<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> prefix(n, 0), suffix(n, 0);
        prefix[0] = nums[0];
        suffix[n - 1] = nums[n - 1];
        for (int i = 1; i < n; i++)
            prefix[i] = prefix[i - 1] + nums[i];
        if (x > prefix[n - 1])
            return -1;
        for (int i = n - 2; i >= 0; i--)
            suffix[i] = suffix[i + 1] + nums[i];
        // 2 sum problem: suffix[i] + prefix[j] == x such that j + (n - i) is
        // minimized
        int ans = INT_MAX;
        int lb = lower_bound(prefix.begin(), prefix.end(), x) - prefix.begin();
        if (x == prefix[lb])
            ans = lb + 1;
        for (int i = n - 1; i >= 0; i--) {
            if (suffix[i] > x)
                break;
            if (suffix[i] == x) {
                ans = min(ans, n - i);
                break;
            }
            int pref_sum = x - suffix[i];
            lb = lower_bound(prefix.begin(), prefix.end(), pref_sum) - prefix.begin();
            if (lb < i && pref_sum == prefix[lb])
                ans = min(ans, lb + 1 + (n - i));
        }
        return ans == INT_MAX ? -1 : ans;
    }
};
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