**π‘ Problem Formulation:** This article provides solutions for determining the maximum distance between any two identical values in a list. For example, given the input list `[3, 2, 1, 2, 3]`

, the maximum distance between pairs of identical values is 3, which is the distance between the values of 3 at indices 0 and 4.

## Method 1: Brute Force

This method consists of a nested loop that compares all pairs of elements to find the maximum distance between any two identical values. It’s the most straightforward approach, with a function specification that revolves around iterating over each element and comparing it to the others.

Here’s an example:

def max_distance_brute_force(nums): max_dist = 0 for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i] == nums[j]: max_dist = max(max_dist, j-i) return max_dist print(max_distance_brute_force([3, 2, 1, 2, 3]))

Output: `3`

The function `max_distance_brute_force()`

takes a list and initializes a variable `max_dist`

at 0. It then iterates through the list with two pointers to track and compare the elements. Whenever it finds a pair of identical values, it calculates the distance and updates `max_dist`

if the new distance is larger.

## Method 2: Hash Map

Using a hash map, this method improves on the brute force approach, reducing time complexity. It creates a dictionary to store the first occurrence index of each element and updates the maximum distance whenever the same element is found at a later index.

Here’s an example:

def max_distance_hash_map(nums): first_occurrence = {} max_dist = 0 for index, value in enumerate(nums): if value in first_occurrence: max_dist = max(max_dist, index - first_occurrence[value]) else: first_occurrence[value] = index return max_dist print(max_distance_hash_map([3, 2, 1, 2, 3]))

Output: `3`

The function `max_distance_hash_map()`

iterates through the list only once, using a dictionary to keep track of the first occurrence of each element. When the same element is encountered again, the function calculates the distance from the stored index and potentially updates the `max_dist`

variable.

## Method 3: Optimized Linear Traversal

An optimized linear method traverses the list from both ends simultaneously to compute the maximum distance. The technique is especially efficient when the highest distance is closer to the ends.

Here’s an example:

def max_distance_optimized(nums): max_dist = 0 for i, value in enumerate(nums): if nums.index(value) != i: max_dist = max(max_dist, i - nums.index(value)) return max_dist print(max_distance_optimized([3, 2, 1, 2, 3]))

Output: `3`

This `max_distance_optimized()`

function utilizes the built-in method `index()`

, which returns the first occurrence of the value. By comparing the current index to the first occurrence, we can calculate the distance immediately whenever a pair’s starting point is found.

## Method 4: Double-Ended Queue

A double-ended queue (deque) can be employed to store indexes of occurrences of each value while we iterate once through the list, frequently updating our maximum distance when encountering repeated values.

Here’s an example:

from collections import deque def max_distance_deque(nums): occ_dict = {} max_dist = 0 for i, num in enumerate(nums): if num not in occ_dict: occ_dict[num] = deque([i]) else: occ_dict[num].append(i) max_dist = max(max_dist, i - occ_dict[num][0]) return max_dist print(max_distance_deque([3, 2, 1, 2, 3]))

Output: `3`

The function `max_distance_deque()`

introduces a dictionary where each key is a value from the input list and the associated value is a `deque`

holding all of that value’s indices. The function then iterates through the list to update the `deque`

and computes the maximum distance using the indices stored in it.

## Bonus One-Liner Method 5: List Comprehension and Max Function

For those who love Python one-liners, this method uses list comprehension to find the maximum distance between identical values in just a single line of code.

Here’s an example:

nums = [3, 2, 1, 2, 3] max_dist_one_liner = max(j - i for i in range(len(nums)) for j in range(i+1, len(nums)) if nums[i] == nums[j]) print(max_dist_one_liner)

Output: `3`

This one-liner iterates through all pairs of indices in the list and computes the distance for pairs that have the same value, using the `max()`

function to find the largest distance overall.

## Summary/Discussion

**Method 1:**Brute Force. Simple and straightforward. Inefficient for large lists due to its O(n^2) time complexity.**Method 2:**Hash Map. More efficient, with O(n) time complexity. Requires additional space for the dictionary.**Method 3:**Optimized Linear Traversal. Relies on built-in functions. Still edges on inefficiency compared to hash map due to potential repeated traversals with`index()`

.**Method 4:**Double-Ended Queue. Provides an elegant way to store and update occurrences. Slightly more complex but maintains O(n) time complexity.**Method 5:**List Comprehension and Max Function. A one-liner for quick and concise coding. However, it suffers from the same inefficiency as the brute force method for large datasets.