<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 findForOneArray(self, arr):
        count = arr[0]
        for i in range(1, len(arr)):
            count += max(arr[i] - arr[i - 1], 0)
        return count

    def minimumOperations(self, nums, target):
        n = len(nums)
        for i in range(n):
            target[i] -= nums[i]  # 1st point, conversion

        single_array = []
        answer = 0
        segments = []

        for i in range(n):
            if len(single_array) == 0:
                single_array.append(target[i])
            elif single_array[-1] < 0 and target[i] < 0:
                single_array.append(target[i])
            elif single_array[-1] >= 0 and target[i] >= 0:
                single_array.append(target[i])
            else:
                segments.append(single_array)
                single_array = [target[i]]

        if len(single_array) > 0:
            segments.append(single_array)

        for segment in segments:
            for i in range(len(segment)):
                segment[i] = abs(segment[i])

        for segment in segments:
            answer += self.findForOneArray(segment)

        return answer

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