Minimum ASCII Delete Sum for Two Strings

Identifying Problem Isomorphism

This involves finding the minimum sum of ASCII values of characters to delete to make two strings equal. This can be mapped approximately to the LeetCode problem titled “Edit Distance” (LeetCode #72).

Reasoning:

In “Edit Distance”, the task is to compute the minimum number of operations (insertions, deletions, or substitutions of characters) required to transform one string into another.

“Minimum Delete Sum” also deals with transforming one string into another, but the operations allowed are just deletions, and we’re required to minimize not just the number of deletions but the sum of the ASCII values of the deleted characters.

While the two problems are not exactly the same, they do share a common theme: both are about transforming one string into another with a certain cost associated with the operations. Hence, the dynamic programming approach used to solve the “Edit Distance” problem can be adapted to solve the “Minimum Delete Sum” problem. You are only allowed to delete characters and that the cost of each operation is the ASCII value of the character being deleted rather than 1.

10 Prerequisite LeetCode Problems

For “712. Minimum ASCII Delete Sum for Two Strings”, the following are a good preparation:

“72. Edit Distance” - It helps in understanding the concept of converting one string into another with the minimum number of operations.
“1143. Longest Common Subsequence” - This problem is relevant as it is about finding the longest common subsequence, a technique that can be used to solve the minimum ASCII delete sum problem.
“392. Is Subsequence” - It practices how to identify subsequences within a string which will be beneficial for finding the common substrings.
“516. Longest Palindromic Subsequence” - Although the concept of palindrome doesn’t apply directly, the dynamic programming solution is very similar to that of the current problem.
“583. Delete Operation for Two Strings” - This problem is almost the same, the only difference is in the way the cost of the operation is calculated.
“650. 2 Keys Keyboard” - Involves a similar concept of minimum operations to reach a certain string state.
“647. Palindromic Substrings” - It helps understand the concept of substrings which is crucial to this problem.
“115. Distinct Subsequences” - It involves the calculation of distinct subsequences in a string.
“1312. Minimum Insertion Steps to Make a String Palindrome” - This problem’s concept of performing operations to reach a certain string state is similar to the problem at hand.
“161. One Edit Distance” - Understanding the minimal distance or operations required to equalize two strings is important for solving the minimum ASCII delete sum problem.

These cover string manipulation, dynamic programming, and substrings, which are key to solving the “Minimum ASCII Delete Sum for Two Strings” problem.

Problem Classification

This problem belongs to the domain of “Strings” and “Dynamic Programming”.

Here’s a categorization based on ‘What’ components:

Input: The problem takes in two strings s1 and s2. The lengths of the strings are constrained to be within 1 and 1000. The strings consist of lowercase English letters.
Output: The problem requires you to output a single integer, which represents the lowest ASCII sum of deleted characters to make the two input strings equal.
Operation: The problem statement requires an operation of deleting characters from the input strings. It mentions that deleting a character adds its ASCII value to a sum, and that you are to minimize this sum.

This problem can be classified as a Dynamic Programming problem.

We have to find the minimum cost of a sequence of decisions (which characters to delete) to achieve a certain condition (making the two strings equal). At each step, we need to make the best decision based on the state of the strings so far. Since the optimal solution can be constructed from optimal solutions to subproblems (substrings), this problem can be solved using dynamic programming.

This problem also has overlapping subproblems property, which means the same subproblems are solved again and again. Hence, dynamic programming can be used to store the results of subproblems and avoid redundant calculations, thereby improving efficiency.

This is a “Dynamic Programming” problem in the domain of “Strings”.

Clarification Questions

While asking clarification questions for a problem, we should focus on the potential ambiguities or edge cases that the problem statement might not have covered. In this case, the problem statement seems quite clear, but we might ask:

What should be returned if one or both the strings are empty?
If there are multiple ways to achieve the lowest ASCII sum of deleted characters, do we need to provide all possible solutions or any one of them?
Are there any constraints on the runtime or memory usage of the solution?
Will the strings only consist of lowercase English letters or are uppercase letters and non-alphabetic characters possible?
If both the strings are already equal, do we assume the ASCII sum of deleted characters as 0, as no deletion operation is needed?
In case of repeated characters in a string, do we consider deletion of each instance separately?
Can the order of characters in the strings be altered or does it have to remain consistent post-deletion?

The aim of asking these questions is to get a clearer understanding of the problem and the expected behavior of the solution in various scenarios.

Problem Analysis and Key Insights

Analyzing the problem statement reveals a few key insights:

The task requires comparing the characters in two strings and deciding which characters to delete to make the strings equal with the least ASCII value loss.
Since the only operation allowed is deletion, we need to find common substrings in the two strings because only these can be preserved while making the strings equal. In essence, this problem can be rephrased as finding the common substring(s) with the maximum total ASCII value.
There is an implicit assumption that the order of the characters in the strings matters, because the problem specifies “strings” not “sets of characters”. This means that a solution may involve dynamic programming to handle the ordered nature of the characters in the strings.
The problem is defined for lowercase English letters. This means the ASCII values we are dealing with will fall within a specific, known range (97 to 122).
The problem constraints specify that the length of the strings will not exceed 1000. This implies that an algorithm with a time complexity of O(n^2) or better may be required to ensure the solution is efficient enough.

Remember, these insights are preliminary in nature based on the problem statement and may be further refined or supplemented once we start devising a solution approach or writing pseudocode.

Problem Boundary

The scope of the problem “Minimum ASCII Delete Sum for Two Strings” includes:

Input: The problem accepts two inputs, both are strings s1 and s2, and each string’s length is between 1 and 1000 characters inclusive. The characters of the strings are lowercase English letters.
Output: The problem requires to output an integer value which represents the minimum total ASCII value of deleted characters to make two input strings equal.
Constraints: The constraints put forth by the problem are clear. The function needs to handle all cases where the lengths of s1 and s2 are within the range of 1 to 1000. The characters in the strings are lowercase English letters. There are no special or non-alphanumeric characters involved.
Problem Solving: The task involves identifying and deleting characters in order to equalize the strings, whilst minimizing the total ASCII value of the deleted characters. The key challenge is deciding which characters to delete in each string in a way that minimizes the total ASCII value.
Functionality: The implemented function should be able to correctly and efficiently process inputs according to the constraints and produce the correct output.

The scope does not include validating the input as it’s guaranteed to follow the constraints (i.e., string length and lowercase English letters). The problem also doesn’t involve any interactive input or output, database, or network operations.

Establishing the boundary of the problem “Minimum ASCII Delete Sum for Two Strings” involves understanding the problem’s constraints and specifications in detail. Here’s how we can do it:

Inputs: The inputs to the problem are two strings s1 and s2, each with a length between 1 and 1000 characters, inclusive. We do not need to consider cases where the strings are empty or exceed 1000 characters because they’re outside the problem’s constraints. The characters in the strings are guaranteed to be lowercase English letters, so we don’t need to consider cases with uppercase letters, numbers, or special characters.
Outputs: The output is an integer value that represents the minimum total ASCII value of characters that must be deleted to make the two input strings equal. We should ensure our solution can handle the entire range of potential output values (which would be from 0, if no characters need to be deleted, up to 1,000,000, if each string is 1000 characters long and we need to delete all characters from one string, assuming each character has a maximum ASCII value of 1000).
Functionality: The function should be able to accurately compute the minimum total ASCII value of deleted characters for any valid inputs. It should handle all edge cases within the constraints, such as strings of length 1, strings that are already equal, and strings that have no common characters.
Performance: Considering the constraints, the function should be efficient enough to execute within a reasonable time for strings with up to 1000 characters. This implies the need for a solution with a time complexity better than O(n^2), where n is the length of the string.

By understanding these aspects, we set the boundaries of the problem, allowing us to focus our solution design and testing efforts within these boundaries.

Distilling the Problem to Its Core Elements

Fundamental Concept: The problem “Minimum ASCII Delete Sum for Two Strings” is fundamentally a problem of string manipulation and dynamic programming. It is a variant of the Longest Common Subsequence problem. We need to find the subsequence common to both strings that has the maximum sum of ASCII values, and then subtract this from the total ASCII values of both strings.
Simplest Description: You have two words, and you want to make them the same by removing some letters. However, each letter has a cost to remove, equivalent to its ASCII value. The goal is to find out the least amount of total cost to make both words the same.
Core Problem and Simplification: The core problem here is to determine the minimum total cost (in terms of ASCII values) required to make two given strings identical by only deleting characters. A simpler version of the problem could be: “Given two strings, find out the minimum number of characters that need to be removed to make the two strings identical.”
Key Components: The key components of this problem are:
- The input strings s1 and s2
- The process of removing characters from both strings to make them identical
- The calculation of the total cost of removed characters
Minimal Set of Operations:
- Calculate the ASCII value of each character in both strings
- Identify the characters to be removed, based on the condition of making the strings identical with the minimum total ASCII value
- Keep track of the total cost of removed characters
- Return the minimum total cost.

Visual Model of the Problem

Visualizing this problem can be best achieved by using a 2D table or grid, a common tool for dynamic programming problems.

Consider the example where s1 is “sea” and s2 is “eat”. Create a grid where the rows represent the characters in s1 (plus an empty character at the beginning), and the columns represent the characters in s2 (plus an empty character at the beginning).

You can fill in this grid with the ASCII deletion costs. Each cell in the grid, (i, j), represents the minimum ASCII deletion sum to make s1[0...i] and s2[0...j] equal. If s1[i] is equal to s2[j], then the cost at this cell is the same as the cost at cell (i-1, j-1) because you don’t have to delete these characters. But if s1[i] is not equal to s2[j], then the cost is the minimum cost of deleting either s1[i] or s2[j], which is min(grid[i-1][j] + ASCII value of s1[i], grid[i][j-1] + ASCII value of s2[j]).

The result will be the value at the bottom right cell of the grid, which represents the minimum ASCII deletion sum to make the whole s1 and s2 equal.

Here is how the table looks like after being filled in:

  _ e a t
_ 0 101 198 314
s 115 216 313 429
e 216 115 212 328
a 313 212 115 229

In this table, 229 represents the minimum ASCII deletion sum to make “sea” and “eat” equal.

Problem Restatement

Sure, let’s rephrase the problem statement:

The task at hand involves two input strings, s1 and s2, which contain lowercase English letters. The goal is to transform these two strings so that they become the same, by only deleting characters. Deleting a character comes with a cost, which is the ASCII value of the character. The objective is to minimize this total deletion cost.

A couple of things to remember:

Both strings need to be equal at the end.
The strings become equal by only deleting characters, not by swapping or replacing.
Each deleted character adds to the total cost, equal to its ASCII value.
The length of both s1 and s2 will be between 1 and 1000 characters.

Therefore, the core problem is a minimization task, where we’re trying to find the most cost-effective way of deleting characters to make the two strings equal.

Abstract Representation of the Problem

Definitely, we can present this problem in an abstract manner as follows:

We have two sequences A and B of alphanumeric elements. We are tasked to make both sequences identical by applying a single type of operation: removal of an element. Every removal operation incurs a cost, which is defined by a given function on the element. The goal is to determine the smallest possible total cost to make both sequences identical.

In this abstracted version of the problem:

Sequences A and B correspond to strings s1 and s2 in the original problem.
Alphanumeric elements correspond to lowercase English letters.
The removal operation corresponds to deleting characters from the strings.
The cost function corresponds to the ASCII value of a character.

This abstract representation focuses on the structural nature of the problem, i.e., dealing with sequences and transformation operations, rather than the specific context involving strings and ASCII values.

Terminology

Yes, there are a few specialized terms that are relevant to this problem:

String: A sequence of characters. In this problem, we are given two strings that we need to manipulate.
ASCII value: ASCII (American Standard Code for Information Interchange) is a standard that assigns numerical values to characters (letters, digits, punctuation, etc.). In this problem, we use the ASCII values of characters to determine the cost of deleting each character.
String deletion: In the context of this problem, deletion refers to removing characters from the given strings. The goal is to delete characters in such a way that the remaining characters in both strings are the same, and the sum of the ASCII values of the deleted characters is as low as possible.
Dynamic Programming: This is a method for solving a complex problem by breaking it down into simpler subproblems and storing the solutions to each subproblem to avoid solving the same subproblem multiple times. For this problem, dynamic programming could be a potential strategy for solving it efficiently.

These concepts are integral to both the understanding of the problem and the formulation of a solution. Understanding the structure of strings and the concept of ASCII values is necessary to grasp the requirements of the problem, and knowledge of string operations and dynamic programming could be beneficial in devising a solution.

Problem Simplification and Explanation

Sure, let’s break down the problem and create an analogy.

You are given two pieces of text (the strings). Each letter in the text has a particular cost associated with it, which is its ASCII value (the ASCII value represents the numerical equivalent of the character in computer systems).

Your task is to make both pieces of text identical by removing some letters from each piece of text. Every time you remove a letter, you have to pay its cost. The goal is to make the texts identical while minimizing the total cost.

Consider this analogy:

You have two piles of different building blocks, and each block has a price tag. You can only remove blocks from the piles. The objective is to make the piles identical (i.e., having the same blocks), but you need to pay every time you remove a block. You want to achieve this with the smallest possible total cost.

The key concepts are:

Strings: These are the piles of blocks or pieces of text you are dealing with.
ASCII value: This is the cost or price of each block or character.
String deletion: This represents the action of removing blocks or characters from the piles or pieces of text.

By understanding these elements and how they relate to each other, you can begin to plan a strategy for solving the problem efficiently.

Constraints

Sure, here are some observations and characteristics of the problem that could be used to our advantage in formulating a solution:

Lowercase English letters: The input strings consist of lowercase English letters only. There are only 26 possible characters, and each one has a unique ASCII value. This can be used to count the frequency of each letter in both strings.
ASCII value as cost: The ASCII value of a character represents the cost of deleting that character. Because ASCII values for English letters are fixed and known, you can calculate the total cost directly from the count of each letter that needs to be removed.
Length of strings: The length of the strings is at most 1000. This size is small enough to allow solutions that iterate over each character in the strings.
Problem of commonality and difference: The core of the problem is about finding common parts of the two strings and evaluating the cost of eliminating the differences. This relates to a common class of problems in computer science dealing with string similarity and difference, which can often be tackled by dynamic programming or greedy algorithms.
Minimization problem: We need to minimize the ASCII sum of deleted characters. This kind of problem often benefits from sorting or other methods of organizing the data to ensure we’re always making the optimal choice.

These are some of the key observations that can help guide the development of an efficient algorithm for this problem.

Analyzing the constraints of a problem can provide insights into the nature of the problem and the kind of solutions that might be feasible. Let’s analyze the constraints for this problem:

String length (1 <= s1.length, s2.length <= 1000): The constraints on the length of the strings tell us that the strings can be quite large. This suggests that an algorithm with time complexity worse than O(n^2) might not be practical. However, the size is still relatively manageable, suggesting that an approach using dynamic programming or other quadratic time complexity solutions could be viable.
Lowercase English letters: The problem specifies that the strings consist of lowercase English letters. This restricts the possible character set to 26 characters, which could allow us to use data structures like arrays of fixed size (26) for efficient frequency counting.
ASCII Value Calculation: The problem is centered around ASCII values, but since the characters are limited to lowercase English letters, the ASCII values range from 97 (‘a’) to 122 (‘z’). This specific range might not lead to any optimization, but it helps understand the problem better and ensures that the sum calculation won’t exceed the integer limit in most programming languages.
Minimization Problem: The aim is to minimize the ASCII sum of deleted characters to make the strings equal. This indicates the need for an optimization strategy, which often points towards approaches like dynamic programming or greedy algorithms.

From these constraints and the nature of the problem, it’s clear that the problem revolves around string manipulation, ASCII value calculation, and optimization to minimize the sum. It also suggests that we might need to use data structures for efficient counting and retrieval, and techniques like dynamic programming for optimal results.

Case Analysis

Sure, let’s take a look at some additional examples, and analyze the edge cases, boundary conditions and the general case.

Example 1:

Input: s1 = "a", s2 = "b"
Output: 195

This is the smallest case where the length of s1 and s2 is 1. Because the two characters are different, both need to be deleted to make the strings equal. The ASCII values of ‘a’ and ‘b’ are 97 and 98, respectively, so the minimum sum is 195.

Example 2:

Input: s1 = "abcd", s2 = "dcba"
Output: 0

In this case, both strings contain the same characters, so no deletions are needed. Hence, the minimum sum is 0.

Example 3:

Input: s1 = "aaa", s2 = "bbb"
Output: 585

In this case, all characters in both strings are the same, but different between the two strings. So, all characters need to be deleted. The ASCII value of ‘a’ is 97 and that of ‘b’ is 98. Therefore, the minimum sum is 397 + 398 = 585.

Example 4:

Input: s1 = "abc", s2 = "def"
Output: 597

This case involves strings with no common characters. So all characters need to be deleted. The ASCII values of ‘a’, ‘b’, ‘c’, ’d’, ’e’, ‘f’ are 97, 98, 99, 100, 101, 102 respectively. The minimum sum is 97 + 98 + 99 + 100 + 101 + 102 = 597.

Example 5:

Input: s1 = "abcdeabcde", s2 = "abcde"
Output: 490

This case includes repeated characters. The optimal way is to make both strings “abcde”. So, we need to delete 5 characters ‘abcde’ from s1. Therefore, the minimum sum is 97 + 98 + 99 + 100 + 101 = 490.

Through these examples, we can observe that the problem involves evaluating the minimum ASCII value of characters to delete to make the two strings equal. It includes the handling of different types of strings, including those with common characters, completely different characters, repeated characters, and empty strings.

Sure, let’s examine some additional examples and categorize them into edge cases, boundary cases, and typical cases. In each case, I will explain the reasoning behind the expected output and what aspects of the problem they highlight.

Edge Cases

Case of smallest inputs:
```
Input: s1 = "a", s2 = "b"
Output: 195
```
This is the smallest possible case with the length of s1 and s2 being 1. Since ‘a’ and ‘b’ are different, we need to remove both to make the strings equal. The ASCII value of ‘a’ is 97, and ‘b’ is 98. So, the sum is 195.
Case of same characters:
```
Input: s1 = "a", s2 = "a"
Output: 0
```
Here, the strings are already equal, so no deletions are needed. Hence, the sum is 0.

Boundary Cases

Case of same characters but different frequencies:
```
Input: s1 = "aa", s2 = "a"
Output: 97
```
Here, both strings have the same character but in different frequencies. We need to delete one ‘a’ from s1 to make the frequencies equal. Hence, the sum is 97.
Case of entirely distinct characters:
```
Input: s1 = "abcd", s2 = "efgh"
Output: 780
```
In this case, there are no common characters between the strings. Therefore, all characters have to be deleted. The sum is the ASCII value of all these characters, which is 780.

Typical Cases

Case of some common characters:
```
Input: s1 = "abc", s2 = "bcd"
Output: 194
```
Here, the strings share some common characters (‘b’ and ‘c’), so we only need to delete the non-common characters (‘a’ from s1 and ’d’ from s2). The sum of the ASCII values of ‘a’ and ’d’ is 194.
Case of equal strings:
```
Input: s1 = "abcd", s2 = "abcd"
Output: 0
```
Here, the strings are already equal, so no deletions are needed. Hence, the sum is 0.

These examples cover a wide range of the input space and highlight different aspects of the problem. We need to handle various scenarios like different frequencies of the same characters, entirely distinct characters, some common characters, and equal strings. Each case will help ensure our solution is robust and handles all possible scenarios.

Visualizing the problem and cases can be done using a table or diagram where you represent each string and the deletions made. Here’s a simple way to do it:

Let’s take the example s1 = "abc" and s2 = "bcd". You could visualize it as such:

s1 =  "a" "b" "c" 
s2 =  ""  "b" "c" "d"

And for the output, we have 194 which is a sum of ASCII values of ‘a’ (97) and ’d’ (97). So the visualization would be something like:

s1 =  "a"(97) "b" "c" ""
s2 =  ""      "b" "c" "d"(97)

Sum of ASCII of deleted characters = 97 + 97 = 194

You could create similar visualizations for all the cases. The deleted characters are denoted by empty strings ("") and the ASCII values are noted alongside.

These visualizations provide a more intuitive understanding of the problem and the process of arriving at the solution. The key here is to understand the principle of alignment and matching, and the operations (deletions) we perform when characters do not match.

Analyzing different cases gives us several key insights:

Case of identical strings (e.g., s1 = “abc”, s2 = “abc”): In this case, we don’t need to delete any character as both strings are already equal. The sum of ASCII values of deleted characters is 0.
Case of completely different strings (e.g., s1 = “abc”, s2 = “def”): In this scenario, we need to delete all characters from both strings to make them equal. The sum of ASCII values of deleted characters is the sum of ASCII values of all characters in both strings.
Case of partially matching strings (e.g., s1 = “abc”, s2 = “bcd”): Here, we only need to delete the characters that do not match. In this example, we delete ‘a’ from s1 and ’d’ from s2. The sum of ASCII values of deleted characters is the sum of ASCII values of ‘a’ and ’d’.
Case of one string being a subset of another (e.g., s1 = “abc”, s2 = “abcd”): In such a scenario, we only need to delete the extra characters in the larger string. Here, we would delete ’d’ from s2. The sum of ASCII values of deleted characters is the ASCII value of ’d’.
Edge Case of empty strings (e.g., s1 = “”, s2 = “”): Both strings are already equal, so no deletion is required. The sum of ASCII values of deleted characters is 0.

These cases help us understand the problem better and guide us in coming up with an algorithm to solve the problem. The strategy is to delete the least possible characters, and when multiple choices are available, prefer deleting lower ASCII characters.

Identification of Applicable Theoretical Concepts

This problem can be solved by leveraging the concepts of Dynamic Programming (DP) and String Manipulation. Here’s how:

String Manipulation: The problem involves comparing two strings, determining their differences, and making certain modifications (deletions). In particular, we need to evaluate ASCII values, count characters, and remove them. These operations fall under the umbrella of string manipulation, a fundamental topic in computer science.
Dynamic Programming (DP): DP is a powerful technique for solving optimization problems by breaking them down into simpler subproblems and reusing solutions to those subproblems to build up the solutions to larger problems. The problem at hand asks for the minimum sum of ASCII values of deleted characters to make two strings equal, which is an optimization problem.

The problem can be modeled as a classic DP problem similar to the Longest Common Subsequence (LCS). The LCS of two strings is the longest sequence of characters that appear left-to-right in both strings (though not necessarily in a contiguous block). In our case, instead of keeping track of the longest length, we want to track the minimum sum of ASCII values of deleted characters.

So, the key concept here is to transform the problem of finding the minimum ASCII sum of deleted characters to make two strings equal into the problem of finding the LCS of the two strings, which is a well-known problem with established solutions. In this way, we can apply the existing DP approach of the LCS problem to simplify our problem and make it more manageable.

Simple Explanation

Think of it as a game of cards, where you and a friend each have a deck of cards. Now, both of you want your deck of cards to look exactly the same. But there are rules to this game. You can’t add any new cards or swap cards with your friend. The only thing you can do is to remove a card from your deck and throw it away.

Now, here’s the tricky part: each card has a value, and throwing away a card costs you points equal to the card’s value. Your goal in this game is to make your deck of cards match your friend’s deck by removing as few cards as possible, so you lose the least amount of points.

This problem is just like that game. Instead of decks of cards, you have two strings (like two lines of text). And instead of removing cards, you’re removing letters. And just like each card had a value, each letter in the string has a value too - its ASCII value. The goal is to make the two strings look the same by deleting letters and adding up their ASCII values, and you want to do this in a way that the total ASCII value of all the letters you delete is as small as possible.

Problem Breakdown and Solution Methodology

Absolutely, let’s start from the beginning and break it down.

Understand the Problem: We have two strings, and we need to make them equal by deleting some characters. The cost of deleting a character is its ASCII value, and we want to minimize this cost.
Key Insight: Since we’re trying to make the strings equal while minimizing the cost, we want to keep as many of the same characters as possible in both strings, as they don’t need to be deleted. This leads us to the core of this problem: it’s essentially about finding the common characters between the two strings.
Approach: We’ll find the frequency of each character in both strings, essentially creating a histogram or frequency chart for each string. From these histograms, we can determine which characters are common to both strings and how often they occur.
Calculation: By subtracting the common characters’ frequencies from the original histograms, we’re left with the characters that need to be deleted from each string. We then calculate the ASCII value for these characters and add them up to get the total cost.
Refine: Finally, we should also consider edge cases where one or both strings might be empty. In such cases, the cost would be the sum of the ASCII values of all characters in the non-empty string.

Let’s illustrate this with an example:

If s1 = “sea” and s2 = “eat”, the frequency histograms would look like this:

s1: {’s’: 1, ’e’: 1, ‘a’: 1} s2: {’e’: 1, ‘a’: 1, ’t’: 1}

We see that ’e’ and ‘a’ are common characters. So, we subtract their frequencies from the histograms:

s1: {’s’: 1} s2: {’t’: 1}

Now we calculate the cost of deleting ’s’ from s1 and ’t’ from s2, which are 115 and 116 respectively. So the total cost would be 231.

As for the effects of changes in parameters, if the strings become longer or contain higher frequency characters, the complexity would increase, making the computation more intensive. However, the approach would remain the same. If one or both of the strings were empty, the cost would simply be the sum of the ASCII values of the non-empty string’s characters.

Inference of Problem-Solving Approach from the Problem Statement

Absolutely, here are some of the key terms and how they guide the problem-solving strategy:

Strings: In this problem, we’re dealing with two strings of lowercase English letters. This guides us towards string manipulation strategies like frequency counting and substring deletion.
ASCII Sum: The problem mentions the ASCII value of characters, suggesting that we might need to work with character codes. This encourages the use of Python’s built-in function ord(), which returns an integer representing the Unicode character.
Minimum: The problem requires us to minimize the ASCII sum of deleted characters. This suggests that we might need to optimize our solution to find the least expensive way to make the strings equal.
Deleted Characters: The task is to delete certain characters to make the two strings equal. This guides us towards approaches that identify and eliminate unnecessary characters, hinting at the possible use of a frequency histogram.
Equal Strings: The goal of the problem is to make two strings equal. This implies we should try to retain the common characters in both strings, leading us to create frequency histograms for both strings and finding the common characters.

Each of these terms helps inform our approach, leading us to a frequency counting strategy where we create histograms for each string, identify common characters, calculate the cost of deleting the remaining characters, and strive to minimize this cost.

One approach to visualize the problem is by using a frequency table, which can be quite useful to track the occurrences of each character in the strings. Here’s a simplified step-by-step guide on how to do it:

Create Frequency Tables: For both strings, construct a frequency table that counts the number of occurrences of each character.
For example, given the strings s1 = “sea” and s2 = “eat”:
Character Frequency in ‘sea’ Frequency in ’eat’
e 1 1
a 1 1
s 1 0
t 0 1
Identify Characters to Delete: Compare the two frequency tables. The characters that appear in both strings should be kept. The characters that only appear in one string need to be deleted.
In our example, ’s’ only appears in “sea” and ’t’ only appears in “eat”. So, these characters need to be deleted.
Calculate ASCII Sum: For each character that needs to be deleted, calculate its ASCII value using the ord() function, then multiply by the frequency of the character. Add these values together to get the total ASCII sum of the deleted characters.
In our example, ASCII(’s’) = 115 and ASCII(’t’) = 116. So, the minimum ASCII sum of deleted characters = 115 + 116 = 231.

Character	Frequency in ‘sea’	Frequency in ’eat’
e	1	1
a	1	1
s	1	0
t	0	1

This diagrammatic representation can give you a clear idea about how to approach the problem. For larger strings, the process remains the same, but the frequency tables will be larger.

The problem doesn’t explicitly state that it should be solved using dynamic programming. However, some key indicators suggest that dynamic programming could be a viable approach:

Optimization Problem: Dynamic Programming (DP) is typically applicable when we’re trying to optimize something. In this case, we’re trying to minimize the ASCII sum of deleted characters.
Overlapping Subproblems: DP is most useful when a problem can be broken down into smaller subproblems and these subproblems overlap. Here, the subproblems could be solving for smaller substrings of s1 and s2. If we notice that we are solving the same subproblems multiple times, then we can save computation time by storing the results of these subproblems - a key characteristic of DP.
Optimal Substructure: This problem has an optimal substructure property, which means the optimal solution can be constructed efficiently from the optimal solutions of its subproblems.

While these clues suggest that a DP solution might work, it’s worth noting that identifying the exact technique or data structure to solve a problem can often require both experience and a deep understanding of the problem domain.

Simple Explanation of the Proof

To clarify, the solution involves using a Dynamic Programming (DP) approach to solve the problem. This problem is a classic example of DP, which relies on the fact that a solution to a problem can be built up from solutions to its subproblems.

Now, let’s talk about the intuition behind the DP solution and how we can understand its proof:

Understanding the problem: Our goal is to find the lowest ASCII sum of deleted characters to make the two strings equal. So, we are essentially looking for a common subsequence in the two strings where the sum of the ASCII values of the characters not in the subsequence is as small as possible.
Defining the subproblems: For each pair of indices (i, j) where i <= len(s1) and j <= len(s2), we define dp[i][j] as the minimum sum of ASCII values of deleted characters to make s1[0..i] and s2[0..j] equal.
Formulating a relation among the subproblems: Now, for each pair of indices (i, j), we have two cases:
- If s1[i] == s2[j], we don’t need to delete anything. So, dp[i][j] = dp[i - 1][j - 1] (i.e., we just carry over the result from the previous state).
- If s1[i] != s2[j], we have two options: delete s1[i] or delete s2[j]. We’ll choose the option that results in the smaller sum. So, dp[i][j] = min(dp[i - 1][j] + ascii(s1[i]), dp[i][j - 1] + ascii(s2[j])).
Base cases: When one of the strings is empty, we just need to delete all characters in the other string. So, dp[i][0] = ascii_sum(s1[0..i]) and dp[0][j] = ascii_sum(s2[0..j]).
Building up the solution: We start with the base cases and then use the relationship among the subproblems to fill up the DP table. Finally, dp[len(s1)][len(s2)] gives us the minimum sum of ASCII values of deleted characters to make s1 and s2 equal.

This DP algorithm correctly solves the problem because it systematically explores all possible ways to make the two strings equal and keeps track of the minimum sum of ASCII values of deleted characters along the way.

Remember, dynamic programming is all about breaking the problem down into simpler subproblems and then building up the solution using those subproblems. Understanding the proof is about understanding how the solution to each subproblem contributes to the solution of the larger problem.

Sure, let’s refine the approach into granular, actionable steps:

Initialization: Initialize an empty 2D array dp where dp[i][j] is used to store the minimum sum of ASCII values of deleted characters to make s1[0..i] and s2[0..j] equal. The size of the array should be (len(s1)+1)x(len(s2)+1).
Base Case Filling: Fill the base cases into our DP table. If one of the strings is empty, we need to delete all characters in the other string. So, for every index i, set dp[i][0] equal to the ASCII sum of the first i characters in s1. Similarly, for every index j, set dp[0][j] equal to the ASCII sum of the first j characters in s2.
DP Table Filling: Now, fill the rest of the DP table using the relationship among the subproblems. We iterate over each cell of the DP table using nested loops (i.e., for i from 1 to len(s1) and for j from 1 to len(s2)). For each cell (i, j), if s1[i-1] is equal to s2[j-1], we set dp[i][j] equal to dp[i - 1][j - 1]. If s1[i-1] is not equal to s2[j-1], we set dp[i][j] equal to min(dp[i - 1][j] + ASCII value of s1[i-1], dp[i][j - 1] + ASCII value of s2[j-1]).
Result Calculation: After the DP table is fully filled, the cell dp[len(s1)][len(s2)] contains the minimum sum of ASCII values of deleted characters to make s1 and s2 equal. So, we just return that value as the result.

Within this solution, the following tasks are repeated and could be potentially abstracted into separate functions or methods:

Calculation of the ASCII sum of a substring: This is done for the base cases and for cells where s1[i-1] is not equal to s2[j-1]. A helper function could take a string and two indices as arguments and return the ASCII sum of the characters between those indices.
Comparison and selection between two operations (i.e., deleting a character from s1 or s2): This is done for each cell in the DP table. A helper function could take the DP table and the current indices as arguments and return the smaller value between deleting the current character from s1 and deleting the current character from s2.

Solution Approach and Analysis

Sure, let’s break down the problem-solving process:

Initialization: We start by initializing a 2D array, which we can think of as a table. Each cell (i, j) in this table represents the minimum sum of ASCII values we need to delete from the first i characters of s1 and the first j characters of s2 to make them equal. It’s like having a ledger that keeps track of the minimum cost of making the partial strings equal.
Base Case Filling: In this step, we fill in the base cases in the table. Imagine if one of the strings is empty (has 0 characters). In that case, the minimum cost of making the two strings equal would be the sum of ASCII values of all characters in the other string. This is because we need to delete all characters in the non-empty string to make it equal to the empty string.
DP Table Filling: Now, it’s time to fill the rest of the table. You can think of it as solving smaller puzzles and using the solutions to solve the bigger ones. We iterate through each cell in the table (from top left to bottom right). For each cell, we compare the corresponding characters in s1 and s2 (s1[i-1] and s2[j-1], because our table has an extra row and column for the base cases). If the characters are the same, we don’t need to delete anything, so we just copy the value from the cell diagonally above and to the left (dp[i-1][j-1]). If the characters are different, we find the minimum cost between deleting the current character from s1 and deleting it from s2, and add the respective character’s ASCII value to the cost.
Result Calculation: At the end of this process, the bottom right cell in the table (dp[len(s1)][len(s2)]) contains the minimum sum of ASCII values that we need to delete to make s1 and s2 equal. This is our desired result.

Let’s illustrate this with the example s1 = “sea”, s2 = “eat”:

We start with the empty 2D array (ignoring base cases):

s1/s2 | "" | "e" | "a" | "t" 
-----------------------------
""    |    |     |     |     
"s"   |    |     |     |     
"e"   |    |     |     |     
"a"   |    |     |     |

We fill in the base cases:

s1/s2 | ""  | "e" | "a"  | "t" 
------------------------------
""    | 0   | 101 | 198  | 314   
"s"   | 115 |     |      |     
"e"   | 231 |     |      |     
"a"   | 332 |     |      |

And then fill in the rest of the table:

s1/s2 | ""  | "e"  | "a"  | "t" 
------------------------------
""    | 0   | 101  | 198  | 314   
"s"   | 115 | 216  | 313  | 429  
"e"   | 231 | 115  | 216  | 323  
"a"   | 332 | 216  | 115  | 216

The bottom right cell tells us the minimum sum of ASCII values to delete to make “sea” and “eat” equal is 216.

Identify Invariant

An invariant in the context of algorithm design is a condition or a property that remains unchanged while an algorithm executes. It aids in the correctness and efficiency of the solution.

In this problem, the invariant is the property of the 2D DP table, specifically, each cell (i, j) in the DP table representing the minimum ASCII sum of deleted characters to make the first i characters in s1 and the first j characters in s2 equal.

As we fill out the DP table, we ensure this invariant is maintained. For every cell, we carefully calculate the value based on its neighboring cells (or directly copy from a neighboring cell in case of a character match). As we progress, the property that the value at DP[i][j] gives us the minimum ASCII sum of deleted characters for first i characters of s1 and j characters of s2 stays constant. This is crucial for the correctness of our solution as we depend on these computed values to calculate further cells and eventually, our final answer which resides at DP[s1.length][s2.length].

Identify Loop Invariant

A loop invariant in this problem is a property or condition that remains unchanged for each iteration of the loop.

In the context of our dynamic programming approach, the loop invariant can be described as follows:

For our two nested loops iterating through strings s1 and s2, at the start of each outer loop iteration, the first i-1 rows of our DP table are correctly filled. Similarly, at the start of each inner loop iteration, the first j-1 cells in the i-th row are correctly filled.

Here, ‘correctly filled’ means that the cell DP[i][j] contains the minimum ASCII sum of deleted characters to make s1 up to index i and s2 up to index j equal.

This invariant holds because each cell in the DP table is calculated based on its neighboring cells, which have been previously computed. This property ensures that the computation of each cell is accurate and is based on the correct information. Thus, the loop invariant is crucial for the correctness of the dynamic programming solution to this problem.

An “invariant” and a “loop invariant” are related concepts, but they are not exactly the same. In this problem they are different.

An invariant is a condition that can be relied upon to be true during the execution of a program or during some portion of it. It is a logical assertion, sometimes checked within the code by an assertion call, that is always held to be true during a certain phase of execution.

A loop invariant, on the other hand, is a specific type of invariant. It is a condition that is initially true and is maintained as true after each iteration of a loop. It’s a property that holds before (and after) each iteration. Knowing its invariant(s) is a central part of understanding the effect of a loop.

In the context of this problem, we spoke earlier about the loop invariant. The invariant, in a broader sense, could be that the DP table accurately represents the minimum ASCII sum of deleted characters for all processed substrings of s1 and s2. This condition holds true at all stages of the dynamic programming algorithm after we’ve started filling the DP table, regardless of whether we’re inside a loop or not. The loop invariant is a more specific form of this condition which applies to each iteration of the loops used to fill the DP table.

Thought Process

Step 1: Understand the Problem

The problem is asking us to find the minimum ASCII sum of deleted characters to make two strings equal. In simpler terms, we are required to remove certain characters from both strings so that they become the same. The cost of removing a character is its ASCII value, and we need to minimize this cost.

Step 2: Insights from the Problem Statement

One key insight from the problem statement is that we’re not asked for the resulting strings but the minimum cost of operations to make them equal. This hints at a Dynamic Programming approach where we are optimizing for a certain parameter—in this case, the minimum sum of ASCII values.

Step 3: Approach the Problem

One potential approach to solve this problem is to use dynamic programming to compare the characters of the two strings one by one. We start from the end of both strings and move towards the beginning. At each step, we have two options—either delete the current character from the first string or delete the current character from the second string.

If the current characters of both strings are equal, we move to the next character without deleting anything. If they’re not equal, we calculate the cost of deleting the current character from both strings and choose the one with the minimum cost. We use a dynamic programming table to store these costs for every pair of characters, which can be used for later comparisons.

Step 4: Coding the Solution

Let’s translate this approach into Python3 code:

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class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        dp = [[0 for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]

        for i in range(len(s1) - 1, -1, -1):
            dp[i][len(s2)] = dp[i + 1][len(s2)] + ord(s1[i])

        for i in range(len(s2) - 1, -1, -1):
            dp[len(s1)][i] = dp[len(s1)][i + 1] + ord(s2[i])

        for i in range(len(s1) - 1, -1, -1):
            for j in range(len(s2) - 1, -1, -1):
                if s1[i] == s2[j]:
                    dp[i][j] = dp[i + 1][j + 1]
                else:
                    dp[i][j] = min(dp[i + 1][j] + ord(s1[i]), dp[i][j + 1] + ord(s2[j]))

        return dp[0][0]

The above code initializes a 2D array dp where dp[i][j] represents the minimum cost to make the strings s1[i:] and s2[j:] equal. We start by calculating the costs for the base cases where one of the strings is empty, and then use this information to find the costs for other cases.

Finally, dp[0][0] will contain the minimum sum of ASCII values of deleted characters to make s1 and s2 equal.

Establishing Preconditions and Postconditions

Parameters:
- The method minimumDeleteSum takes two inputs, s1 and s2.
- Both these parameters are of string type.
- These parameters represent the two strings that need to be made equal by deleting characters.
Preconditions:
- Before this method is called, the two strings s1 and s2 should be defined and passed to the method.
- The length of each string should be in the range [1, 1000].
- Both s1 and s2 should consist of lowercase English letters only.
Method Functionality:
- This method calculates and returns the minimum ASCII sum of deleted characters to make the two strings equal.
- It interacts with the inputs by using dynamic programming to keep track of the minimum ASCII sum required for all prefixes of the strings.
Postconditions:
- After the method has been called, it returns an integer which is the minimum ASCII sum of deleted characters.
- The return value indicates the minimum cost to make the two input strings equal.
- The method does not have any side effects as it doesn’t change the state of the input strings.
Error Handling:
- The method assumes that the preconditions are met. If the preconditions are not met, for instance if a string contains characters other than lowercase English letters, it won’t give the correct result.
- This method does not throw any exceptions or return special values if the preconditions are not met. Instead, it operates under the assumption that valid input parameters are provided.

Problem Decomposition

Problem Understanding:
- In this problem, we have two strings, s1 and s2. The task is to delete certain characters from both strings such that the resulting strings are equal, and the sum of the ASCII values of the deleted characters is minimum.
Initial Breakdown:
- The problem can be broken down into the following broad parts:
  1. Handling the base case where the strings are already equal.
  2. Determining which characters to delete to make the strings equal.
  3. Calculating the ASCII sum of the deleted characters.
Subproblem Refinement:
- For determining which characters to delete, we need to consider all possible combinations of deletions. However, this can be made easier using dynamic programming. For this, we create a 2D table that stores the minimum sum of deleted characters for all prefixes of s1 and s2.
Task Identification:
- Initializing the table to handle base cases is a repeated task. Filling up the table using the defined recurrence relation is another repeated task. Both these tasks can be abstracted and performed in a loop.
Task Abstraction:
- The tasks of initializing and filling up the table are abstracted enough. They can be done using a loop and an if-else condition respectively, and they fit into the context of this problem.
Method Naming:
- The task of initializing the table can be named initializeTable, and the task of filling up the table can be named fillTable.
Subproblem Interactions:
- The task of filling up the table is dependent on the initialization of the table. Therefore, initializeTable needs to be performed before fillTable. Once both these tasks are done, the answer can be found in the last cell of the table, representing the minimum sum of deleted characters for the entire strings s1 and s2.

From Brute Force to Optimal Solution

Let’s begin by discussing a brute force approach to this problem.

The simplest and most straightforward approach to this problem would be to generate all possible pairs of substrings of s1 and s2, and for each pair, we check if they are equal. If they are, we then calculate the sum of the ASCII values of the characters we removed and update our answer if this sum is less than the current minimum.

In Python, this approach could be implemented as follows:

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import itertools

def minimumDeleteSum(s1, s2):
    result = float('inf')

    # generate all possible substrings of s1 and s2
    for len1 in range(len(s1) + 1):
        for substring1 in itertools.combinations(s1, len1):
            for len2 in range(len(s2) + 1):
                for substring2 in itertools.combinations(s2, len2):
                    # if the substrings are equal, calculate the ASCII sum of the removed characters
                    if ''.join(substring1) == ''.join(substring2):
                        removed = [c for c in s1 + s2 if c not in substring1 + substring2]
                        ascii_sum = sum(ord(c) for c in removed)
                        result = min(result, ascii_sum)
    return result

This solution, however, is incredibly inefficient. The time complexity is exponential, as for each string, we generate all its possible substrings, which amounts to 2^n possibilities for a string of length n. Furthermore, for each pair of substrings, we have to check if they are equal, and if so, calculate the ASCII sum of the removed characters. This adds additional time complexity. The space complexity is also quite high, as we store all possible substrings and their ASCII sums.

To optimize this solution, we can make use of dynamic programming. The key insight here is that we can break down the problem into smaller subproblems, solve each subproblem once, and reuse these solutions to solve larger subproblems. We can create a 2D table dp, where dp[i][j] represents the minimum sum of ASCII values of deleted characters to make s1[0..i-1] and s2[0..j-1] equal.

We fill up this table using a bottom-up approach, starting from dp[0][0] and ending at dp[m][n], where m and n are the lengths of s1 and s2 respectively. For each cell dp[i][j], if the characters at the current positions in the strings are the same, i.e., s1[i-1] == s2[j-1], then we don’t need to delete this character, so dp[i][j] is the same as dp[i-1][j-1]. Otherwise, we can either delete the character from s1 or from s2, so we take the minimum of dp[i-1][j] and dp[i][j-1], and add the ASCII value of the deleted character.

This dynamic programming approach improves the time complexity to O(mn), as we fill up a 2D table of size mn. The space complexity is also O(m*n), as we use a 2D table to store the solutions to the subproblems. Compared to the brute force approach, this is a significant improvement.

Code Explanation and Design Decisions

Let’s consider the optimized Python code for the problem:

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class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        m, n = len(s1), len(s2)
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]

        # Initialization: dealing with empty strings
        for i in range(1, m+1):
            dp[i][0] = dp[i-1][0] + ord(s1[i-1])
        for j in range(1, n+1):
            dp[0][j] = dp[0][j-1] + ord(s2[j-1])

        # Main DP transition
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j] + ord(s1[i-1]), dp[i][j-1] + ord(s2[j-1]))

        return dp[m][n]

Initial Parameters: The inputs to the method are s1 and s2, which represent two strings. We are to make these strings equal by deleting characters. The ASCII sum of the deleted characters should be minimized.
Primary Loop: The outer loop iterates over the string s1 while the inner loop iterates over the string s2. For every pair of characters in the strings, it tries to find the minimal ASCII sum that can make the substrings ending at these characters equal.
Conditions/Branches: The condition within the nested loop checks whether the current characters in the strings are equal. If they are, no deletion is necessary and thus the minimum sum remains the same as for the previous characters. If they’re not equal, a character has to be deleted. We calculate the sum for both possible deletions and keep the smaller one.
Updates/Modifications: The 2D table dp is updated within the nested loop. Each cell dp[i][j] keeps track of the minimum sum to make the substrings ending at characters s1[i-1] and s2[j-1] equal. This is an ongoing representation of our solution for the given substrings.
Invariant: Throughout the code, the invariant maintained is the property that dp[i][j] stores the minimum sum of ASCII values to make the two substrings s1[0..i] and s2[0..j] equal. This invariant is necessary to fulfill the problem’s requirements, and it’s maintained throughout all the iterations and updates.
Final Output: The final output dp[m][n] represents the minimum sum of ASCII values that we need to delete to make the entire strings s1 and s2 equal. It satisfies the problem’s requirement by providing the minimum possible sum of ASCII values of the deleted characters.

Coding Constructs

High-Level Strategies: This code utilizes a dynamic programming (DP) strategy, which is an approach that solves complex problems by breaking them down into simpler, overlapping subproblems. The solution for each subproblem is stored and reused when needed, resulting in a more efficient solution.
Explaining to a Non-programmer: Imagine you have two stacks of alphabet blocks and you want to make them identical by removing as few blocks as possible. You start from the top of the stacks and if the topmost blocks are the same, you leave them as is. If they’re different, you remove the block that causes the least disruption (in this case, the block with the lower ASCII value). This code essentially follows the same approach but for two strings of characters.
Logical Elements: The key logical constructs here are loops and conditionals. Loops are used to iterate over the characters in the strings and conditionals are used to determine whether characters should be removed or left intact.
Algorithmic Approach: The code begins by initializing a two-dimensional array that stores the minimum ASCII sum needed to make two substrings equal. It then populates the first row and column to handle the scenario where one string is empty. The main part of the code involves two nested loops, one for each string. For each pair of characters, it checks if they’re equal. If they are, no deletion is necessary. If they’re not, it calculates the sum if each character were to be deleted and keeps the smaller one. Finally, it returns the value in the last cell of the array which represents the minimum sum for the entire strings.
Key Steps: The key steps in the code include:
- Initialization of a two-dimensional array that acts as a DP table.
- Populating the first row and column of the DP table.
- Using nested loops to iteratively compute the minimum sum for each pair of characters.
- Making decisions based on comparisons and keeping track of the minimum sum in the DP table.
- Returning the final minimum sum from the last cell of the DP table.
Algorithmic Patterns: The main pattern in this code is dynamic programming. The code maintains a DP table that stores solutions to subproblems, with each subproblem representing the scenario for a substring of the original strings. It uses these stored solutions to solve larger subproblems, thus eliminating redundant computations and optimizing the overall solution.

Language Agnostic Coding Drills

Dissecting Code into Concepts:
a. Variables and Data Types: Understanding and using different data types like integers, strings, and arrays.
b. Looping Constructs: Familiarity with using looping constructs like for-loops to iterate over arrays or other data structures.
c. Conditional Statements: Ability to use conditional (if-else) statements to execute different code blocks based on certain conditions.
d. Array Manipulation: Knowledge of how to create and manipulate arrays, specifically two-dimensional arrays in this case.
e. Dynamic Programming: Comprehending the concept of dynamic programming and how it can be applied to optimize solutions by storing and reusing results of subproblems.
Listing Concepts in Order of Increasing Difficulty:
a. Variables and Data Types: This is usually one of the first concepts learned in programming. Understanding variables and basic data types is fundamental to writing any code.
b. Looping Constructs: Once you have the basics down, learning how to use loops is typically the next step. This allows you to perform repeated actions on sets of data.
c. Conditional Statements: This adds a level of complexity because it introduces decision making into the code. The code can now take different paths based on certain conditions.
d. Array Manipulation: While arrays are one of the more basic data structures, working with two-dimensional arrays, as is required in this problem, can be more challenging due to their nested nature.
e. Dynamic Programming: This is typically considered a more advanced concept as it requires not only a good understanding of the problem at hand but also the ability to identify and express it in terms of overlapping subproblems. It requires strategic planning and careful implementation.
Problem-Solving Approach:
a. Variables and Data Types: Start by identifying the variables needed and their types. In this problem, we need two strings (s1 and s2) and a two-dimensional array (dp) to keep track of the minimum ASCII sum for the substrings.
b. Looping Constructs: We then identify that we need to traverse the strings. We see that we need to iterate over each character in the strings, so we use two nested for-loops.
c. Conditional Statements: Within these loops, we add decision-making logic. If the current characters in the strings are the same, no deletion is necessary; if they’re different, we need to determine which character to delete.
d. Array Manipulation: As part of this decision-making logic, we also realize that we need to update our two-dimensional array. We use the array to store the minimum ASCII sum for each pair of substrings, updating it based on our decisions.
e. Dynamic Programming: After all these steps, we see that our solution involves solving overlapping subproblems. We optimize our solution by storing and reusing results, following a bottom-up dynamic programming approach. We fill in our two-dimensional array in a way that allows us to build up the solution iteratively, reusing results of smaller subproblems to solve larger ones.

Targeted Drills in Python

Python Coding Drills:
a. Variables and Data Types: python # Defining a variable and assigning a string to it s1 = "hello" s2 = "world" # Defining an integer ascii_sum = 0 # Converting characters to ASCII ascii_sum += ord(s1[0])
b. Looping Constructs: python # Simple for loop iterating over a string for i in range(len(s1)): print(s1[i])
c. Conditional Statements: python # Simple if-else conditional statement if s1[0] == s2[0]: print("Characters are equal") else: print("Characters are not equal")
d. Array Manipulation: python # Defining a two-dimensional array dp = [[0 for _ in range(len(s2)+1)] for _ in range(len(s1)+1)] # Accessing and updating values in the two-dimensional array dp[0][0] = 0
e. Dynamic Programming: python # A simple dynamic programming example that calculates Fibonacci numbers fib = [0, 1] + [0]*(n-1) for i in range(2, n+1): fib[i] = fib[i-1] + fib[i-2] print(fib[n])
Problem-Specific Concepts:
One problem-specific concept in this case is the calculation of the minimum ASCII delete sum. This involves a combination of conditional checking, array manipulation, and dynamic programming, which are all general concepts we’ve covered above. Therefore, no additional drills are needed.
Integration of Drills:
We would start by defining our input strings and a two-dimensional array to keep track of the minimum ASCII delete sum for substrings.
We then add our nested for-loop structure to iterate over the strings. Inside these loops, we add our conditional checks to compare the current characters in the strings.
Based on the results of these checks, we update our array. If the characters are the same, we set the value of the current position in the array equal to the value of the previous position. If they’re different, we calculate the minimum ASCII delete sum based on previous results stored in the array.
Finally, we return the value in the bottom right corner of the array, which gives us the minimum ASCII delete sum for the entire strings. This process encapsulates our dynamic programming concept.
Each of these drills plays a crucial role in our overall solution. The variables/data types drill provides us with the necessary tools to store and manipulate our data. The looping constructs allow us to traverse our data. Conditional statements enable decision-making in our code. Array manipulation is key for us to store our results, and dynamic programming allows us to optimize our solution by avoiding recalculating results for subproblems.