Minimum Moves to Equal Array Elements II

Problem Description

Ah, the classic dilemma of life: how do we make everyone in a group as equally miserable as possible? Just kidding! The problem at hand is the “Minimum Moves to Equal Array Elements II.” Imagine you have a group of friends, and they all have different amounts of candy. You want to make sure everyone has the same amount of candy, but you can only move candy from one friend to another. The goal is to minimize the number of moves you make.

In technical terms, given an integer array nums, you need to find the minimum number of moves required to make all elements equal. A move consists of incrementing or decrementing an element by 1.

Code Solution

import statistics

class Solution:
    def minMoves2(self, nums: list[int]) -> int:
        median = int(statistics.median(nums))
        return sum(abs(num - median) for num in nums)

Approach

The approach taken in the code is quite straightforward. First, it calculates the median of the array. Why the median, you ask? Because it minimizes the total distance to all other points in the array. Then, it sums up the absolute differences between each number and the median. Voilà! You have the minimum moves required to make all elements equal.

Time and Space Complexity

Time Complexity: O(n), where n is the number of elements in the array. This is because calculating the median and summing the differences both require a single pass through the array.

Space Complexity: O(1), as we are using a constant amount of space regardless of the input size.

Real-World Example

Imagine you and your friends are at a party, and each of you has a different number of snacks. You want to make sure everyone has the same amount of snacks to avoid any fights over who has more. By redistributing the snacks based on the median amount, you can ensure that the least number of snack passes happen, keeping the peace and the party going!