Minimum Moves to Equal Array Elements II Interview Question for Google

# Minimum Moves to Equal Array Elements

## Problem Description

Given an integer array `nums` of size `n`, the task is to find 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.

For example:

- Input: `nums = [1, 2, 3]`
- Output: `2`
- Explanation: `[1, 2, 3] -> [2, 2, 3] -> [2, 2, 2]` (2 moves)

- Input: `nums = [1, 10, 2, 9]`
- Output: `16`

## Naive Approach

A naive approach would be to try every possible target value (i.e., every value between the min and max element of the array) and calculate the moves needed to convert all elements to that target value. Then, return the minimum of these moves.

```python
def min_moves_naive(nums):
    min_val = min(nums)
    max_val = max(nums)
    min_moves = float('inf')

    for target in range(min_val, max_val + 1):
        moves = 0
        for num in nums:
            moves += abs(num - target)
        min_moves = min(min_moves, moves)

    return min_moves

This approach has a time complexity of O(n * (max - min)), where n is the length of the array and max and min are the maximum and minimum values in the array. This can be inefficient when the range (max - min) is large.

Optimal Approach

The optimal approach involves finding the median of the array. The minimum moves required will be achieved when all elements are made equal to the median. This is because the median minimizes the sum of absolute differences.

Here's the optimized solution:

def min_moves_optimal(nums):
    nums.sort()
    median = nums[len(nums) // 2]
    moves = 0
    for num in nums:
        moves += abs(num - median)
    return moves

Big(O) Runtime Analysis

Sorting: nums.sort() has a time complexity of O(n log n), where n is the number of elements in the array.
Calculating Moves: The loop that calculates moves iterates through each element once, resulting in O(n).

Therefore, the overall time complexity is O(n log n), dominated by the sorting step.

Big(O) Space Usage Analysis

Sorting: Most sorting algorithms (like Timsort, which is commonly used in Python) have a space complexity of O(n) in the worst case or O(log n) in the average case if done in place.
Other Variables: The other variables (median, moves) take constant space, O(1).

Therefore, the overall space complexity is O(1) if the sorting is done in place and O(n) otherwise.

Edge Cases

Empty Array: If the input array is empty, it might be appropriate to return 0 or raise an exception, depending on the specific requirements.
Single Element Array: If the array contains only one element, the number of moves needed is 0.
Array with Duplicate Elements: The algorithm works correctly even if the array contains duplicate elements.
Large Numbers: The problem statement mentions that the answer will fit in a 32-bit integer, but if the numbers are extremely large, intermediate calculations might overflow. Handling large numbers might require using appropriate data types or modular arithmetic.

Here's how to handle the empty array edge case:

def min_moves_optimal(nums):
    if not nums:
        return 0  # Or raise an exception
    nums.sort()
    median = nums[len(nums) // 2]
    moves = 0
    for num in nums:
        moves += abs(num - median)
    return moves