<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) {
      const int sum = accumulate(nums.begin(), nums.end(), 0);  
      const int target = sum - x;
      if (target == 0) {
        return nums.size();
      }
      // find the longest subarray whose sum is target.
      const int len = LongestSubarray(nums, target);
      if (len == 0) {
        return -1;
      }
      return nums.size() - len;
    }
private:
  int LongestSubarray(const vector<int>& nums, int target) {
    unordered_map<int, int> m; // prefix sum v.s. first index.
    int sum = 0;
    m[0] = -1;
    int ans = 0;

    for (int i = 0; i < nums.size(); ++ i) {
      sum += nums[i];
      const auto it = m.find(sum - target);
      if (it != m.end()) {
        ans = max(ans, i - it->second);
      }
      if (m.count(sum) == 0) {
        m[sum] = i;
      }
    }

    return ans;
  }
};
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