<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>& a, int x) {
        // let the sum of array = s
        // now we want x-sum = 0
        // sum = x
        // remaining sub array sum a - x
        // so we have to find the largest subarray length
        // with sum a - x so answer is n - len(largestSubarray)

        // sum of subarray = pref[r] - pref[l-1] = s - x
        ios_base::sync_with_stdio(false);
        cin.tie(0);

        
        int n = a.size();
        int s = 0;
        for(int i=0;i<n;i++) s+=a[i];
        vector<int> pref(n+1);
        map<int,int> mp;
        mp[0] = 0;
        for(int i=0;i<n;i++) pref[i+1] = pref[i] + a[i];
        int sz = -1;
        for(int i=1;i<=n;i++)
        {
            int finder = pref[i] - s + x;
            if(mp.find(pref[i]) == mp.end()) mp[pref[i]] = i;
            if(mp.find(finder) != mp.end())
            {
                // found finder 
                // so r to l is a valid subarray
                int r = i;
                int l = mp[finder];
                sz = max(sz,r-l);
            }
            
        }
        cout << sz << endl;
        if(sz == -1) return -1;
        else return n - sz;
    }
};
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