public:
//     int minOperations(vector<int>& nums, int x) {
//         int sum = 0, n = nums.size();
//         for (int i : nums) sum += i;
//         int target = sum - x;
//         int curr_sum = 0, max_len = 0;
//         int start_idx = 0;
//         bool found = false;
//         for (int end_idx = 0; end_idx < n; end_idx++) {
//             curr_sum += nums[end_idx];
//             while (start_idx <= end_idx && curr_sum > target) {
//                 curr_sum -= nums[start_idx];
//                 start_idx += 1;
//             }
//             if (curr_sum == target) {
//                 found = true;
//                 max_len = max(max_len, end_idx - start_idx + 1);
//             }
//         }
//         return found ? n - max_len : -1;
//     }
// };
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