Minimum Operations to Reduce X to Zero

[juneharold]

Use two pointers. Step 1: Calculate the total sum of nums.

Step 2: Initialize two pointers left and right to 0. Initialize an integer current to represent the sum from nums[0] to nums[left-1] and from nums[right+1] to nums[last]. Initialize an integer mini to record the minimum length that sums up to x

Step 3: Iterate right form 0 to the end of nums. In each iteration:

• Update current.
• If current is smaller than x, move left to left.
• If current is equal to x, update the maximum length.

Step 4: Return the result.

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        current = sum(nums)
        n = len(nums)
        mini = inf
        left = 0

        for right in range(n):
            # sum([0,..,left) + (right,...,n-1]) = x
            current -= nums[right]
            # if smaller, move `left` to left
            while current < x and left <= right:
                current += nums[left]
                left += 1
            # check if equal
            if current == x:
                mini = min(mini, (n-1-right)+left)

        return mini if mini != inf else -1