<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) {
          static int N = [] () {
            ios::sync_with_stdio(false); 
            cin.tie(0); 
            return 0;
        } ();
        unordered_map<int,int> mp;
        int sum=0;
        mp[0]=0;
        for(int i=0;i<nums.size();i++){
            sum+=nums[i];
            mp[sum]=i;
        }//mapping the prefix sums
        if(x>sum) return -1;//edge cases
        if(x==sum) return nums.size();
        int cur=0;
        int len=-1;
        sum-=x;//finding the largest subarray of rest =sum-x
        for(int i=0;i<nums.size();i++){
           cur+=nums[i];
            if(sum==cur){
                len=max(len,i+1);
            }   
            if(mp.count(cur-sum)){
                len=max(len,i-mp[cur-sum]);
            }
        }
        return len==-1?-1:nums.size()-len;//returning the remaining len of rest for the target sum x
    }
};
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