<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) 
    {
        map<int,int> mpp; //is map ke andar apan sum from left to right and index store krenge
        
        mpp.insert({0,-1});
        int sum=0;
        for(int i=0; i<nums.size(); i++)
        {
            sum+=nums[i];
            mpp.insert({sum,i});
        }   

        //Abh right to left answer ki khoj me serach kro 
        int ans=INT_MAX;
        sum=0;

        if(mpp.find(x)!=mpp.end())
        {
            ans=mpp[x]+1;
        }

        for(int i=nums.size()-1; i>=0; i--)
        {
            sum+=nums[i];

            // if(sum == x)
            // {
            //     int l= nums.size()-i;
            //     ans = min(ans,l);
            // }

            if(mpp.find(x-sum)!=mpp.end() && mpp[x-sum]<i)
            {
                int l = mpp[x-sum] + 1 + nums.size()-i; 
                ans = min(ans , l);
            }
        }

        return (ans == INT_MAX)? -1: ans;
    }
};
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