3229. Minimum Operations to Make Array Equal to Target

Problem Link

Description

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You are given two positive integer arrays nums and target, of the same length.

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

Return the minimum number of operations required to make nums equal to the array target.

 

Example 1:

Input: nums = [3,5,1,2], target = [4,6,2,4]

Output: 2

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..3] by 1, nums = [4,6,2,3].
- Increment nums[3..3] by 1, nums = [4,6,2,4].

Example 2:

Input: nums = [1,3,2], target = [2,1,4]

Output: 5

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..0] by 1, nums = [2,3,2].
- Decrement nums[1..1] by 1, nums = [2,2,2].
- Decrement nums[1..1] by 1, nums = [2,1,2].
- Increment nums[2..2] by 1, nums = [2,1,3].
- Increment nums[2..2] by 1, nums = [2,1,4].

 

Constraints:

• 1 <= nums.length == target.length <= 105
• 1 <= nums[i], target[i] <= 108

Solution

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Python3

class Solution:
    def minimumOperations(self, nums: List[int], target: List[int]) -> int:
        N = len(nums)
        diff = [x - t for x, t in zip(nums, target)]
        res = abs(diff[0])
 
        for i in range(1, N):
            if diff[i - 1] >= 0 and diff[i] >= 0:
                res += max(diff[i] - diff[i - 1], 0)
            elif diff[i - 1] <= 0 and diff[i] <= 0:
                res += max(abs(diff[i]) - abs(diff[i - 1]), 0)
            else:
                res += abs(diff[i])
 
        return res