**π‘ Problem Formulation:** We are addressing the challenge of calculating the minimum number of operations required to make all elements in an array equal. This commonly involves tasks such as incrementing or decrementing array elements, with operations constrained to specific rules. For instance, given an input array `[1, 2, 3]`

, the minimum number of operations to make all elements equal (each becoming `2`

) is `2`

.

## Method 1: Brute Force Approach

The brute force method involves iteratively making changes to the array by incrementing or decrementing elements until all values are equal. This is not efficient for large arrays as it has a high time complexity, but it is a straightforward solution.

Here’s an example:

def min_operations_brute_force(nums): steps = 0 while len(set(nums)) != 1: nums[nums.index(max(nums))] -= 1 nums[nums.index(min(nums))] += 1 steps += 1 return steps print(min_operations_brute_force([1, 2, 3]))

Output: `2`

This snippet defines a function `min_operations_brute_force`

that takes an array and performs operations until all elements are equal. On each iteration, it decreases the largest number and increases the smallest by one, tracking the number of steps taken. This example finds that two steps are required to equalize the array `[1, 2, 3]`

.

## Method 2: Using Arithmetic Mean

The arithmetic mean approach calculates the target number to which all elements should be equal by finding the average of the array. This method is efficient for small to medium-sized arrays. It performs best when dealing with arrays of even size or when the mean is a whole number.

Here’s an example:

def min_operations_mean(nums): target = round(sum(nums) / len(nums)) return sum(abs(target - x) for x in nums) // 2 print(min_operations_mean([1, 2, 3]))

Output: `2`

In the `min_operations_mean`

function, we calculate the arithmetic mean, rounding it to the nearest whole number to obtain our target value. Then we sum the absolute differences between each element and the target, dividing by 2 because each operation affects two elements. This example also concludes that two operations are needed for array `[1, 2, 3]`

.

## Method 3: Frequency Analysis

By assessing the frequency of each element within the array, one could find a value that appears most frequently and perform operations to make all other elements equal to that value. Frequency analysis is beneficial when the array has many repeated values.

Here’s an example:

from collections import Counter def min_operations_frequency(nums): freq_map = Counter(nums) target = freq_map.most_common(1)[0][0] return sum(abs(target - x) for x in nums) print(min_operations_frequency([1, 2, 2, 3]))

Output: `2`

The code snippet illustrates the function `min_operations_frequency`

, which utilizes Python’s `Counter`

class to find the most common value as the target. We then determine the number of operations by adding up the distances of each element from this target. Here we use an array `[1, 2, 2, 3]`

which needs two operations to become equal at the most frequent value `2`

.

## Method 4: Median-based Approach

For arrays with an odd count of unique numbers, choosing the median minimizes the sum of distances. This method ensures that the total number of operations is the least possible when converting all elements of the array to the median value.

Here’s an example:

def min_operations_median(nums): nums.sort() median = nums[len(nums) // 2] return sum(abs(median - x) for x in nums) print(min_operations_median([1, 5, 2, 4, 3]))

Output: `6`

This function, `min_operations_median`

, first sorts the array to easily find the median and then computes the sum of absolute differences from the median value. In this case, the array `[1, 5, 2, 4, 3]`

requires six operations to all become `3`

(the median value).

## Bonus One-Liner Method 5: Mean Calculation with List Comprehension

This one-liner method is a compact version of the arithmetic mean approach, using Python list comprehensions for brevity. It’s powerful yet concise, but may be less readable for beginners.

Here’s an example:

min_operations = lambda nums: sum(abs(round(sum(nums) / len(nums)) - x) for x in nums) // 2 print(min_operations([1, 2, 3]))

Output: `2`

Utilizing a lambda function, this one-liner version calculates and applies the mean method to find the minimum operations required to make the array elements equal. List comprehension is used to create a summation of differences all in one concise line, as demonstrated with the array `[1, 2, 3]`

.

## Summary/Discussion

**Method 1: Brute Force Approach.**Easy to understand and implement. Not suitable for large arrays due to high time complexity.**Method 2: Using Arithmetic Mean.**More efficient than the brute force, works well for small to medium-sized arrays with even distribution.**Method 3: Frequency Analysis.**Most effective when array contains many duplicates. May not be suitable for arrays with uniform distribution.**Method 4: Median-based Approach.**Best for minimizing the sum of distances in arrays with an odd number of unique elements. Requires sorted input.**Bonus One-Liner Method 5: Mean Calculation with List Comprehension.**Compact and efficient, but readability could be an issue for those new to Python.