<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:
    def minOperations(self, nums: List[int], x: int) -> int:
        """
        Find longest subarray that sums to target (sum(nums) - x)
        """
        if sum(nums) == x:
            return len(nums)

        target = sum(nums) - x
        max_len = -1
        curr_sum = 0

        # { curr_sum : earliest index at which curr_sum occurs }
        d = { 0 : -1 }

        for i, n in enumerate(nums):
            curr_sum += n
            other = curr_sum - target

            # sum of subarray from nums[d[other] + 1] to nums[i] == target
            # len of this subarray is i - d[other]
            if other in d:
                max_len = max(max_len, i - d[other])
            
            # store the earliest index at which curr_sum occurs
            if curr_sum not in d:
                d[curr_sum] = i
        
        return len(nums) - max_len if max_len != -1 else -1

        
        
38100