<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>

#include <vector>
#include <unordered_map>
#include <algorithm>

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int n = nums.size();
        int res = n + 1;
        int t_sum = 0;
        for (int i : nums) {
            t_sum += i;
        }
        int k = t_sum - x;
        if (k < 0) return -1;
        unordered_map<int, int> ps_ar;
        int pre_sum = 0;
        ps_ar[0] = -1;
        
        for (int i = 0; i < n; i++) {
            pre_sum += nums[i];
            if(pre_sum==x) res=min(res,i+1);
            ps_ar[pre_sum] = i;
        }
        
        int suf_sum = 0;
        
        
        for (int j = n - 1; j >= 0; j--) {
            suf_sum += nums[j];
            if(suf_sum == x) res=min(res,n-j);
            int req = x - suf_sum;
            
            if (ps_ar.find(req) != ps_ar.end()) {
                int idx = ps_ar[req];
                res = min(res, (n - j) + (idx + 1));
            }
        }
        
        return res == n + 1 ? -1 : res;
    }
};

149685