Count Binary Substrings

Identifying Problem Isomorphism

“Count Binary Substrings” can be approximately mapped to “Longest Palindromic Substring”.

Reasoning:

Both of these problems involve analyzing substrings of a string for certain properties.

“Count Binary Substrings” involves counting all the substrings that have equal number of 0’s and 1’s and all 0’s (or 1’s) comes before 1’s (or 0’s), which is a form of balance in the structure of the substrings.

“Longest Palindromic Substring” involves identifying a substring where the characters read the same backward as forward, which is also a form of balance in the structure of the substring.

Despite the different context (binary strings vs palindromes), both problems share a common approach in that they involve scanning through the string and maintaining a form of state or count as they go along.

“Longest Palindromic Substring” is a more complex problem as it involves not just counting the palindromic substrings, but also keeping track of the longest one, which adds an extra layer of complexity to the problem.

The problem “Count Binary Substrings” involves concepts of strings and counting sequences. Here are some simpler problems to prepare for this problem:

1. 344. Reverse String: This problem will get you comfortable with basic string manipulations.

2. 28. Implement strStr(): This problem gives a good understanding of searching for a substring.

3. 387. First Unique Character in a String: You learn how to keep track of characters in a string.

4. 125. Valid Palindrome: Helps with understanding of strings and sequence of characters.

5. 14. Longest Common Prefix: More practice on strings.

6. 20. Valid Parentheses: This problem gives a good understanding of matching pairs, which is somewhat similar to the problem of matching binary substrings.

7. 680. Valid Palindrome II: This problem involves string manipulation and handling multiple cases.

8. 13. Roman to Integer: Although not directly related, this problem helps to build logic for dealing with sequential characters in a string.

9. 38. Count and Say: This problem requires counting sequences in a string, which is a key skill for solving the Count Binary Substrings problem.

10. 151. Reverse Words in a String: This problem gives a good understanding of dealing with substrings.

Practicing these problems provide a good foundation for handling strings and sequences, which are key skills for solving the Count Binary Substrings problem.

This problem can be solved using a two-pointer or “run-length encoding” approach.

The idea is to keep track of consecutive characters (either ‘0’ or ‘1’) and their counts. When the count of the current group equals the count of the previous group, we know that we have a valid substring where 0’s and 1’s are grouped consecutively and in equal number.

Here is a Python implementation of this idea:

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class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        prev, curr, result = 0, 1, 0
        for i in range(1, len(s)):
            if s[i-1] != s[i]:
                result += min(prev, curr)
                prev, curr = curr, 1
            else:
                curr += 1
        return result + min(prev, curr)

In the above code:

• prev and curr are variables to keep track of the count of the current group of consecutive characters and the count of the previous group.
• result is used to count the number of valid substrings.
• We iterate through the string, and whenever we encounter a change in character (from ‘0’ to ‘1’ or from ‘1’ to ‘0’), we update result by adding the minimum count between prev and curr. The reason for adding the minimum count is because the valid substrings are formed by pairing equal number of consecutive 0’s and 1’s.
• We then update prev and curr for the next iteration.
• After the loop, we add the minimum count between prev and curr to result one more time because there could be a valid substring that ends at the last character of the string.

The time complexity of this solution is O(N) as we make one pass through the string, and the space complexity is O(1) as we only use a constant amount of space to store our variables.

This solution uses several fundamental concepts:

1. String Iteration: We iterate over the string s character by character using a loop. This is a basic concept in dealing with any string-related problem.

2. Counters: We use two counters: prev to count the group of same characters that we have just finished processing, and curr to count the current group of same characters.

3. Condition Checking: We use an if condition to check if the current character is different from the previous character. This is a simple but powerful concept used in a wide range of problems.

4. Counter Update: When we encounter a different character, we update the counters prev and curr. This ensures that prev holds the count of the group of characters that we have just finished processing, and curr starts to count the new group of same characters.

5. Result Accumulation: In every iteration, if prev and curr are equal, it means we have found a valid substring. So, we increment result by the minimum of prev and curr.

6. Return Statement: Finally, we return the result as the total number of valid substrings. This result is computed by the accumulation of values when we found each valid substring during our string iteration.

These are the building blocks of the solution. The solution is a specific arrangement of these concepts to solve the problem at hand. Each of these concepts is generally useful in programming and problem solving.

Certainly, let’s delve into the broader principles underlying this solution.

1. Understanding the problem’s requirements: The core objective is to find substrings in the binary string that have an equal number of consecutive 0’s and 1’s. To satisfy this, we need a way to track occurrences of both ‘0’ and ‘1’ in the string.

2. Observation: The key insight is that a valid substring occurs whenever a sequence of consecutive ‘0’s is followed by a sequence of consecutive ‘1’s, or vice versa, of the same length. This observation is crucial to formulate an effective solution.

3. Tracking Character Groups: To act on this insight, we need to keep track of consecutive character groups and their lengths. prev and curr serve this purpose by counting the length of the previous group and the current group respectively.

4. Leveraging Change: We identify a change in the character from ‘0’ to ‘1’ or from ‘1’ to ‘0’ as the critical juncture to act. This is when a group of similar characters ends and a new group starts. At this juncture, we add the minimum of prev and curr to the result, which effectively counts the number of valid substrings that ended at this juncture.

5. Accumulating Results: The problem calls for a count of valid substrings, so we need to accumulate these results over the entirety of the string.

6. End Condition: The last character doesn’t trigger a character change, but it might end a valid substring. Therefore, after the loop ends, we add the minimum of prev and curr to the result one more time to account for this possibility.

These principles represent the larger “why” behind the code. The code is a specific implementation of these principles, using basic programming constructs to track character groups, identify when those groups change, and accumulate a count of valid substrings.

Sure, here is the purpose of the code at a higher abstraction level:

The code is to count the number of binary substrings in a string. A binary substring is a substring of a string that consists of only 0s and 1s. The code works by tracking the number of consecutive 0s and 1s in the string. When the current character is the same as the previous character, the number of consecutive 0s or 1s is incremented. When the current character is different from the previous character, the number of consecutive 0s or 1s is decremented. The code then adds the minimum of the prev and curr variables to the result variable. The result variable tracks the number of binary substrings that have been found.

The purpose of the code is to count the number of binary substrings in a string. This can be useful for a variety of tasks, such as:

• Analyzing the structure of a binary file.
• Counting the number of occurrences of a particular pattern in a binary file.
• Identifying the boundaries of different sections of a binary file.

The code is efficient because it only needs to iterate through the string once. This makes the code efficient because it does not need to iterate through the string multiple times.

The code is also easy to understand because it uses simple building blocks. The building blocks are easy to understand because they are well-known and well-documented. This makes the code easy to understand because it does not use any complex or obscure concepts.

Absolutely, you’ve hit the nail on the head. This solution and the underlying principles it’s based on could be leveraged to perform the tasks you’ve mentioned.

• Analyzing the structure of a binary file: The code essentially identifies and counts substrings in a binary string. In the context of binary files, these substrings could represent distinctive patterns or structures.

• Counting the number of occurrences of a particular pattern in a binary file: While this code identifies equal-length groups of 0’s and 1’s, similar principles could be used to identify other patterns in binary strings (and thus binary files).

• Identifying the boundaries of different sections of a binary file: The shift from a group of 0’s to a group of 1’s, or vice versa, denotes a boundary. Similar approaches could be used to identify section boundaries in binary files based on predefined patterns.

All of these applications fall under the broader domain of data or file analysis, and string manipulation techniques like this one are often essential tools for these tasks.

The basic building block of the solution is Grouping Digits

Yes, you are correct. The basic building block of the solution is grouping digits. The code works by grouping the consecutive 0s and 1s in the string. When the current character is the same as the previous character, the number of consecutive 0s or 1s is incremented. When the current character is different from the previous character, the number of consecutive 0s or 1s is decremented. The code then adds the minimum of the prev and curr variables to the result variable. The result variable tracks the number of binary substrings that have been found.

The grouping of digits is a fundamental concept in computer science. It is used in a variety of tasks, such as:

• Counting the number of occurrences of a particular pattern in a string.
• Identifying the boundaries of different sections of a binary file.
• Parsing and analyzing data.

The grouping of digits is a powerful tool that can be used to solve a variety of problems. In the case of the problem of counting binary substrings, the grouping of digits allows the code to efficiently track the number of binary substrings in a string.

I hope this helps! Let me know if you have other requests or questions.

Absolutely correct! The solution’s core building block involves grouping consecutive identical digits in the binary string. Here’s a breakdown of this process:

1. Exercise 1 - Counting Consecutive Digits:

   First, create a function to count the number of consecutive identical digits in a string. This function should output the digit and its count.

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   def count_consecutive_digits(s): digit = s[0] count = 0 for ch in s: if ch == digit: count += 1 else: print(f"{digit} appears {count} times consecutively.") digit = ch count = 1 print(f"{digit} appears {count} times consecutively.") count_consecutive_digits("00110011")

2. Exercise 2 - Grouping Digits into Lists:

   Now, adapt the function to store each group of consecutive digits in a list. The function should return a list of these groups.

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   def group_consecutive_digits(s): digit = s[0] count = 0 groups = [] for ch in s: if ch == digit: count += 1 else: groups.append((digit, count)) digit = ch count = 1 groups.append((digit, count)) return groups print(group_consecutive_digits("00110011"))

3. Exercise 3 - Comparing Groups:

   Create a function that uses the output from Exercise 2. It should iterate over the groups and for each pair of adjacent groups, check if their counts are equal. If they are, print the pair of groups.

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   def compare_groups(s): groups = group_consecutive_digits(s) for i in range(len(groups) - 1): group1, group2 = groups[i], groups[i+1] if group1[1] == group2[1]: print(f"Groups {group1} and {group2} have equal counts.") compare_groups("00110011")

These exercises will guide you to build the main parts of the final solution. After completing these exercises, you will understand how to group consecutive identical digits and compare the counts of these groups, the basic building blocks of the final solution.

Another basic building block of this solution is Count Transition

Yes, you are correct. Another basic building block of the solution is counting transitions. The code works by counting the number of times the current character is different from the previous character. When the current character is different from the previous character, the code increments the count variable. When the current character is the same as the previous character, the code decrements the count variable. The code then adds the minimum of the count variable and 1 to the result variable. The result variable tracks the number of binary substrings that have been found.

Counting transitions is a fundamental concept in computer science. It is used in a variety of tasks, such as:

• Detecting changes in data.
• Identifying the boundaries of different sections of a binary file.
• Parsing and analyzing data.

Counting transitions is a powerful tool that can be used to solve a variety of problems. In the case of the problem of counting binary substrings, counting transitions allows the code to efficiently track the number of binary substrings in a string.

The grouping of digits and counting transitions are two of the basic building blocks of the solution to count binary substrings. These building blocks are combined to form a solution that is both efficient and easy to understand.

Indeed, another crucial part of the solution is tracking transitions between 0s and 1s, which is effectively counting the boundary or “transition” points between groups of consecutive digits. This gives us the number of potential substrings starting points.

Here are the exercises related to this concept:

1. Exercise 1 - Counting Transitions:

   Write a function to count and print the number of transitions in a binary string.

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   def count_transitions(s): transitions = sum(s[i-1] != s[i] for i in range(1, len(s))) print(f"Number of transitions: {transitions}") count_transitions("00110011") # Output: Number of transitions: 3

2. Exercise 2 - Identifying Transitions:

   Modify the function to not only count the transitions, but also to mark and print the transition points in the binary string.

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   def identify_transitions(s): transitions = [s[i-1] != s[i] for i in range(1, len(s))] transition_points = [i for i, transition in enumerate(transitions, start=1) if transition] print(f"Transition points: {transition_points}") identify_transitions("00110011") # Output: Transition points: [2, 4, 6]

3. Exercise 3 - Transition Groups:

   In this exercise, modify the function to group the transitions and print out the group lengths. This is essentially counting the number of 0s and 1s between each transition point.

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   def transition_groups(s): groups = [] prev_char = None for char in s: if char != prev_char: groups.append(1) else: groups[-1] += 1 prev_char = char print(f"Transition groups: {groups}") transition_groups("00110011") # Output: Transition groups: [2, 2, 2, 2]

By completing these exercises, you gain an understanding of how to track transitions in the binary string, a key concept for the final solution.

Clarification Questions

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

Identifying Problem Isomorphism

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class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        pre_cont_occ, cur_cont_occ = 0, 1
        counter = 0

        for idx in range(1, len(s)):
            if s[idx] == s[idx-1]:
                cur_cont_occ += 1

            else:
                counter += min(pre_cont_occ, cur_cont_occ)
                pre_cont_occ = cur_cont_occ
                cur_cont_occ = 1

        counter += min(pre_cont_occ, cur_cont_occ)

        return counter

Problem Classification

Domain Classification: The problem statement falls under the domain of “String Manipulation” and “Counting”, which is a common topic in computer science and programming, especially in problems related to pattern recognition and data analysis.

‘What’ Components:

1. Input: The problem gives us an input, which is a binary string s.

2. Desired Output: The goal is to find the number of non-empty substrings in the given string that have the same number of 0’s and 1’s.

3. Constraints: There are a few conditions that the substrings must meet: a. All the 0’s and all the 1’s in these substrings are grouped consecutively. b. If the same substring occurs multiple times, each occurrence is counted separately.

Problem Classification: This problem can be classified as a “Counting and Enumeration” problem with an aspect of “Pattern Recognition”. It involves scanning the string to identify certain patterns (substrings that contain equal numbers of consecutive 0’s and 1’s) and count them. It’s also a “String Manipulation” problem as it requires understanding and manipulation of string data type.

Language Agnostic Coding Drills

Distinct Coding Concepts:

1. Variable Initialization: Variables are created to keep track of certain values, in this case pre_cont_occ, cur_cont_occ, and counter.

2. Iterating through a string: This involves looping through each character in the given string, s, using a for-loop.

3. String Indexing: Accessing elements of a string by their indices, for comparison purposes.

4. Conditional statements (if-else): Conditional statements are used to perform different actions based on different conditions. In this problem, they’re used to decide whether to increment cur_cont_occ or to update pre_cont_occ, cur_cont_occ, and counter.

5. Mathematical operations: This includes addition and the use of the min function to add the smaller of pre_cont_occ and cur_cont_occ to counter.

Increasing Difficulty:

1. Variable Initialization - This is the most basic concept where variables are declared to hold values.

2. String Indexing - This concept involves accessing specific elements of a string using their index, which requires understanding of how indexing works.

3. Iterating through a string - This involves looping over each element in a string, slightly more complex as it requires understanding of loops.

4. Mathematical operations - While basic arithmetic operations are easy, the use of the min function adds a layer of complexity.

5. Conditional statements (if-else) - This is the most complex of the identified concepts, as it requires a strong understanding of logical conditions and flow control in programming.

Problem-Solving Approach:

The given problem is solved by:

1. Initializing variables to keep track of the number of consecutive occurrences of the current character (cur_cont_occ), the previous character (pre_cont_occ), and the total number of valid substrings (counter).

2. Iterating through the string. If the current character is the same as the previous one, increment cur_cont_occ. Otherwise, add the smaller of pre_cont_occ and cur_cont_occ to counter, then set pre_cont_occ equal to cur_cont_occ and cur_cont_occ equal to 1.

3. Once the string is completely iterated, there may be a final valid substring that ends at the last character of the string. To account for this, add the smaller of pre_cont_occ and cur_cont_occ to counter one last time.

4. Finally, return counter as the total number of valid substrings in the original string.

Targeted Drills in Python

Drill 1: Variable Initialization

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x = 10
y = "Hello"
z = []

This drill is to demonstrate how to initialize different types of variables. In Python, you don’t need to declare the type of the variable. It gets determined by the value it is assigned.

Drill 2: String Indexing

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s = "Python"
print(s[0])  # prints "P"
print(s[-1])  # prints "n"

This drill shows how to access individual characters of a string using their index.

Drill 3: Iterating through a string

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s = "Python"
for character in s:
    print(character)

This drill shows how to iterate through each character in a string. We use a for-loop in Python to achieve this.

Drill 4: Mathematical operations

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x = 10
y = 20
print(x + y)  # prints 30
print(min(x, y))  # prints 10

This drill shows how to perform simple mathematical operations in Python, such as addition and finding the minimum of two numbers.

Drill 5: Conditional statements (if-else)

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x = 10
if x > 10:
    print("Greater")
elif x < 10:
    print("Smaller")
else:
    print("Equal")

This drill shows how to use conditional statements to control the flow of the program based on different conditions.

Problem-specific Drills:

Drill 6: String Comparison

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s = "Hello"
if s[0] == s[1]:
    print("Equal")
else:
    print("Not Equal")

This drill shows how to compare two characters in a string. This concept is important for our problem because we need to compare adjacent characters in the binary string.

Integration of the Drills:

To solve the given problem, we can integrate these drills in the following way:

1. Start by initializing the variables pre_cont_occ, cur_cont_occ, and counter to keep track of the consecutive occurrences and the count of valid substrings.

2. Begin iterating through the string (Drill 3). On each iteration, compare the current and previous characters (Drill 6 and Drill 2).

3. If they’re equal, increase cur_cont_occ by 1 (Drill 1 and Drill 4). If they’re not, add the minimum of pre_cont_occ and cur_cont_occ to counter, and update pre_cont_occ and cur_cont_occ (Drill 4 and Drill 5).

4. Once the iteration is over, add the minimum of pre_cont_occ and cur_cont_occ to counter again, to account for a possible valid substring ending at the last character.

5. Finally, return counter as the result. This is the total count of valid substrings in the original binary string.

Sure, here are some smaller exercises that you can do to practice the different concepts used to solve the problem of counting binary substrings:

Exercise 1:

• Write a function that takes a string as input and returns the number of consecutive 0s in the string.

Exercise 2:

• Write a function that takes a string as input and returns the number of consecutive 1s in the string.

Exercise 3:

• Write a function that takes a string as input and returns the minimum of two integers.

Exercise 4:

• Write a function that takes a string as input and returns the number of binary substrings in the string.

By practicing these exercises, you will gain a deeper understanding of the concepts used to solve the problem of counting binary substrings. You will also be able to write more efficient and elegant code to solve this problem.

Here are some additional tips for breaking down the solution into smaller exercises:

• Start by identifying the key concepts that are used in the solution.
• For each key concept, write a function that implements the concept.
• Once you have written the functions, you can combine them to solve the problem.
• It is important to break the problem down into small, manageable pieces. This will make the problem easier to understand and solve.

I hope this helps!

Sure, here is the code in Python3 for the exercises:

Exercise 1:

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def count_consecutive_zeros(str):
    count = 0
    for char in str:
        if char == "0":
            count += 1
        else:
            break
    return count

Exercise 2:

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def count_consecutive_ones(str):
    count = 0
    for char in str:
        if char == "1":
            count += 1
        else:
            break
    return count

Exercise 3:

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def min(a, b):
    if a < b:
        return a
    else:
        return b

Exercise 4:

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def count_binary_substrings(str):
    prev = 0
    curr = 1
    result = 0
    for i in range(1, len(str)):
        if str[i-1] != str[i]:
            result += min(prev, curr)
            prev, curr = curr, 1
        else:
            curr += 1
    return result + min(prev, curr)

I hope this helps!

Absolutely, let’s break the problem down into a series of smaller exercises:

1. Exercise 1 - Character Comparison:

   Write a function to iterate over a string and print a message whenever the current character is different from the previous one.

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   def char_comparison(s): for i in range(1, len(s)): if s[i-1] != s[i]: print(f"Character changed at index {i} from {s[i-1]} to {s[i]}")

2. Exercise 2 - Counting Consecutive Characters:

   Modify the previous function to count the number of consecutive identical characters. Reset the count every time the character changes.

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   def count_consecutive_chars(s): count = 1 for i in range(1, len(s)): if s[i-1] != s[i]: print(f"{s[i-1]} repeated {count} times") count = 1 else: count += 1

3. Exercise 3 - Tracking Previous and Current Counts:

   Adapt the previous function to keep track of both the current character count and the previous character count.

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   def track_prev_and_curr_counts(s): prev, curr = 0, 1 for i in range(1, len(s)): if s[i-1] != s[i]: print(f"Previous count: {prev}, Current count: {curr}") prev, curr = curr, 1 else: curr += 1

4. Exercise 4 - Counting Valid Substrings:

   Building on the previous function, instead of printing counts, increment a result variable whenever the count of the current group equals the count of the previous group.

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   def count_valid_substrings(s): prev, curr, result = 0, 1, 0 for i in range(1, len(s)): if s[i-1] != s[i]: result += min(prev, curr) prev, curr = curr, 1 else: curr += 1 return result

These exercises progressively build up the concepts used in the original problem and provide a step-by-step pathway to the final solution. By tackling each exercise, you can gain a better understanding of the individual components that make up the solution.

Q&A

Can anyone say what is wrong in this logic as i’m getting worng result for some inputs

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var countBinarySubstrings = function (s) {
  let balance = 0;
  let count = 0;

  outer: for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      if (s[j] === "0") {
        balance--;
      } else {
        balance++;
      }
      if (balance === 0) {
        count++;
        continue outer;
      }
    }
    balance = 0;
  }
  return count;
};

Your logic has a small flaw. The balance variable is not sufficient to account for the requirement that the number of 0’s and 1’s are equal and also that they are grouped consecutively. The balance variable will be 0 if there are an equal number of 0’s and 1’s, regardless of their ordering. Therefore, your function will incorrectly count substrings where the 0’s and 1’s are not grouped consecutively.

Let’s look at an example, s = "1001". Your function will count this as a valid substring because it has an equal number of 0’s and 1’s. However, the 0’s and 1’s are not grouped consecutively, which violates the problem constraints.

To fix the problem, you can keep track of two groups: the current group of identical digits and the previous group. As you iterate over the string, if you encounter a different digit, you can compare the sizes of the two groups. If the previous group’s size is greater than or equal to the current group’s size, you can increment your counter.

Here’s a revised version of your function that uses this approach:

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var countBinarySubstrings = function(s) {
    let prev = 0, curr = 1, count = 0;
    
    for (let i = 1; i < s.length; i++) {
        if (s[i] == s[i - 1]) {
            curr++;
        } else {
            prev = curr;
            curr = 1;
        }
        
        if (prev >= curr) {
            count++;
        }
    }
    
    return count;
};

In this function, prev and curr represent the sizes of the previous and current groups of identical digits. count is the number of valid substrings. As you iterate over the string, if you encounter a digit that is the same as the previous digit, you increment curr. If you encounter a different digit, you set prev to curr and curr to 1. Then, if prev is greater than or equal to curr, you increment count.

This approach ensures that you only count substrings where the 0’s and 1’s are grouped consecutively.

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 public int CountBinarySubstrings(string s)
        {
            int count = 0;
            int consecutiveCount = 1;
            int lastConsecutiveCount = 0;

            for (int i = 1; i < s.Length; i++)
            {
                if (s[i] != s[i - 1])
                {
                    lastConsecutiveCount = consecutiveCount;
                    consecutiveCount = 1;
                    count++;
                }
                else
                {
                    consecutiveCount++;
                    if (consecutiveCount <= lastConsecutiveCount) count++;
                }
            }

            return count;
        }

Python equivalent:

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class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        count = 0
        consecutive_count = 1
        last_consecutive_count = 0

        for i in range(1, len(s)):
            if s[i] != s[i - 1]:
                last_consecutive_count = consecutive_count
                consecutive_count = 1
                count += 1
            else:
                consecutive_count += 1
                if consecutive_count <= last_consecutive_count:
                    count += 1

        return count

In this Python function, we use the same logic as your C# function to count the binary substrings. We iterate over the string s, keeping track of the current consecutive count and the last consecutive count. When we encounter a character that is different from the previous one, we update the last consecutive count and reset the current consecutive count to 1. If the current character is the same as the previous one, we increment the current consecutive count. If the current consecutive count is less than or equal to the last consecutive count, we increment our counter. At the end, we return the count of valid substrings.

The solution is computing things on the fly, or as it’s also known, in a “just in time” manner.

The key idea here is to compute and update the consecutive_count and last_consecutive_count as we traverse the string. At each step, the function decides whether to increment the count based on the current and previous run of identical characters. We don’t need any pre-computation or post-processing steps - everything is done in a single pass through the string.

This approach is also efficient, as we only make one pass through the input string, making the time complexity of this solution O(n), where n is the length of the string.

This pattern of problem-solving is often referred to as a “Two Pointer” or “Two Counter” approach, as it keeps track of two quantities (consecutive_count and last_consecutive_count), updating them as it iterates through the input.

So, in the spirit of what you’re asking, it could be named “Just-In-Time Two Counter” or “Just-In-Time Two Pointer”. However, note that these are not established terminology, but rather an attempt to name the approach in a way that captures its essence based on your question.