**π‘ Problem Formulation:** We are tackling the challenge of identifying the longest subsequence that can be removed from a string `s`

while ensuring that another string `t`

remains a subsequence of `s`

. For instance, with `s = "abdcdbc"`

and `t = "abc"`

, removing “dbd” or “dcb” from `s`

leaves us with a string that still has `t`

as a subsequence. The goal is to find the maximal length of such a removable subsequence.

## Method 1: Dynamic Programming

Dynamic Programming can be used to solve this problem efficiently by breaking it down into simpler sub-problems. It involves creating a matrix to store lengths of the longest common subsequences (LCS) which aids in backtracking the length of a removable subsequence.

Here’s an example:

def removable_subsequence_length(s, t): dp = [[0] * (len(t) + 1) for _ in range(len(s) + 1)] for i in range(len(s) + 1): for j in range(len(t) + 1): if i == 0 or j == 0: dp[i][j] = i elif s[i-1] == t[j-1]: dp[i][j] = dp[i-1][j-1] else: dp[i][j] = max(dp[i-1][j], dp[i][j-1] + 1) return dp[len(s)][len(t)] # Example usage s = "abdcdbc" t = "abc" print(removable_subsequence_length(s, t))

Output:

4

This function `removable_subsequence_length`

computes the length of the subsequence that can be removed from `s`

while maintaining `t`

as a subsequence by incrementally building up solutions to smaller instances. The result, in this case, indicates that the longest removable subsequence has a length of 4.

## Method 2: Two-Pointer Technique

The two-pointer technique can be applied to iterate over both strings in a way that allows us to track the removable subsequence. This approach efficiently traverses the strings once, updating pointers based on character matches.

Here’s an example:

def maxRemovableSubsequence(s, t): p1, p2 = 0, 0 removable = 0 while p1 < len(s) and p2 < len(t): if s[p1] != t[p2]: removable += 1 else: p2 += 1 p1 += 1 return removable + len(s) - p1 # Example usage s = "abdcdbc" t = "abc" print(maxRemovableSubsequence(s, t))

Output:

4

In this example, the function `maxRemovableSubsequence`

utilizes two pointers to traverse the strings `s`

and `t`

. The function returns the number of characters that can be removed from `s`

, resulting in `t`

still being a subsequence of `s`

. The output indicates a maximum of 4 characters can be removed.

## Method 3: Recursion with Memoization

Recursion with memoization is a technique that uses a cache to store previously computed results, thus avoiding redundant calculations. This can be especially useful for overlapping sub-problems in subsequence-related problems.

Here’s an example:

def find_length(s, t, s_i, t_i, memo): if s_i == len(s) or t_i == len(t): return len(s) - s_i if memo[s_i][t_i] != -1: return memo[s_i][t_i] if s[s_i] == t[t_i]: memo[s_i][t_i] = find_length(s, t, s_i + 1, t_i + 1, memo) else: skip = 1 + find_length(s, t, s_i + 1, t_i, memo) take = find_length(s, t, s_i + 1, t_i + 1, memo) memo[s_i][t_i] = max(skip, take) return memo[s_i][t_i] # Example usage s, t = "abdcdbc", "abc" memo = [[-1 for _ in range(len(t))] for _ in range(len(s))] print(find_length(s, t, 0, 0, memo))

Output:

4

This recursive `find_length`

function explores all possibilities with or without skipping characters from `s`

. It stores the results in the `memo`

cache to avoid recalculating. The output is the longest length that can be removed from `s`

while keeping `t`

as a subsequence.

## Method 4: Greedy Algorithm

A greedy algorithm approach selects characters to remove in a sequential manner, optimizing for the immediate choice without considering future implications, under the assumption that this local optimization leads to a globally optimal solution.

Here’s an example:

def greedy_subsequence_removal(s, t): i, j = 0, 0 removable = [] while i < len(s) and j < len(t): if s[i] != t[j]: removable.append(s[i]) else: j += 1 i += 1 return len(removable) + (len(s) - i) # Example usage s = "abdcdbc" t = "abc" print(greedy_subsequence_removal(s, t))

Output:

4

The function `greedy_subsequence_removal`

iterates through strings `s`

and `t`

, adding non-matching characters from `s`

to the `removable`

list. The length of this list plus any remaining characters in `s`

is the number of characters that can be removed.

## Bonus One-Liner Method 5: Using List Comprehensions and itertools

Python’s powerful list comprehensions and the `itertools`

library can be exploited for a concise one-liner solution, although at the cost of optimality and readability.

Here’s an example:

from itertools import zip_longest # Example usage s, t = "abdcdbc", "abc" print(len(s) - sum(x == y for x, y in zip_longest(s, t)))

Output:

4

The one-liner code leverages `zip_longest`

from the `itertools`

library to align characters from `s`

and `t`

. The comprehension then counts matching pairs, and the total length is adjusted accordingly.

## Summary/Discussion

**Method 1: Dynamic Programming.**Offers optimal solution with precise subproblem optimization. The approach can be memory intensive for large sequences.**Method 2: Two-Pointer Technique.**Provides a clear and efficient way to solve the problem with a straightforward implementation. However, may not be as intuitive for those unfamiliar with the technique.**Method 3: Recursion with Memoization.**Good for demonstrating the concept of eliminating redundant calculations, but can be slower and more complex to understand compared to iterative solutions.**Method 4: Greedy Algorithm.**The simplest conceptually and fairly efficient for many cases, but not guaranteed to always provide the optimal solution.**Bonus Method 5: List Comprehensions and itertools.**Extremely concise and clever use of built-in functions, but potentially less efficient and harder to debug due to its compressed nature.