Buddy Strings

Identifying Problem Isomorphism

“Buddy Strings” can be mapped to “Valid Anagram”.

“Buddy Strings” requires you to check whether it is possible to achieve two equal strings by swapping exactly one pair of characters in the first string. The strings consist of lowercase English letters.

“Valid Anagram” requires you to determine if two given strings are anagrams of each other. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

The connection between these problems is that both involve comparisons between two strings, focusing on the frequencies and positions of their characters.

“Valid Anagram” is simpler as it doesn’t involve the swapping condition. It simply requires a comparison of character frequencies, whereas “Buddy Strings” necessitates checking for two specific conditions related to character swaps.

10 Prerequisite LeetCode Problems

“Buddy Strings” involves some intricacies of string manipulation and understanding of string properties. Here are some problems to solve before tackling “Buddy Strings”:

“Reverse String” (LeetCode 344): This problem helps understand basic string manipulation techniques.
“Valid Palindrome” (LeetCode 125): This problem provides practice for inspecting elements from both ends of a string.
“Valid Anagram” (LeetCode 242): This problem introduces the concept of character frequencies in a string, which is important for “Buddy Strings”.
“Longest Substring Without Repeating Characters” (LeetCode 3): The concept of characters and their positions in a string becomes important in this problem.
“First Unique Character in a String” (LeetCode 387): This problem extends on the concept of character frequencies.
“Jewels and Stones” (LeetCode 771): Here you practice counting characters in a string.
“Find All Anagrams in a String” (LeetCode 438): This problem combines the concepts of character frequencies and string windows.
“Isomorphic Strings” (LeetCode 205): This problem introduces the concept of character mapping, which is also used in “Buddy Strings”.
“Permutation in String” (LeetCode 567): This problem continues to practice character frequencies and string windows.
“Palindrome Permutation” (LeetCode 266): This problem involves understanding and manipulating character frequencies.

Problem Classification

The domain of this problem is string manipulation and comparison. We are tasked with manipulating one string (via swapping two characters) and comparing the resultant string with a given target string.

‘What’ Components:

Two input strings, ’s’ and ‘goal’.
A swap operation that can be performed on ’s’. The operation involves exchanging two characters at different indices.
A comparison operation where the modified ’s’ is compared with ‘goal’ to check for equality.
A Boolean return type, returning ’true’ if the manipulated ’s’ equals ‘goal’, and ‘false’ otherwise.

This problem can be classified as a string manipulation and comparison problem. It involves the operations of swapping (a form of modification) and comparison to check equality. The goal is to determine if a single swap operation can transform the first string into the second. Given the constraints, the problem has to be solved in a way that is efficient for large strings, up to length 2 * 104.

The problem does not specify whether multiple swap operations can be performed, or if we are restricted to only one. From the problem statement, it’s assumed we are restricted to a single swap operation. This adds an additional constraint and complicates the problem by requiring a specific modification (single swap) that leads to a match, rather than any modification. The constraints section does not provide additional clarity on this, but it does limit the input size and specifies the character set, which will influence the method used to solve the problem.

The problem can be seen as analogous to correcting a spelling mistake in a word. Specifically, if a spelling mistake was made by swapping two letters in a word (like typing “hte” instead of “the”), this problem would be about determining whether you could correct the mistake by swapping two letters back.

This is a good example of how you can use real-world analogies to help understand and reason about abstract problems. It also shows how versatile problem-solving strategies can be, as the same solution concept used to solve this problem could potentially be applied in various other contexts, like error correction in data transmission, or even in genetic algorithms where swapping two genes could lead to a better solution.

Clarification Questions

What are the clarification questions we can ask about this problem?

Visual Model of the Problem

Visualizing this problem can be done by considering the strings ’s’ and ‘goal’ as sequences of boxes or slots that contain individual characters. These boxes are ordered and indexed from 0 to n-1 (where n is the length of the strings), and each box can contain any lowercase letter.

You can start by placing the strings ’s’ and ‘goal’ side by side for comparison.
For example, let’s take s = "ab" and goal = "ba". You can visualize it as:
s: | a | b | goal: | b | a |
For the swapping operation, imagine choosing two boxes in the ’s’ string and swapping their contents.
In our example, swapping the characters at index 0 and 1 would result in s = "ba":
s: | b | a | goal: | b | a |
After the swap operation, we compare the transformed ’s’ with ‘goal’. If they are identical, we return true. Otherwise, we return false.
In this case, the strings are identical after the swap, so we return true.

Remember, the problem is asking if there exists a valid swap operation that can make ’s’ and ‘goal’ identical. If such a swap operation exists, the visualization would help identify the indices that need to be swapped. This can be done by visually comparing the strings and identifying the indices where the characters differ.

Problem Restatement

I’ll rephrase the problem statement to understand it better:

We have two strings, ’s’ and ‘goal’, both of them contain lowercase letters. The lengths of these strings are between 1 and 2 * 104, inclusive.

The main task here is to check if it’s possible to make string ’s’ identical to ‘goal’ by performing exactly one swap operation. A swap operation is defined as selecting any two distinct positions in the string ’s’ and exchanging the characters at these positions.

If we can make ’s’ identical to ‘goal’ with a single swap, we should return true. If it’s not possible to do so, then return false.

A couple of examples to make the problem clearer:

For ’s’ = “ab” and ‘goal’ = “ba”, we can swap ‘a’ and ‘b’ in ’s’ to make it equal to ‘goal’. Hence, return true.
For ’s’ = “ab” and ‘goal’ = “ab”, any swap will result in ’s’ not equal to ‘goal’. Hence, return false.
For ’s’ = “aa” and ‘goal’ = “aa”, no swap is needed as both are already identical. In this case, it’s considered that we can swap the same character with itself, and return true.

It’s important to note that the positions of the characters to be swapped must be distinct. This means we cannot select the same position for both characters in the swap operation unless the character is the same in both strings.

Abstract Representation of the Problem

Given two sequences of elements, we can perform exactly one operation that involves swapping the position of any two distinct elements in the first sequence. The goal is to determine whether we can make the first sequence identical to the second sequence by performing this operation.

The operation is considered valid if it involves swapping the same element with itself. If it’s possible to achieve the goal with a single operation, we return a positive outcome; otherwise, we return a negative outcome. The lengths of the sequences fall within a certain range, and each sequence only contains a specified set of elements.

The specific details, like the sequences being strings of lowercase letters and the operation being a swap of characters, can be filled in depending on the concrete problem we want to solve, but the abstract problem remains the same. This abstraction helps us recognize that we might be able to apply the same or similar strategies to solve other problems with the same structure.

Terminology

In the context of this problem, here are a few important terms and concepts:

Swap: In computer programming, a swap is an operation that interchanges the values of two variables. Swapping two characters at different indices is a central operation in this problem.
String: A string is a sequence of characters. In most programming languages, strings are a fundamental data type, and many standard operations and manipulations can be performed on them, such as indexing, slicing, and swapping characters.
Equality of Strings: Two strings are considered equal if they are of the same length and contain the same characters in the same order. This is an important concept for this problem as we need to compare if the string obtained after swapping is equal to the target string.
Index: In the context of arrays or strings, an index is a numerical representation of a position of an element or character within the array or string. This problem requires swapping two characters in a string, which involves the use of indices.

Understanding these basic concepts is essential to correctly implement a solution to the problem.

Problem Simplification and Explanation

We are given two words, let’s say they are “codes” and “decos”. Now our task is to see if we can make these two words match by swapping exactly two letters in the first word.

The key concept involved here is the idea of a “swap”. Think of each letter in the word as a person sitting in a chair. A swap would be like having two people switch chairs with each other.

The other main concept is string equality. This is the idea that two words match if they have the exact same letters in the exact same positions.

So, the problem is like having two rows of chairs with people (letters) sitting in them. Can you make the rows identical by having two people in the first row switch seats?

A couple of conditions make this more complex though. One, the rows have to be the same length (the words have to be the same number of letters long). And two, if the two rows are already identical, you can’t switch two identical people (letters) in the first row to make them match, as that would be cheating!

The interaction of these key concepts forms the basis of our problem. Our task is to figure out a way to decide if a swap is possible and necessary to make the two words (strings) match.

Constraints

Let’s explore these constraints:

“1 <= s.length, goal.length <= 2 * 10^4”: The length of the strings is limited to 20,000 characters. This means our solution needs to work efficiently even for the largest input size. So, solutions with time complexity worse than O(n) may not work in reasonable time. This encourages us to find a linear solution.
“s and goal consist of lowercase letters”: The input strings only contain lowercase English alphabet letters. This means that we are working with a fixed and known set of characters (26 to be exact). We can leverage this information by using a data structure like an array or a dictionary that can efficiently track and compare the frequency of each character in the two strings.

The specific condition that we can exploit here is that we are allowed to perform exactly one swap operation. So we need to compare each character at the same position in both strings. If the characters are different, we should remember these positions. If we have exactly two such positions and swapping characters at these positions in the first string makes the two strings equal, then we can conclude that one swap operation is enough. If there are more than two positions where the characters are different, it’s impossible to make the strings equal with one swap.

Therefore, we can use a linear scan to compare each character in the same position of both strings, and use a small array or dictionary to track the positions of different characters, and then determine the possibility of a swap operation based on the comparison result.

Analyzing the constraints provides several key insights:

Bounded Lengths: The constraint that the lengths of s and goal are within a specific range (from 1 to 2*10^4) indicates that we must be careful about time complexity. We cannot afford to have solutions that are worse than O(n), as they may not run within a reasonable time for large inputs. So, we must focus on finding a linear-time solution.
Limited Characters: The strings s and goal only consist of lowercase letters. This is useful to know because we can use data structures like arrays or dictionaries to efficiently track and compare the frequency of each character in the two strings. This drastically reduces the problem space and simplifies the problem by limiting the types of inputs that we can receive.
Single Swap: The fact that we’re allowed to swap only two letters in the string implies we should look for scenarios where swapping two letters in s would make it equal to goal. This can help to further narrow down our problem solving approach and focus on the positions of different characters between s and goal.

By using these insights, we can come up with an efficient approach to solve the problem. We need to do a linear scan of both strings, tracking the positions of different characters, and finally analyze the tracked positions to determine whether a single swap operation can make the two strings equal.

Case Analysis

Considering small values of the variables and exploring edge cases often help to gain insights about a problem. Here are some examples:

Case: s and goal are identical.
- s = “a”, goal = “a”
- As both strings are the same, we need two identical characters to swap with each other. As there’s only one character, and it’s the same in both strings, we can’t make a swap. So, the result is false.
Case: s and goal have different characters.
- s = “a”, goal = “b”
- Here, we can’t make a swap that will make s equal to goal because the characters are different. So, the result is false.
Case: s and goal are reversed.
- s = “ab”, goal = “ba”
- In this case, swapping the two characters in s results in goal. So, the result is true.
Case: s and goal have two of the same characters.
- s = “aa”, goal = “aa”
- Here, swapping the two ‘a’ characters in s still results in “aa”, which equals goal. So, the result is true.
Case: s and goal have more than two characters, and are not identical.
- s = “abc”, goal = “acb”
- Here, by swapping ‘b’ and ‘c’ in s, we can make s equal to goal. So, the result is true.
Case: s and goal have more than two characters, and are identical.
- s = “abc”, goal = “abc”
- Here, since s and goal are already the same, and there are no duplicate characters in s, we cannot make a valid swap. So, the result is false.
Case: s and goal have more than two characters, and are identical but have at least one character repeating.
- s = “aab”, goal = “aab”
- Here, since s and goal are already the same, and there are duplicate characters in s, we can swap the two ‘a’s and still have the same string. So, the result is true.

Through these examples, we can see that the problem can be distilled down to checking if s and goal have the same characters and if there’s at least one character in s that appears more than once if s and goal are identical.

The problem statement does not explicitly specify that the two strings, s and goal, must be of the same length. However, if the lengths of the strings s and goal are not equal, we can immediately return false, since it would be impossible for a single swap operation within one string to make two strings of different lengths equal.

So, an additional edge case to consider is:

Case: s and goal are of different lengths.

s = “abc”, goal = “ab”
Here, since the lengths of s and goal are not the same, we can immediately return false. No amount of swapping within s would make s equal to goal.

Considering the lengths of the input strings is indeed a crucial aspect of this problem.

In addition to the cases already discussed, here are some more edge and boundary cases to consider:

Case: The strings s and goal are of length 1.

s = “a”, goal = “a”
As there is only one character, no swaps are possible. But since the strings are already identical, the output should be true.
s = “a”, goal = “b”
As there is only one character, no swaps are possible. Since the characters are different, the output should be false.

Case: The strings s and goal have all identical characters but are arranged in different orders, and there are more than two characters present in each string.

s = “abc”, goal = “cba”
While each string has the same characters, it’s impossible to make the two strings identical with only one swap in s. So, the output should be false.

Case: The strings s and goal are identical, but there are no duplicate characters present in each string.

s = “abc”, goal = “abc”
Since no duplicates are present, no swap operation can be performed in s without making s different from goal. So, the output should be false.

By considering these edge and boundary cases, we can better ensure our solution handles all possible inputs and avoids bugs. It’s also important to note that these cases could vary based on how the problem statement defines a “swap”, or how it handles cases where s and goal are already identical.

Identification of Applicable Theoretical Concepts

Here are some concepts and methodologies that might be applicable to this problem:

Sorting and Comparison: The order of characters in a string doesn’t matter when determining whether one string can be formed from another by swapping two characters. Sorting both strings and comparing them can give us an answer.
Hashing or Frequency Count: We could create a frequency table or hash map for the characters in the strings. The frequency of each character in both strings should be the same if one can be transformed into the other by swapping two characters.
Count of Mismatches: We can iterate through both strings simultaneously and keep track of the positions where they differ. If the strings can be made identical by swapping two characters, they should have exactly two mismatches, and the characters involved in the mismatches should be able to form pairs when swapped.
Greedy Algorithms: At a high level, we want to make the least number of changes to achieve our goal. This aligns with the principle of greedy algorithms.

Each of these approaches uses different aspects of computer science, algorithmic theory, or mathematics to make the problem more manageable or simplify its solution. They don’t all have to be applied together, and the best one to use depends on the specifics of the problem and the constraints given.

Problem Breakdown and Solution Methodology

Let’s approach this problem in a systematic way using one of the concepts I mentioned earlier - Count of Mismatches and Hashing:

Initialize Variables: Start by initializing two variables, mismatch_count and mismatch_pairs, to 0 and an empty list, respectively. These will keep track of the number of character mismatches between the two strings and the pairs of characters involved in these mismatches.
Iterate Over Strings: Next, iterate over the characters in both strings simultaneously. If the characters at the same position in both strings do not match, increment mismatch_count by 1 and add the pair of mismatched characters to mismatch_pairs.
Check Mismatches: After the iteration, check if mismatch_count is either 0 or 2. If it’s not, return False because it means that more than two swaps are required to make the strings identical, which is not allowed.
Check Pairings: If mismatch_count is 2, check if the characters involved in the two mismatches can form pairs when swapped. This means that the first character of the first mismatch should be the same as the second character of the second mismatch and vice versa. If this condition is met, return True.
Check Frequency Count: If mismatch_count is 0, it means that both strings are already identical. But there’s a catch - we still need to check if there is at least one character that appears more than once in the strings. This is because a swap is still needed, even if it doesn’t change the string. We can use a frequency count or hash map to check this condition.

In a metaphorical sense, consider that you’re trying to correct a typo in a word (string s) to match it with the correct word (string goal). You’re only allowed to swap two characters in the wrong word. As you check each letter from the start, you note the positions where the typo word differs from the correct word. If there are more than two such positions, it’s impossible to correct the typo with just one swap. If there are exactly two, you verify that swapping the letters at these positions will give you the correct word. If there are no differences, it’s like having a repeated letter typed twice in a row - it’s still a typo, but can be corrected by swapping the repeated letter with itself.

Example:

Let’s take the example s = “abcc”, goal = “abcc”. Even though s and goal are identical, the answer is True because we can swap the two ‘c’ in s to still get “abcc”.

During the iteration, mismatch_count remains 0 because there are no mismatches. So, we move to the frequency count check. Here, we find that ‘c’ appears more than once, so we return True.

If we change s to “abcd”, now the answer is False because even though mismatch_count is 0, there are no characters in s that appear more than once, so no valid swap exists.

So, changes in the problem’s parameters, like the characters in the strings and their frequencies, directly affect the solution.

Inference of Problem-Solving Approach from the Problem Statement

Inferring the use of frequency count and greedy algorithms comes from understanding the problem statement thoroughly and having familiarity with various problem-solving strategies, patterns, and algorithmic paradigms. In this problem, the following cues may lead us towards considering these approaches:

Character Comparison and Counting: The problem asks us to compare two strings and find if they can be made equal by swapping two characters. This directly points to comparing the characters of the two strings. Also, the condition that we need to check for a character appearing more than once in case of no mismatches indicates that we need some sort of count for the characters. These clues suggest that frequency count might be a useful approach.
Optimal Decision Making: The problem is asking us to determine whether we can make the strings equal with just a single swap operation. This is essentially an optimization problem - we’re trying to achieve our goal with a minimum number of operations. Greedy algorithms are a common approach for such problems. Here, the greedy choice is to perform a swap operation whenever we find a pair of mismatched characters that would make the strings identical.

However, it’s essential to understand that recognizing these patterns and inferring possible solutions often comes with practice and experience. The more problems you solve, the more patterns you’ll recognize, and the quicker you’ll be able to infer potential solutions.

Here’s a stepwise refinement of our approach:

Step 1 - Input Validation: We should first validate the inputs to ensure the lengths of the two strings are the same, as it’s impossible to make them identical if they aren’t.
Step 2 - Character Frequency Counting: We can initialize an array of size 26 (for 26 lowercase English letters) to store the frequency of each character in string s. We then traverse both strings from the beginning. For every character in s, we increment the count in the corresponding position in our frequency array. For every character in goal, we decrement the count in the corresponding position in our frequency array.
Step 3 - Mismatch Counting and Duplicate Checking: We also keep track of the number of positions where s and goal have different characters (i.e., mismatches). Simultaneously, we check if there’s any character in s that appears more than once (this can be done by checking if any count in our frequency array goes above 1).
Step 4 - Condition Checking and Returning Result: If the number of mismatches is 2 and all the counts in our frequency array are 0 (which means every character in s has a corresponding character in goal), we return true. If the number of mismatches is 0 and we’ve found at least one character in s that appears more than once, we also return true. In all other cases, we return false.

This stepwise approach will allow us to solve the problem effectively. Please note that this solution assumes that the strings only contain lowercase English letters, as stated in the problem. The actual implementation may need to handle edge cases and additional constraints based on the specific requirements of your application or environment.

Here’s a more granular breakdown of the problem:

Step 1 - Initialize Variables:

Create a frequency array of 26 integers all initialized to 0. This will be used to store the frequency of each letter from ‘a’ to ‘z’.
Initialize two integers, mismatchCount and duplicateCount, to 0. These will be used to track the number of mismatching characters between the two strings and the number of duplicate characters in the string s, respectively.

Step 2 - Iterate through Strings:

Iterate over the strings s and goal. These loops can be combined into a single loop if the lengths of s and goal are guaranteed to be the same (which is validated in step 1).
For each character c in s, increment the corresponding index in the frequency array (i.e., frequency[c - 'a']++). If the count at this index now exceeds 1, increment duplicateCount.
For each character c in goal, decrement the corresponding index in the frequency array (i.e., frequency[c - 'a']--).
If at any point, the characters at the current position in s and goal are not the same, increment mismatchCount.

Step 3 - Analyze Frequency Array and Counts:

If mismatchCount is exactly 2 and all the counts in the frequency array are 0, return true. This means that there are exactly two positions that can be swapped in s to make it match goal.
If mismatchCount is 0 and duplicateCount is greater than 0, return true. This means that s and goal are already identical, and s contains at least one character that appears more than once, so we can swap these duplicate characters without changing s.

Step 4 - Return False:

If neither of the conditions in step 3 are met, return false. This means it’s not possible to make s identical to goal by swapping exactly two characters in s.

This is a more detailed breakdown of the problem, giving a step-by-step walkthrough of how we can solve it. It can be used as a blueprint for writing the code solution.

There are a few different parts of this problem that can be isolated and solved independently. Here are some examples:

Checking if two strings are of equal length: You can create a function that takes two strings as input and returns a boolean indicating whether they are of equal length. This can be done independently of the rest of the problem.
Creating a frequency array from a string: You can write a function that takes a string as input and returns a frequency array representing the counts of each character in the string. This function can be used in various other problems and contexts, making it an independent subproblem.
Counting the number of mismatches between two strings: You could write a function that takes two strings of equal length as input and returns the number of positions at which the strings have different characters. This is another example of a subproblem that can be solved independently.
Determining if a string contains any duplicate characters: You can write a function that takes a string as input and returns a boolean indicating whether the string contains any duplicate characters. This can be done independently of the rest of the problem.

By breaking down the problem into these smaller, independent parts, you can simplify the process of developing a solution. Each of these parts can be tested and debugged independently before being integrated into the final solution.

There are a couple of repeatable patterns in our solution:

Iterating over the characters in a string: In various steps of our solution, we iterate over each character in the input strings. This is a common pattern in problems dealing with strings.
Counting the frequency of elements: Constructing a frequency array or dictionary is a common pattern that is used in a wide range of problems, not just this one. It helps us keep track of the number of times each element (in this case, character) appears in a given set (in this case, a string).
Comparing elements in two arrays (or strings): When we compare the two strings character by character to find mismatches, we are employing a common pattern of element-wise comparison of two arrays (or strings).

These patterns are common in many different problems, so spotting them can help you recognize when you can apply a similar approach. Recognizing these patterns can be a stepping stone to devising an algorithm to solve the problem.

Solution Approach and Analysis

Let’s break down the approach for the problem:

We have two strings, s and goal, and our task is to check if we can make these two strings equal by swapping exactly two characters within the string s.

Step 1: Basic Checks
First, we need to check if both strings have the same length, because if they don’t, we can return false immediately.
For example, if s = "abc" and goal = "abcd", no amount of swapping in s will make it equal to goal because they have different lengths.
Step 2: Character Frequency Comparison
Next, we check if both strings have the same frequency of characters. If they don’t, we can return false right away.
Why? Let’s say s = "abc" and goal = "abd". Here, s contains c while goal contains d, and no swapping in s can produce a d in it.
We can calculate the frequency of characters using a frequency table or an array.
Step 3: Finding Mismatches
If the strings pass the above two checks, we move forward and compare the strings character by character.
We are looking for positions where characters in s and goal are different. These are our potential swap candidates.
For example, if s = "abcd" and goal = "adcb", positions 1 and 2 have different characters in s and goal.
Step 4: Deciding Based on Mismatches
We now look at the number of mismatches.
- If there are no mismatches, it means the strings are already equal and we return true.
- If there are exactly 2 mismatches, it could mean that swapping the mismatched characters in s could make it equal to goal, and we return true.
- If the number of mismatches is anything other than 0 or 2, we return false as we can’t make s equal to goal by swapping exactly two characters.

Following these steps should solve the problem as per the requirements. Changes in the problem’s parameters, like increasing the length of the strings, would primarily affect the time complexity of the solution, but the approach itself would remain the same.

Let’s see an example to illustrate this:

s = "abcd", goal = "adcb"
Both strings have the same length.
The frequency of characters in both strings is the same (a: 1, b: 1, c: 1, d: 1).
On comparing s and goal character by character, we find mismatches at position 1 and 2 (b in s vs d in goal and c in s vs b in goal). There are exactly 2 mismatches.
Hence, we can swap b and c in s to make it equal to goal. So, the function would return true.

Thought Process

The problem is about checking whether we can make two strings equal by swapping exactly two characters within one string.

Cues from the Problem Statement

The problem is about string manipulation.
We have the possibility of swapping two characters in the string s.
The goal is to make string s equal to string goal.
A crucial hint is that we can swap exactly two characters.

Insights Generated

The problem can be solved by comparing the two strings and finding out the characters that need to be swapped. However, before that, we need to ensure that both the strings are of the same length and have the same set of characters.

Here is a Python solution that follows the thought process:

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def are_swappable(s, goal):
    # Check if the lengths are different
    if len(s) != len(goal):
        return False

    # Check if the frequency of characters is different
    if sorted(s) != sorted(goal):
        return False

    # Check the number of mismatched characters
    mismatches = sum([i != j for i, j in zip(s, goal)])

    # Return true only if the number of mismatches is 0 or 2
    return mismatches in [0, 2]

The are_swappable function checks whether two strings can be made identical by swapping exactly two characters in one of the strings. It first checks whether the two strings have the same length and the same set of characters. It then counts the number of positions where the two strings have different characters. The function returns True if and only if this count is 0 or 2, indicating that the strings are either already identical or can be made identical by swapping two characters.

From Brute Force to Optimal Solution

A brute force solution to this problem would involve iterating through every pair of characters in string s and swapping them, then checking if the resulting string is equal to the string goal. If it is, we return True, otherwise, we continue checking other pairs. If we finish checking all possible pairs and none of them make the strings equal, we return False.

Here’s a Python code for the brute force approach:

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def are_swappable(s, goal):
    if len(s) != len(goal): return False

    s_list = list(s)
    for i in range(len(s)):
        for j in range(i+1, len(s)):
            s_list[i], s_list[j] = s_list[j], s_list[i]  # swap
            if "".join(s_list) == goal:
                return True
            s_list[i], s_list[j] = s_list[j], s_list[i]  # swap back
    return False

The inefficiencies in the brute force approach are apparent when considering the time complexity. The nested loop structure leads to a time complexity of O(n^2), where n is the length of the string s. For large inputs, this would be highly inefficient.

Optimization Steps:

Instead of attempting to swap every pair of characters, we can just compare the strings and count the positions where the characters are not the same. This will result in a linear time complexity of O(n).
Furthermore, by using a frequency count for the characters in both strings, we can ensure that the strings contain the same characters. If not, we can immediately return False.

By applying these optimizations, we get a solution that has a time complexity of O(n) and a space complexity of O(1), assuming a fixed-size alphabet (which is reasonable in this problem, as we are only dealing with lowercase English letters). Here is the optimized solution in Python:

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def areAlmostEqual(s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
    
    if s == goal:
        return len(set(s)) < len(s)
    
    diffs = [(a, b) for a, b in zip(s, goal) if a != b]
    
    return len(diffs) == 2 and diffs[0] == diffs[1][::-1]

# Test the function
s = "ab"
goal = "ab"
print(areAlmostEqual(s, goal)) # Output: False

Coding Constructs

High-level Problem-Solving Strategies: The code uses a technique of comparing the frequency of characters in both strings and checking the number of mismatches between the two strings. It is essentially a counting mechanism paired with a comparison operation. The strategy hinges on the key insight that a single pair of swaps would either result in 2 characters changing places or no change if the characters being swapped are the same.
Purpose for a Non-programmer: This code checks whether we can make two strings identical by just swapping two characters once in one of the strings. It’s like jumbling up the letters of a word slightly, and then checking if we can unjumble them to match another word by only swapping two letters.
Logical Elements: The logical constructs used here are condition checks (if statements), sorting operation, and a list comprehension combined with a sum function to count the number of differences. It also uses a comparison operator to check the equality of sorted strings and the number of mismatches.
Algorithmic Approach in Plain English: First, we check if both strings have the same length, if not, we know they can’t be made identical. Then we check if both strings have the exact same set of characters (not considering order), if not, they can’t be made identical. Finally, we check how many characters are in the wrong position. If it’s exactly two or zero, then a single swap can make the strings identical, so we return True. If not, we return False.
Key Steps and Operations: The key steps include checking string lengths, sorting the strings and comparing them, and counting the number of positions where characters differ in the two strings. These operations ensure that the two strings can be made identical through a single swap.
Algorithmic Patterns: The algorithm uses elements of counting (frequency analysis), sorting, and comparison, a common pattern in many string manipulation problems. The pattern of checking preconditions (like length and character set) before performing more complex operations is also quite common in algorithmic problem solving.

Language Agnostic Coding Drills

Distinct Concepts in the Code:
- Variable Assignment: This is the most basic concept where we assign values to variables for future use. It’s used in this code to store the two input strings.
- Conditional Statements (If-Else): Conditional statements are used to control the flow of execution based on certain conditions. In this code, they’re used to check preconditions like the length of strings and character sets.
- Sorting: Sorting is a common operation in many algorithms. In this code, sorting is used to order the characters in the strings, which is a prerequisite for the subsequent comparison.
- List Comprehension: List comprehension is a concise way to create lists. Here, it is used to create a list of differences between corresponding characters of the two strings.
- Built-in Functions (sum and zip): Built-in functions offer ready-made solutions for common tasks. sum is used to count the number of differences, and zip is used to iterate over two sequences in parallel.
Coding Concepts/Drills in Order of Increasing Difficulty:
- Variable Assignment: This is the most fundamental concept in programming, where we assign values to identifiers for use in subsequent code. Level: Beginner.
- Conditional Statements (If-Else): Conditional statements require understanding of boolean expressions and control flow. They’re slightly more complex than simple assignment. Level: Beginner.
- Sorting: While using a sorting function is simple, understanding how sorting algorithms work and their complexity requires a deeper understanding of programming and data structures. Level: Intermediate.
- List Comprehension: List comprehension requires understanding of sequences, iteration, and optionally conditionals, making it slightly more complex than basic control flow and data structures. Level: Intermediate.
- Built-in Functions (sum and zip): Using built-in functions effectively requires a good understanding of the language’s standard library and the various data types and structures it operates on. Level: Intermediate to Advanced.
Problem-Solving Approach:
- Understand the Problem: The first step is to understand what the problem is asking for. This involves recognizing that the problem is asking whether we can make two strings identical by swapping two characters in one string.
- Identify Pre-conditions: Before we proceed with more complex operations, we identify and check some preconditions. If the two strings are not of equal length, or they do not consist of the same set of characters, they cannot be made identical by swapping characters.
- Order the Characters: To compare the two strings, we sort their characters. This makes it easier to identify differences.
- Identify Differences: With the characters ordered, we can now identify differences. We create a list that marks where the strings differ.
- Count Differences: We count the number of differences. If it’s exactly two or zero, then a single swap operation can make the two strings identical, and we return True. Otherwise, we return False.

This process involves several basic and intermediate coding concepts, each building on the previous ones. Understanding the problem leads to recognizing the need for conditional statements, which in turn leads to the need for sorting, which then leads to using list comprehension and built-in functions to identify and count differences.

Targeted Drills in Python

Let’s break down our problem-solving approach into the core concepts and create drills for each one of them:

Variable Assignment:

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s = "abcd"
goal = "bcda"

This is the most basic concept, where we assign values to variables.

Conditional Statements (If-Else):

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x = 10
y = 20
if x < y:
    print("x is less than y")
else:
    print("x is not less than y")

Conditional statements are used to control the flow of execution based on certain conditions.

Sorting:

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unsorted_list = [3, 1, 4, 1, 5, 9]
sorted_list = sorted(unsorted_list)
print(sorted_list)  # Output: [1, 1, 3, 4, 5, 9]

Sorting is used to order elements of a sequence.

List Comprehension:

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squares = [x ** 2 for x in range(10)]
print(squares)  # Output: [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

List comprehension is a concise way to create lists.

Built-in Functions (sum and zip):

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nums1 = [1, 2, 3, 4, 5]
nums2 = [6, 7, 8, 9, 10]
paired = list(zip(nums1, nums2))
sum_of_nums1 = sum(nums1)
print(paired)  # Output: [(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
print(sum_of_nums1)  # Output: 15

sum is used to calculate the sum of a sequence, and zip is used to iterate over two or more sequences in parallel.

Problem-Specific Concept: For our problem, the specific concept would be identifying differences between two sequences. Here’s a simple drill to illustrate this concept:

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s1 = "abcd"
s2 = "bcda"
differences = [x != y for x, y in zip(s1, s2)]
print(differences)  # Output: [True, True, True, True]

This drill is essential for our problem because it provides the means to identify which characters in the strings differ. This information is crucial for determining whether the strings can be made identical with a single swap operation.

Integrating the Drills: Now that we have drills for each of the core concepts, we can integrate them to solve our problem. We start with variable assignment, then use conditional statements to check our preconditions. If the preconditions are met, we use sorting to order our strings, and list comprehension along with zip to identify differences. Finally, we use sum to count the number of differences, and conditional statements once again to determine whether a single swap operation can make the strings identical.