# 1658. Minimum Operations to Reduce X to Zero

You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.

Return *the **minimum number** of operations to reduce* `x` *to **exactly*** `0` *if it's possible, otherwise, return* `-1`.

**Example 1:**

```
Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
```

**Example 2:**

```
Input: nums = [5,6,7,8,9], x = 4
Output: -1
```

**Example 3:**

```
Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
```

**Constraints:**

* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 104`
* `1 <= x <= 109`

从给定数组左边和右边取连续的数，问最少的取数个数能让和等于x。问题可以转化为找最长子数组使得和为sum - x =>动态窗口。也可以看成左右此消彼长的过程，两个指针先让左边的尽量取直到取到x，之后再从右边开始取，每次取一个数后将左边指针退到左右总和不超过x的位置，以此类推直到右指针往前走完所有可能的位置，是形式稍微特别些的动态窗口。

```python
class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        N = len(nums)
        l, r, s, cnt, res = 0, N - 1, 0, 0, float('inf')
        while l < N and s + nums[l] <= x:
            s += nums[l]
            l += 1
            cnt += 1
            if s == x: res = min(res, cnt)
        if l == N and s < x: return -1
        l -= 1
        while r >= 0:
            cnt += 1
            s += nums[r]
            while l >= 0 and s > x:
                cnt -= 1
                s -= nums[l]
                l -= 1
            if s == x: res = min(res, cnt)
            r -= 1
        return -1 if res == float('inf') else res
    
```