**π‘ Problem Formulation:** Given a string, the challenge is to find the length of the longest substring that contains exactly *k* distinct characters. For instance, if the input is “`abcba`

” and *k* is 2, the longest substrings with 2 distinct characters are “`abc`

” and “`bcb`

“, both with a length of 3. Hence, the desired output is 3.

## Method 1: Brute Force Approach

Using the brute force method involves generating all possible substrings of the given string and then checking each substring for exactly *k* distinct characters. While this method is straightforward, it is not efficient for large strings due to its O(n^3) time complexity.

Here’s an example:

def longest_substring_brute_force(s, k): max_length = 0 for start in range(len(s)): for end in range(start + 1, len(s) + 1): if len(set(s[start:end])) == k: max_length = max(max_length, end - start) return max_length print(longest_substring_brute_force("abcba", 2))

Output: `3`

This code defines a function `longest_substring_brute_force()`

that receives a string and an integer *k*. It initializes `max_length`

to 0 and iteratively explores all substrings, updating `max_length`

if a substring of length *k* distinct characters is found. However, due to the nested loops, the performance can degrade sharply as the size of the input grows.

## Method 2: Sliding Window Technique

This method improves the efficiency of the search by using a sliding window to maintain a dynamic set of characters. The window is expanded and shrunk to always include up to *k* distinct characters, tracking the maximum length sub-string found. This approach has an O(n) time complexity.

Here’s an example:

def longest_substring_sliding_window(s, k): char_map = {} left, max_length = 0, 0 for right in range(len(s)): char_map[s[right]] = char_map.get(s[right], 0) + 1 while len(char_map) > k: char_map[s[left]] -= 1 if char_map[s[left]] == 0: del char_map[s[left]] left += 1 max_length = max(max_length, right - left + 1) return max_length print(longest_substring_sliding_window("abcba", 2))

Output: `3`

In the `longest_substring_sliding_window()`

function, we create a dictionary `char_map`

to count character occurrences. The variables `left`

and `right`

represent the window edges. The loop expands the window and adjusts `char_map`

. If the distinct character count exceeds *k*, the left edge of the window moves up until the condition is met, continuously checking for the maximum length. This method is practical for larger datasets due to its linear time complexity.

## Method 3: Optimized Sliding Window with OrderedDict

This method builds upon the sliding window technique but uses Python’s `collections.OrderedDict`

to efficiently keep track of the order in which characters appear. This allows us to remove the oldest entry quickly when we exceed *k* distinct characters, potentially improving performance in some scenarios.

Here’s an example:

from collections import OrderedDict def longest_substring_ordered_dict(s, k): char_map = OrderedDict() max_length = 0 for char in s: if char in char_map: del char_map[char] char_map[char] = None if len(char_map) > k: char_map.popitem(last=False) max_length = max(max_length, len(char_map)) return max_length print(longest_substring_ordered_dict("abcba", 2))

Output: `3`

The function `longest_substring_ordered_dict()`

illustrates an optimized strategy for the sliding window method using an `OrderedDict`

. This approach leverages the ordered nature of `OrderedDict`

to quickly discard the oldest character when needed. The function ensures that at most *k* distinct characters are kept in the `char_map`

, maintaining the maximum length of the substring throughout the iteration.

## Method 4: Frequency Table with Two Pointers

This method uses a frequency table to track character counts and two pointers to manage the window’s boundaries. One key difference is the explicit use of the frequency table to ensure quick updates and lookups of character counts within the window.

Here’s an example:

def longest_substring_two_pointers(s, k): char_freq = {} left = result = 0 for right, char in enumerate(s): char_freq[char] = char_freq.get(char, 0) + 1 while len(char_freq) > k: char_freq[s[left]] -= 1 if char_freq[s[left]] == 0: del char_freq[s[left]] left += 1 result = max(result, right - left + 1) return result print(longest_substring_two_pointers("abcba", 2))

Output: `3`

The `longest_substring_two_pointers()`

function uses a dictionary `char_freq`

to count the characters and two pointers (`left`

and `right`

) to adjust the window size. Similar to the sliding window technique, it iterates over the string, updating counts and checking the number of distinct characters. When the character limit is exceeded, it reduces the count of the leftmost character and shifts the left pointer rightwards, constantly updating the result.

## Bonus One-Liner Method 5: Using itertools and Set

A more Pythonic, albeit less efficient, one-line solution using `itertools`

. This method leverages Python’s comprehensive standard library to express the solution very succinctly at the cost of potentially increased time complexity.

Here’s an example:

from itertools import combinations def longest_substring_oneliner(s, k): return max((len(s[a:b]) for a, b in combinations(range(len(s) + 1), 2) if len(set(s[a:b])) == k), default=0) print(longest_substring_oneliner("abcba", 2))

Output: `3`

The `longest_substring_oneliner()`

expression here makes use of Python’s generator expressions and the `combinations`

function from the `itertools`

module to generate all possible substrings. Due to the combination of `range`

and `set`

operations, it checks for the distinct character condition and finds the maximum length. As noted, this method is concise but may not be practical for long strings.

## Summary/Discussion

**Method 1: Brute Force Approach.**Simple to understand and implement. Not suitable for large strings due to its O(n^3) time complexity.**Method 2: Sliding Window Technique.**Significantly more efficient with linear time complexity. Suitable for most practical scenarios.**Method 3: Optimized Sliding Window with OrderedDict.**Adds a potential performance boost for certain cases. Efficiently manages character order within the window.**Method 4: Frequency Table with Two Pointers.**Similar benefits to the sliding window technique. The explicit frequency table can simplify the understanding and manipulation of character counts.**Method 5: Using itertools and Set.**Offers a Pythonic one-liner. Best used when code conciseness is preferred over performance, especially with shorter strings.