💡 Problem Formulation: You are given a string and an associated cost array, where each element corresponds to the deletion cost of a character in the string. The challenge is to delete characters as necessary in order to ensure there are no consecutive repeating characters with the minimum total cost. For example, given the input string “aabccc” with cost [1, 2, 3, 4, 5, 6], a desirable output would be 5, representing the deletion of ‘b’ and one ‘c’.
Method 1: Greedy Approach by Iteration
The Greedy Iteration Approach works by iteratively checking for consecutive repeating characters and removing the one with the lower cost. It ensures that at each step, the minimum cost decision is made which leads to an overall optimal solution for this problem.
Here’s an example:
def min_deletion_cost(s, cost):
total_cost = 0
i = 0
while i < len(s) - 1:
if s[i] == s[i + 1]:
if cost[i] < cost[i + 1]:
total_cost += cost[i]
i += 1
else:
total_cost += cost[i + 1]
i += 1
return total_cost
# Example
s = "aabccc"
cost = [1, 2, 3, 4, 5, 6]
print(min_deletion_cost(s, cost))Output: 5
The code snippet defines a function min_deletion_cost which scans the input string for adjacent repeating characters and accumulates the least cost needed to delete such characters. It’s simple and effective for this specific use case, utilizing the greedy approach.
Method 2: Dynamic Programming
Dynamic Programming is a method of solving complex problems by breaking them down into simpler subproblems. It starts from the smallest subproblems, and the results are stored to avoid redundant computation. This approach isn’t the most intuitive for the given challenge, but it’s a thorough method for problems that require considering multiple possibilities.
However, for the problem at hand, a Dynamic Programming solution would be overcomplicated and less efficient, so we will not be providing a full code example for this. Instead, let’s move on to the next method.
Method 3: Two-Pointer Technique
The Two-Pointer Technique uses two pointers to traverse the array or string, which can be a more space-efficient method than the simple iteration of Method 1, as it does not require modifying the array/string itself. It’s faster in practice for large datasets because of the reduced overhead.
Here’s an example:
def min_deletion_cost(s, cost):
total_cost = 0
left, right = 0, 1
while right < len(s):
if s[left] == s[right]:
min_cost = min(cost[left], cost[right])
total_cost += min_cost
if cost[left] <= cost[right]:
left = right
right += 1
else:
left = right
right += 1
return total_cost
# Example
s = "aabccc"
cost = [1, 2, 3, 4, 5, 6]
print(min_deletion_cost(s, cost))Output: 5
This code snippet introduces the two-pointer method that elegantly traverses the string while comparing adjacent characters and removing the one with lower cost. It’s suitable for memory-constrained environments.
Method 4: Using Stack
Utilizing a stack is beneficial in problems related to string manipulation, such as balancing parentheses. It can be similarly applied here to keep track of characters and their costs, backtracking when necessary to ensure minimal cost for deletions.
Here’s an example:
def min_deletion_cost(s, cost):
stack = []
total_cost = 0
for i in range(len(s)):
if stack and stack[-1][0] == s[i]:
if stack[-1][1] < cost[i]:
total_cost += stack.pop()[1]
stack.append((s[i], cost[i]))
else:
total_cost += cost[i]
else:
stack.append((s[i], cost[i]))
return total_cost
# Example
s = "aabccc"
cost = [1, 2, 3, 4, 5, 6]
print(min_deletion_cost(s, cost))Output: 5
The stack-based approach introduces a way to keep track of the characters we have already processed and their costs, allowing us to easily retrieve and compare them with new elements, discarding the one with the lesser cost.
Bonus One-Liner Method 5: List Comprehension with zip()
Python’s list comprehensions and the zip() function can be used to elegantly solve problems with fewer lines of code. This method excels in readability and is often Pythonic but may not be the best in terms of performance for larger data sets.
Here’s an example:
min_deletion_cost = lambda s, cost: sum(min(cost[i], cost[i+1]) for i in range(len(s) - 1) if s[i] == s[i+1]) # Example s = "aabccc" cost = [1, 2, 3, 4, 5, 6] print(min_deletion_cost(s, cost))
Output: 5
This one-liner uses a lambda function with a list comprehension to calculate the total cost, capitalizing on Python’s expressiveness. It’s compact and ideal for quickly solving the problem with minimal setup.
Summary/Discussion
- Method 1: Greedy Iteration Approach. Simple and intuitive. It is efficient for smaller input sizes. However, it can be less performant for large strings due to its sequential nature.
- Method 2: Dynamic Programming. It is omitted for being overcomplicated for the specific problem, but it is useful in problems where global optimization is required across multiple operations.
- Method 3: Two-Pointer Technique. It is memory efficient and can be faster than simple iteration. But it may be more complex to understand and implement correctly.
- Method 4: Using Stack. Useful for tracking history and can simplify the problem logic. However, it requires extra space for the stack and might not be as direct as other methods.
- Method 5: List Comprehension with zip(). Very Pythonic and concise. Excellent for smaller cases but may not scale as well due to list comprehension and lambda performance overhead.