<p>You are given two positive integer arrays <code>nums</code> and <code>target</code>, of the same length.</p>

<p>In a single operation, you can select any subarray of <code>nums</code> and increment each element within that subarray by 1 or decrement each element within that subarray by 1.</p>

<p>Return the <strong>minimum</strong> number of operations required to make <code>nums</code> equal to the array <code>target</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [3,5,1,2], target = [4,6,2,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>We will perform the following operations to make <code>nums</code> equal to <code>target</code>:<br />
- Increment&nbsp;<code>nums[0..3]</code> by 1, <code>nums = [4,6,2,3]</code>.<br />
- Increment&nbsp;<code>nums[3..3]</code> by 1, <code>nums = [4,6,2,4]</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,3,2], target = [2,1,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p>We will perform the following operations to make <code>nums</code> equal to <code>target</code>:<br />
- Increment&nbsp;<code>nums[0..0]</code> by 1, <code>nums = [2,3,2]</code>.<br />
- Decrement&nbsp;<code>nums[1..1]</code> by 1, <code>nums = [2,2,2]</code>.<br />
- Decrement&nbsp;<code>nums[1..1]</code> by 1, <code>nums = [2,1,2]</code>.<br />
- Increment&nbsp;<code>nums[2..2]</code> by 1, <code>nums = [2,1,3]</code>.<br />
- Increment&nbsp;<code>nums[2..2]</code> by 1, <code>nums = [2,1,4]</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length == target.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], target[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution:
    def minimumOperations(self, nums: List[int], target: List[int]) -> int:
        # Step 1: Compute the difference between each element in target and nums
        # This difference array represents how much each element of nums needs to be changed to match target
        # We store these differences in the vector `vec`
        #<step_1>
        vec = [target[i] - nums[i] for i in range(len(nums))]
        #</step_1>

        # Step 2: Handle base cases for arrays of length 1 or 2
        # If there's only one element, we need to return the absolute value of that element in vec
        # For two elements, we need to handle positive and negative differences separately
        #<step_2>
        n = len(vec)
        if n == 1:
            return abs(vec[0])
        elif n == 2:
            if (vec[0] >= 0 and vec[1] >= 0) or (vec[0] < 0 and vec[1] < 0):
                return max(abs(vec[0]), abs(vec[1]))
            else:
                return abs(vec[0]) + abs(vec[1])
        #</step_2>

        # Step 3: Initialize the total number of operations and the starting difference
        # We start by adding the absolute value of the first element to the total operations count
        # We also determine if the first element is positive or negative
        #<step_3>
        x = abs(vec[0])
        last = vec[0]
        f = last >= 0
        #</step_3>

        # Step 4: Iterate through the difference vector to compute the minimum operations
        # We consider transitions between positive and negative differences and how they change
        # We update the total operations based on these transitions
        #<step_4>
        for i in range(1, n - 1):
            if f:
                if vec[i] != last:
                    if last < vec[i] and vec[i] > vec[i + 1]:
                        x += abs(vec[i] - last)
                        last = vec[i]
                        f = last >= 0
                    elif last > vec[i] and vec[i] < vec[i + 1]:
                        if vec[i] < 0:
                            x += abs(vec[i])
                        last = vec[i]
                        f = last >= 0
            else:
                if vec[i] != last:
                    if last < vec[i] and vec[i] > vec[i + 1]:
                        if vec[i] > 0:
                            x += abs(vec[i])
                        last = vec[i]
                        f = last >= 0
                    elif last > vec[i] and vec[i] < vec[i + 1]:
                        x += abs(vec[i] - last)
                        last = vec[i]
                        f = last >= 0
        #</step_4>

        # Step 5: Handle the last element in the difference vector
        # We need to check the final transition and add any additional operations required
        #<step_5>
        if f:
            if vec[-1] > last:
                x += vec[-1] - last
            elif vec[-1] < 0:
                x += abs(vec[-1])
        else:
            if vec[-1] < last:
                x += last - vec[-1]
            elif vec[-1] > 0:
                x += vec[-1]
        #</step_5>

        return x

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