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Leetcode 462. Minimum Moves to Equal Array Elements II

Problem Statement:

Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.

In one move, you can increment or decrement an element of the array by 1.

You have to determine the minimum moves required to make all elements in the array equal.

Clarifying Questions:

  1. Q: Can the array contain negative numbers?
    • A: Yes, the array can contain negative numbers.
  2. Q: What is the maximum size of the array?
    • A: The size of the array can be up to $10^5$ based on typical constraints in similar problems.
  3. Q: Should we consider the possibility of integer overflow?
    • A: Typical constraints suggest that the values will be within a range that won’t cause overflow with conventional operations.

Strategy:

  1. Intuition:
    • To minimize the number of moves required to make all the array elements equal, it is optimal for all array elements to converge to the median of the array, rather than the mean or any other value. This is because the median minimizes the total absolute deviation from all elements in a distribution.
  2. Steps:
    • Sort the array.
    • Find the median of the sorted array.
    • Compute the total number of moves required by summing the absolute differences between each element and the median.
  3. Why the Median?
    • The median splits the data into two halves such that the sum of absolute deviations from the median is minimal. This is a well-known property in statistics and ensures that the deviations are minimized.

Code:

import java.util.Arrays;

public class MinimumMoves {
    public int minMoves2(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int median = nums[n / 2];
        int moves = 0;
        
        for (int num : nums) {
            moves += Math.abs(num - median);
        }
        
        return moves;
    }

    public static void main(String[] args) {
        MinimumMoves solution = new MinimumMoves();
        int[] nums = {1, 2, 3};
        System.out.println(solution.minMoves2(nums));  // Output: 2

        nums = new int[]{1, 10, 2, 9};
        System.out.println(solution.minMoves2(nums));  // Output: 16
    }
}

Time Complexity:

  1. Sorting the Array: The Arrays.sort() method sorts the array in $O(n \log n)$ time.
  2. Finding the Median & Computing Moves:
    • Finding the median is an $O(1)$ operation (after sorting).
    • Computing the total number of moves involves iterating through the array once, which is $O(n)$.

Thus, the overall time complexity is $O(n \log n)$ due to the sorting step, which is the most computationally expensive operation.

This approach ensures that the solution is both optimal and efficient for large-sized inputs typically encountered in such problems.

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