Program to Find Minimum Number of Deletions to Balance a List from Both Ends in Python

Rate this post

πŸ’‘ Problem Formulation: In this article, we discuss how to find the minimum number of deletions required from both ends of a list to achieve balance. If a list has an equal number of a certain element at both ends, it is considered balanced. For instance, given the input list [1, 3, 1, 2, 1], the desired output would be 1 because deleting the single element 2 balances the list.

Method 1: Two-pointer Approach

The two-pointer approach involves initializing pointers at both ends of the list that move towards the center, assessing balance as they go. When imbalance is detected, the algorithm opts to delete from the end that contributes to imbalance and counts the deletions. This method requires only a single iteration through the list and offers O(n) time complexity where n is the length of the list.

Here’s an example:

def min_deletions_to_balance(lst):
    left, right = 0, len(lst) - 1
    deletions = 0
    while left < right:
        if lst[left] != lst[right]:
            deletions += 1
            if lst[left] == lst[0]:
                right -= 1
            else:
                left += 1
        else:
            left += 1
            right -= 1
    return deletions

print(min_deletions_to_balance([1, 3, 1, 2, 1]))

Output: 1

This snippet defines a function min_deletions_to_balance() that computes the minimum number of deletions for balance by incrementing pointers from both ends. When elements at these pointers don’t match, the algorithm deletes the non-matching element and carries on until the list is balanced or the pointers cross each other.

Method 2: Recursion with Memoization

Recursion with memoization involves a more straightforward approach by checking every possible deletion combination with recursive calls. The memoization aspect stores previously computed minima to reduce the number of calculations. This approach, although intuitive, may be less efficient due to increased function calls and is generally O(n^2) because of memoization.

Here’s an example:

def min_del_balance_helper(lst, start, end, memo):
    if start >= end:
        return 0
    if (start, end) in memo:
        return memo[(start, end)]
    if lst[start] == lst[end]:
        memo[(start, end)] = min_del_balance_helper(lst, start+1, end-1, memo)
    else:
        memo[(start, end)] = 1 + min(min_del_balance_helper(lst, start+1, end, memo),
                                        min_del_balance_helper(lst, start, end-1, memo))
    return memo[(start, end)]

def min_deletions_to_balance(lst):
    memo = {}
    return min_del_balance_helper(lst, 0, len(lst) - 1, memo)

print(min_deletions_to_balance([1, 3, 1, 2, 1]))

Output: 1

In this code, we define a helper function min_del_balance_helper() for our recursive approach which uses a memo dictionary to store and fetch previously computed results per start and end index combinations to avoid redundant computations. The main function min_deletions_to_balance() initializes the memo and starts the process.

Method 3: Dynamic Programming

Dynamic Programming (DP) is another systematic approach for optimization problems like this, where we construct a table to hold the answers to subproblems. This method offers an optimized and bottom-up approach as opposed to recursion and is generally more efficient.

Here’s an example:

def min_deletions_to_balance(lst):
    n = len(lst)
    dp = [[0]*n for _ in range(n)]
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if lst[i] == lst[j]:
                dp[i][j] = dp[i + 1][j - 1]
            else:
                dp[i][j] = 1 + min(dp[i + 1][j], dp[i][j - 1])
    return dp[0][n - 1]

print(min_deletions_to_balance([1, 3, 1, 2, 1]))

Output: 1

The min_deletions_to_balance() function starts by initializing a DP table and then fills it by considering all possible sub-lists, updating minimum deletions needed based on matching conditions. This manner of processing ensures no repeated work, making it more efficient for longer lists.

Method 4: Greedy Algorithm

This method involves a greedy strategy, which continuously deletes the ‘outlying’ element that causes an imbalance. It’s an intuitive way to achieve a balanced list, proceeding with local optimal choices. Depending on the structure of the list, this may or may not lead to a globally optimal solution.

Here’s an example:

def min_deletions_to_balance(lst):
    deletions = 0
    while lst and lst[0] != lst[-1]:
        if lst.count(lst[0]) > lst.count(lst[-1]):
            lst.pop()
        else:
            lst.pop(0)
        deletions += 1
    return deletions

print(min_deletions_to_balance([1, 3, 1, 2, 1]))

Output: 1

In the min_deletions_to_balance() function, we directly manipulate the list by popping elements from either end based on which side contains more of the same element as the end element. This greedy approach focusses on retaining the majority element equally mirrored at both ends.

Bonus One-Liner Method 5: List Slicing with Min Deletions

For a touch of Pythonic elegance, we apply a one-liner that utilizes list comprehension and slicing to count the minimum deletions required. This method leverages the expressiveness of Python and can be less readable but concise.

Here’s an example:

min_deletions_to_balance = lambda lst: len(lst) - max((lst[:i] == lst[:i-1:-1]).count(True) * 2 for i in range(len(lst)//2+1))

print(min_deletions_to_balance([1, 3, 1, 2, 1]))

Output: 1

The one-liner defines min_deletions_to_balance as a lambda function that calculates the maximum length of a balanced sub-list within the original list by comparing the forwards and reversed slices of the list. The difference in length between the original and maximum balanced sub-list equates to the minimum deletions required.

Summary/Discussion

  • Method 1: Two-pointer Approach. Straightforward and efficient for linear lists. The speed makes it practical for larger datasets. Less intuitive than some other methods and requires careful pointer management.
  • Method 2: Recursion with Memoization. Intuitive, mirroring human-like problem-solving, but can be slower due to recursion overhead. Memoization enhances efficiency significantly but not as much as DP or iterative solutions.
  • Method 3: Dynamic Programming. Offers a precise and potentially fastest solution with O(n^2) time and space complexity. It does require additional memory for the DP table but is very efficient for large inputs.
  • Method 4: Greedy Algorithm. Provides a more straightforward approach with easier implementation but risks not finding the optimal solution in all cases.
  • Bonus Method 5: One-Liner. Showcases Python’s ability for concise expressions. Ideal for those familiar with Python’s list comprehension and slicing. Not the most readable or maintainable for complex problems.