<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 maxSubArrayLen(vector<int>& nums, int k) {

        //[1,-1,5,-2,3]. k=3
        //[1,0, 5, 3,6]

        // -2,-1,2,1   k=1
        // -2, -3,-1,0
        // -2  -4, -2,-1
        unordered_map<long long, int>mmap;
        mmap[0]=-1;
        int res=0;
        long long sum = 0;
        for(int i=0;i<nums.size();i++){
            sum+=nums[i];
            if(mmap.find(sum-k) != mmap.end()){
                    res= max(res, i-mmap[sum-k]);
            }
            if(mmap.find(sum) == mmap.end()){
                    mmap[sum]=i;
            }
        }
        return res;
    }

    int minOperations(vector<int>& nums, int x) {
        
        int sum=0;
        for(int i=0;i<nums.size();i++){
            sum+=nums[i];
        }
        int rem = sum -x; //required max len subarray with sum ==rem
        if(rem==0){
            return nums.size();
        }
        int length = maxSubArrayLen(nums, rem);
        if(length==0)   return -1;
    
        return nums.size()-length;
    }
};
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