Minimum Operations to Make Array Equal to Target

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Problem Description

Ah, the classic dilemma of life: how to make two arrays equal without losing your sanity! The problem at hand, “Minimum Operations to Make Array Equal to Target,” is like trying to convince your stubborn friend to change their mind about pineapple on pizza. You know it’s wrong, but they just won’t budge!

In this LeetCode challenge, you are given two arrays, nums and target. Your mission, should you choose to accept it, is to determine the minimum number of operations required to make nums equal to target. An operation consists of incrementing or decrementing the elements of nums in such a way that they match the corresponding elements in target.

Imagine you’re trying to get your kids to clean their room. You can either bribe them with candy (increment) or threaten to take away their video games (decrement). Either way, you’re going to need a strategy!

Code Solution


class Solution:
    def minimumOperations(self, nums: list[int], target: list[int]) -> int:
        ans = abs(nums[0] - target[0])

        for (prevNum, prevTarget), (currNum, currTarget) in (
            itertools.pairwise(zip(nums, target))
        ):
            currDiff = currTarget - currNum
            prevDiff = prevTarget - prevNum
            if currDiff >= 0 and prevDiff >= 0:
                ans += max(0, currDiff - prevDiff)
            elif currDiff <= 0 and prevDiff <= 0:
                ans += max(0, abs(currDiff) - abs(prevDiff))
            else:
                ans += abs(currDiff)

        return ans

Approach Explanation

The code begins by calculating the absolute difference between the first elements of nums and target. Then, it iterates through the pairs of elements in both arrays using itertools.pairwise. For each pair, it calculates the difference between the current and previous elements. Depending on whether these differences are positive or negative, it updates the total number of operations needed to make the arrays equal.

In simpler terms, it’s like keeping track of how many times you have to bribe or threaten your kids to get them to clean their room!

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n), where n is the length of the input arrays. We traverse the arrays once.
Space Complexity	O(1), as we are using a constant amount of space for variables.

Real-World Example

Let’s say you’re organizing a family dinner, and you have two lists: one for what everyone wants to eat (nums) and another for what you actually have in the fridge (target). You need to figure out how many items you need to buy or throw away to make the lists match. This problem is just like that, but with numbers instead of food!