Find in Mountain Array

We’re given a mountain array and a target value. We need to find the minimum index at which the target exists in the array. Since we have limited access to the array (up to 100 calls to MountainArray.get), we need to be mindful of how we access the elements.

Here’s an approach using binary search, which allows us to find the result within the constraint:

  1. Find the Peak: Use binary search to find the index of the peak element in the mountain array (the highest value).

  2. Search Left of the Peak: Again, use binary search to search for the target in the increasing part of the array (from index 0 to the peak). If the target is found, return the index.

  3. Search Right of the Peak: If the target is not found in the left part, use binary search to search for the target in the decreasing part of the array (from the peak to the end of the array). If the target is found, return the index.

  4. Return -1 if Not Found: If the target is not found in either part, return -1.

Here’s the code:

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class Solution:
    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
        # Find the peak using binary search
        left, right = 0, mountain_arr.length() - 1
        while left < right:
            mid = left + (right - left) // 2
            if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
                left = mid + 1
            else:
                right = mid
        peak = left

        # Search left of the peak
        left, right = 0, peak
        while left <= right:
            mid = left + (right - left) // 2
            val = mountain_arr.get(mid)
            if val == target:
                return mid
            elif val < target:
                left = mid + 1
            else:
                right = mid - 1

        # Search right of the peak
        left, right = peak, mountain_arr.length() - 1
        while left <= right:
            mid = left + (right - left) // 2
            val = mountain_arr.get(mid)
            if val == target:
                return mid
            elif val < target:
                right = mid - 1
            else:
                left = mid + 1

        return -1

Key Takeaways:

  • We use binary search to find the peak and to search both the left and right parts of the mountain array.
  • This approach ensures that we do not exceed the constraint of 100 calls to MountainArray.get.
  • The time complexity is (O(\log n)), where (n) is the length of the mountain array.

Identifying Problem Isomorphism

“Find in Mountain Array” can be mapped to “Peak Index in a Mountain Array”.

In “Find in Mountain Array”, you are given a ‘mountain array’ (an array that increases and then decreases) and a target value, and you have to find the minimum index where the target value exists.

In “Peak Index in a Mountain Array”, you are given a ‘mountain array’ and you have to find the index of the peak of the mountain.

Both share the structure of the ‘mountain array’, and they involve finding an index in that array. The specific tasks are different, making this an approximate mapping. “Peak Index in a Mountain Array” is simpler, as it only asks for the peak of the array, rather than a specific target value.

10 Prerequisite LeetCode Problems

Find in Mountain Array involves binary search, array manipulation, and understanding of peak element concept. Here are 10 problems to prepare for this:

  1. “704. Binary Search” - Basic binary search algorithm.

  2. “33. Search in Rotated Sorted Array” - Understand how to modify binary search when dealing with a rotated array.

  3. “34. Find First and Last Position of Element in Sorted Array” - Binary search related problem.

  4. “153. Find Minimum in Rotated Sorted Array” - Another problem on a rotated array, but now we want to find the smallest element.

  5. “162. Find Peak Element” - Helps you understand how to find a peak in an array which is useful for the mountain array problem.

  6. “35. Search Insert Position” - Another binary search problem.

  7. “74. Search a 2D Matrix” - Binary search in a 2D matrix.

  8. “658. Find K Closest Elements” - Binary search to find closest elements.

  9. “852. Peak Index in a Mountain Array” - This is the most similar problem to 1095. It helps understand the concept of mountain array.

  10. “981. Time Based Key-Value Store” - A more complex binary search problem.

Problem Classification

The problem falls under the domain of “Array Manipulation” and “Binary Search.” It deals with finding a specific target value in a specialized type of array called a “Mountain Array.”

What

  1. Array: An array called mountainArr is given.
  2. Array Length: The array length is between 3 and 10^4.
  3. Array Values: The array values are between 0 and 10^9.
  4. Target: A target value that needs to be found in the array.
  5. Mountain Array Conditions: The array is a mountain array, meaning it first strictly increases and then strictly decreases.
  6. Access Limitation: You can only access the array using the MountainArray interface. Also, you can’t make more than 100 calls to MountainArray.get().
  7. Output: The minimum index where the target value is found, or -1 if it doesn’t exist in the array.

Further Classification

  • Search Problem: You are required to find a particular target in the array.
  • Optimization Problem: You need to do it in less than 100 calls to MountainArray.get(), meaning a brute-force solution won’t suffice.
  • Constraint Handling: The array cannot be accessed directly, and you have to use a given interface for it.

In summary, this is an “Optimized Search Problem in a Constrained Environment,” falling mainly in the categories of Array Manipulation and Binary Search.

Visual Model of the Problem

Visualizing this problem can help in better understanding its constraints and what’s being asked. Here are some ways to do it:

Graphical Representation

  1. Array as a Mountain: Draw an x-y graph. The x-axis represents the indices of the array, and the y-axis represents the values at those indices. Sketch the array as a mountain, with a peak at some index i.

    y
|
|                     /
|                    / \
|                   /   \
|                  /     \
|                 /       \
|                /         \
------------------------------
               x

  2. Highlight the Peak: Specifically point out the peak of the mountain. This is the highest point in the array.

    y
|
|                     /\
|                    /  \
|                   /    \
|                  /      \
|                 /        \
|                /          \
------------------------------
               x

  3. Target Values: Mark the possible positions of the target value, if it exists, on both the increasing and decreasing sides of the mountain.

    y
|
|                T  / \  T
|                    / \
|                   /   \
|                  /     \
|                 /       \
|                /         \
------------------------------
               x

    Here, “T” marks possible positions for the target value.

Conceptual Representation

  1. Three Phases: Divide the mountain array into three phases: increasing sequence, peak, and decreasing sequence. The target could be in either the increasing or the decreasing sequence, but not at the peak (unless the target value matches the peak).

  2. Search Spaces: Identify that you have two distinct search spaces to look for the target: the increasing sequence and the decreasing sequence.

  3. Constraints: Make a note or bubble about the constraints like “max 100 API calls” to keep them in mind.

By doing this, you can get a clearer picture of where you need to look for the target and how to optimize your search, given the constraints.

Problem Restatement

You’re given an array that forms a “mountain,” meaning the elements first strictly increase and then strictly decrease after reaching a peak. Your task is to find the smallest index at which a given target value exists in this mountain array. The catch is, you can’t access the array directly but can only use two functions: one that returns an element at a specific index, and another that tells you the length of the array. Also, you’re limited to a maximum of 100 calls to fetch an element from the array. If the target value doesn’t exist, you should return -1.

Abstract Representation of the Problem

In abstract terms, you’re dealing with a sequence of elements organized in a unimodal fashion — first strictly ascending until a peak, and then strictly descending. You have a target element that you’re searching for in this sequence. You are not allowed to access the sequence directly but can query the value of an element at a particular index. You are also restricted by the number of queries you can make to the sequence. The objective is to find the lowest index at which the target element occurs in this sequence, if it exists at all. If it doesn’t, return a sentinel value indicating “not found.”

Terminology

  1. Mountain Array: A specialized term for an array that first strictly ascends to a peak element and then strictly descends. This structure is crucial because it restricts where the target element can be found and influences the search strategy.

  2. Interface: In this problem, you interact with the Mountain Array through a set of predefined methods (MountainArray.get(k) and MountainArray.length()). Understanding this constraint is essential because you can’t directly access the array’s elements, which affects both the algorithm’s complexity and implementation.

  3. Binary Search: Though not explicitly mentioned, understanding binary search is key to solving this problem efficiently. Due to the Mountain Array’s structure, a modified form of binary search can be employed to find the target element while minimizing the number of queries made to the array.

  4. Time and Space Complexity: These are not terms within the problem but are key to evaluating the efficiency of your solution. Because you’re limited to a specific number of calls to MountainArray.get, you need to consider these complexities.

  5. Sentinel Value: In computing, this refers to a specific value that indicates a condition like ’not found.’ In this problem, -1 serves as the sentinel value when the target element is not present in the Mountain Array.

Each of these terms and concepts plays a role in defining the problem’s constraints or in shaping the solution strategy.

Problem Simplification and Explanation

At its core, this problem is a search problem. You’re looking for a specific number, called the “target,” in a list of numbers arranged in a special way. This list, called a “Mountain Array,” looks like a hill: numbers go up to a peak, and then they come down.

Key Concepts:

  1. Mountain Array: Think of this as a hill. You hike uphill until you reach the peak, and then you descend. The numbers in the array behave the same way; they increase to a highest point and then decrease.

  2. Target: This is the number you’re looking for on your hike.

  3. Limited Access: You can’t see the entire hill at once. You have to check one point at a time to see if it’s the spot you’re looking for.

  4. Interface: You’re walking blindfolded, and a friend tells you how high you are at any point. You can only know the height at a given point and how long the hike (array length) is. You can’t see ahead or behind.

  5. Efficiency: You have to find the target spot by asking your friend as few questions as possible. Imagine your friend will only respond to 100 questions about height before getting tired.

Metaphor:

Imagine you’re playing a game of “hot and cold” on a hill. You’re blindfolded, and your friend is guiding you. You’re trying to find a particular rock (the “target”). Your friend will only tell you the height of the ground where you’re standing and is willing to do this only a limited number of times. Your job is to figure out where that special rock is by moving wisely and asking your friend about the ground’s height as few times as possible. If you find it, great! If not, or if you ask too many questions, you lose.

So, you have to be clever about how you move and when you ask your friend for information. You’re also racing against the clock, metaphorically speaking, because you can only ask 100 questions before your friend gets tired.

Constraints

Several characteristics of the problem can be leveraged for an efficient solution:

  1. Array Length: The length of the mountain array is between 3 and 10^4. The upper limit suggests that linear scans should be avoided if possible, as they might result in too many calls to MountainArray.get().

  2. Mountain Structure: The array is not random; it has a specific “mountain” shape. The elements are sorted in increasing order until the peak and sorted in decreasing order afterward. This pattern allows us to use binary search to find the peak and the target, thereby reducing the number of calls needed.

  3. Limited Range of Target: The target value is between 0 and 10^9. This information doesn’t help much with the algorithm, but it does tell us that the numbers can be compared directly without worrying about overflow or underflow.

  4. Bounded Number of Calls: With a limit of 100 calls to MountainArray.get(), any algorithm that makes more than this number of calls is automatically disqualified. This constraint reinforces the need for a highly efficient algorithm, ideally one that works in logarithmic time.

  5. Unique Peak: The mountain array has only one peak, which means once we find it, we can split the problem into two separate binary searches: one for the increasing side and one for the decreasing side of the mountain.

  6. The Target is Integer: Since the target is an integer, we don’t have to worry about precision errors when comparing numbers.

By identifying these characteristics, we can design an algorithm that takes advantage of the mountain structure and limited call constraint to solve the problem efficiently. Specifically, binary search fits the bill, as it can find an element in a sorted array with O(log n) complexity, thereby keeping the number of calls to MountainArray.get() well below 100.

The key insights from analyzing the constraints are:

  1. Efficiency Requirement: Given that only 100 calls to MountainArray.get() are allowed, an efficient algorithm is essential. This rules out any brute-force or linear-time solutions.

  2. Exploit Mountain Structure: The mountain-like structure of the array naturally lends itself to binary search. The array’s shape allows you to apply binary search twice: once for finding the peak and once for finding the target. This can minimize the number of array accesses.

  3. Fixed Array Size: The array size is limited to 10^4 elements, which suggests that our algorithm should aim for logarithmic time complexity. A binary search would be ideal in this situation.

  4. Integer Comparisons: The constraint that elements and targets are integers simplifies the problem a bit by eliminating concerns about floating-point errors during comparisons.

By taking these insights into account, we can tailor our solution to meet the constraints while efficiently solving the problem. Specifically, using binary search will help meet both the efficiency requirement and the call limit.

Case Analysis

Let’s consider the following additional examples and test cases.

Example 1: Small Array

  • Input: array = [1, 2, 1], target = 2
  • Output: 1
  • Explanation: The array has the minimum allowed length (3). The target exists and is the peak.

Key Insights:

  • This example tests the lower boundary condition for array length.
  • The algorithm should handle small arrays efficiently and correctly.

Example 2: All Elements are Same

  • Input: array = [1, 1, 1, 1, 1], target = 1
  • Output: 0
  • Explanation: Every element is the same and equals the target. Return the smallest index.

Key Insights:

  • This example tests whether the algorithm handles arrays where all elements are equal.
  • It tests the algorithm’s ability to return the minimum index.

Example 3: Target Does Not Exist

  • Input: array = [1, 3, 5, 3, 1], target = 4
  • Output: -1
  • Explanation: The target is not in the array.

Key Insights:

  • This example helps confirm that the algorithm correctly identifies when a target does not exist and returns -1 as required.

Example 4: Target Exists in Decreasing Segment

  • Input: array = [1, 2, 3, 5, 4], target = 4
  • Output: 4
  • Explanation: The target exists but is in the decreasing segment of the array.

Key Insights:

  • This tests the algorithm’s ability to search both increasing and decreasing segments efficiently.

Example 5: Target at Edges

  • Input: array = [1, 2, 3], target = 1
  • Output: 0
  • Explanation: The target exists and is at the edge of the array.

Key Insights:

  • This tests if the algorithm considers edge elements.

Example 6: Peak at the Middle

  • Input: array = [1, 2, 3, 2, 1], target = 3
  • Output: 2
  • Explanation: The peak element is exactly at the middle.

Key Insights:

  • This tests if the algorithm identifies peaks that are exactly at the midpoint.

By analyzing these examples, we ensure that the algorithm can handle various edge and boundary conditions, thereby making it robust.

Identification of Applicable Theoretical Concepts

The array has a distinct peak, and elements before the peak are in ascending order while elements after the peak are in descending order. This sorted-but-inverted nature can be exploited by a modified binary search algorithm to find the target or the peak efficiently.

Divide and Conquer

The problem can be broken down into sub-problems of finding the peak and then searching in the ascending and descending segments. Divide and conquer can be a useful strategy here.

Complexity Analysis

Understanding the time and space complexity constraints can help tailor the algorithm for efficiency. Given the constraint that no more than 100 API calls to MountainArray.get are allowed, the algorithm must have a log-linear or better complexity.

Condition Checking

The problem involves various conditions to determine the boundaries of the peak and where the target might exist. Proper condition checks (greater than, less than, equals to) can reduce the number of required operations.

Caching/Memoization

Since API calls are expensive in terms of allowed operations, memoization could be used to store the result of previous calls for re-use, reducing the number of API calls.

By applying these mathematical and algorithmic concepts, the problem becomes more tractable and can be solved in an efficient manner.

Problem Breakdown and Solution Methodology

Steps

  1. Find the Peak
  • The first task is to identify the peak of the mountain. This is the highest point where the sequence switches from increasing to decreasing.
  1. Divide the Mountain
  • Once you have the peak, you can divide the mountain into two sections: the ascending part before the peak and the descending part after the peak.
  1. Search for the Target
  • Use binary search twice: once in the ascending part and then in the descending part to find the target value. Return the first index where you find the target.

Metaphor

Imagine you’re looking for a specific tree on a mountain. First, you go to the highest point (the peak) to get a full view. Then you only explore the sides where that type of tree is more likely to grow (ascending or descending slopes). You don’t traverse the whole mountain; instead, you take shortcuts, eliminating areas where the tree can’t be found.

How Parameters Affect the Solution

  • Array Length: The longer the mountain array, the more time it will take to find the peak and the target. But since we’re using binary search, the time complexity is logarithmic, making it efficient.

  • Target Value: If the target value is larger than the peak or smaller than the minimum value in the array, you can immediately return -1 without searching.

  • API Call Limit: Since the number of API calls is limited to 100, a binary search approach fits well within this constraint. However, if the constraint were stricter, we might have to think of other optimization techniques.

Example Case

Let’s take the array [1,2,3,4,5,3,1] with a target value of 3.

  1. Find the Peak:
  • Use binary search to find the peak, which is 5 at index 4.
  1. Divide the Mountain:
  • The ascending part is [1, 2, 3, 4, 5] (indices 0 to 4).
  • The descending part is [5, 3, 1] (indices 4 to 6).
  1. Search for the Target:
  • Use binary search on [1, 2, 3, 4, 5]. You find 3 at index 2.
  • You could also search in the descending part, but since we found it already, we stop and return 2.

By breaking down the problem in this way and using efficient search strategies, we arrive at a solution that not only finds the target index but also does so efficiently.

Inference of Problem-Solving Approach from the Problem Statement

How did you infer from the problem statement that this problem can be solved using ?

Stepwise Refinement

High Level Steps

  1. High-Level Solution: Implement a binary search algorithm to find the peak, and then use it twice to find the target in the ascending and descending parts of the array.

    1.1 Find the Peak

    • Start with the entire array’s indices as the search space.
    • Calculate the middle index.
    • Compare the values at the middle, middle+1, and middle-1 indices.
    • Narrow down the search space based on the comparison.
    • Repeat until you find the peak.

    1.2 Divide the Array into Two Sections

    • The array indices from the start to the peak index form the ascending section.
    • The array indices from the peak index to the end form the descending section.

    1.3 Binary Search for Target in Ascending Section

    • Start with the ascending section’s indices as the search space.
    • Calculate the middle index.
    • Compare the value at the middle index to the target.
    • Narrow down the search space based on the comparison.
    • Repeat until you either find the target or exhaust the search space.

    1.4 Binary Search for Target in Descending Section

    • Repeat the same process as in 1.3, but adjust the comparison for a descending order.
    • If the target is found, compare its index with any index found in the ascending section and return the minimum.

Granular, Actionable Steps

  • Initialize variables for start, end, and mid-indices for each binary search process.
  • Implement a function for binary search that takes in parameters like the MountainArray object and the target value.
  • Use loops or recursive calls to execute the binary search within the function.
  • Implement a condition to break out of the loop if the target is found or if the search space is empty.
  • Return the index where the target is found or -1 if not found.

Independent Parts of the Problem

  1. Finding the peak is independent of searching for the target.
  2. Searching for the target in the ascending section is independent of searching in the descending section.

Repeatable Patterns

  • The core repeatable pattern here is the binary search algorithm.
    • The same binary search mechanism is being adapted for finding the peak, searching in the ascending section, and searching in the descending section. This forms a modular, repeatable pattern that’s central to the solution.

Solution Approach and Analysis

Step 1: Identify the Peak

  • Think of this as looking for the highest mountain peak in a range. You start at one end and keep moving to higher ground.
  • Method: Implement a binary search. Start, Mid, and End pointers initially point to the beginning, middle, and end of the array.
    1. Calculate the middle index.
    2. If MountainArray.get(mid) < MountainArray.get(mid + 1), the peak is on the right side. Move Start to Mid + 1.
    3. Else, the peak is on the left side, including Mid. Move End to Mid.

Step 2: Search for the Target Before the Peak

  • Imagine this as looking for a specific altitude mark while climbing to the peak. You would only check upwards because the altitudes are increasing.
  • Method: Implement a binary search again, but this time from Start to Peak.
    1. If MountainArray.get(mid) == target, return Mid.
    2. If MountainArray.get(mid) < target, move Start to Mid + 1.
    3. Otherwise, move End to Mid - 1.

Step 3: Search for the Target After the Peak

  • This is like looking for a specific altitude mark while descending. You would only check downwards because the altitudes are decreasing.
  • Method: Implement a binary search from Peak to End.
    1. If MountainArray.get(mid) == target, return Mid.
    2. If MountainArray.get(mid) > target, move Start to Mid + 1.
    3. Otherwise, move End to Mid - 1.

Step 4: Return the Result

  • Return the minimum index if the target is found in both ascending and descending sections. If not found, return -1.

How Parameters Affect the Solution

  • The size of the array affects the number of iterations in the binary search.
  • The location of the target (before or after the peak) will decide which part of the Step 2 or Step 3 is actually needed.

Example Case

  • array = [1,2,3,4,5,3,1], target = 3

    • Step 1: Peak is 5, at index 4.
    • Step 2: Start search from index 0 to 4, find 3 at index 2.
    • Step 3: Start search from index 4 to 6, find 3 at index 5.
    • Step 4: Return the minimum index, which is 2.

Your entire code will consist of these 4 steps, allowing you to find the target in a mountain array efficiently.

Thought Process

Step 1: Problem Statement Insights

  • The mountain array has three main parts: ascending, peak, and descending.
  • The search operation is constrained to 100 calls, suggesting binary search for efficiency.
  • The goal is to find the minimum index where the target resides, either before or after the peak.

Step 2: Finding the Peak

  • We cannot traverse the whole array due to constraint, so binary search is needed to find the peak.

Step 3: Divide and Conquer

  • Once the peak is found, the array naturally divides into two sorted arrays (ascending and descending).

Step 4: Final Plan

  • Find the peak with binary search.
  • Use binary search again on the ascending part to find the target.
  • Use binary search on the descending part to find the target.
  • Return the minimum index if found in both parts, otherwise -1 if not found.

Python Code

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# Assume MountainArray is a given interface
# def MountainArray.get(index: int) -> int:
# def MountainArray.length() -> int:

class Solution:
    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
        # Find the peak of the mountain array
        start, end = 0, mountain_arr.length() - 1
        while start < end:
            mid = start + (end - start) // 2
            if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
                start = mid + 1
            else:
                end = mid
        peak = start
        
        # Binary search to find the target before the peak
        start, end = 0, peak
        while start <= end:
            mid = start + (end - start) // 2
            mid_val = mountain_arr.get(mid)
            if mid_val == target:
                return mid
            elif mid_val < target:
                start = mid + 1
            else:
                end = mid - 1
                
        # Binary search to find the target after the peak
        start, end = peak, mountain_arr.length() - 1
        while start <= end:
            mid = start + (end - start) // 2
            mid_val = mountain_arr.get(mid)
            if mid_val == target:
                return mid
            elif mid_val > target:
                start = mid + 1
            else:
                end = mid - 1
        
        return -1  # Target not found

The cues in the problem statement led us to consider binary search for both finding the peak and the target. The constraints further emphasized the need for an efficient approach. By breaking down the problem into its essential components and constraints, we were able to form a solution that addresses all these aspects effectively.

From Brute Force to Optimal Solution

Brute Force Solution

Description:

  1. Start from the first element and iterate through the array to find the peak.
  2. Continue iterating to find the target element.
  3. Return the first occurrence of the target.

Python Code for Brute Force:

shitgpt generates buggy code

Inefficiencies:

  1. Time Complexity: The brute-force approach has a time complexity of (O(n)) where (n) is the length of the array.
  2. Space Complexity: (O(1))
  3. API Calls: You could end up making (n) calls to mountain_arr.get(), which violates the constraint of no more than 100 calls.

Optimized Solution

Step 1: Find the Peak Efficiently

Instead of linearly searching for the peak, use binary search to find it. This reduces the number of calls to (O(\log n)).

Step 2: Divide and Conquer

After finding the peak, split the problem into two separate binary searches, one for the ascending part and one for the descending part. This further reduces the number of calls needed to find the target.

Step 3: Exit Early

If the target is found in the ascending part, we can return immediately without searching the descending part.

Optimized Python Code:

The code for the optimized solution is the same as the one provided in the previous answer.

Efficiency:

  1. Time Complexity: The time complexity becomes (O(\log n)) due to binary search.
  2. Space Complexity: Still (O(1))
  3. API Calls: Now well within the limit of 100 calls, since it’s (O(\log n)).

By optimizing, we align the solution with the problem constraints, improve the efficiency from (O(n)) to (O(\log n)), and stay within the API call limits.

Coding Constructs

  1. What are the high-level problem-solving strategies or techniques being used by this code?

  2. If you had to explain the purpose of this code to a non-programmer, what would you say?

  3. Can you identify the logical elements or constructs used in this code, independent of any programming language?

  4. Could you describe the algorithmic approach used by this code in plain English?

  5. What are the key steps or operations this code is performing on the input data, and why?

  6. Can you identify the algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax?

Language Agnostic Coding Drills

Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.

  1. Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.

  2. Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.

  3. Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.

Targeted Drills in Python

Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.

  1. Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.

  2. In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.

  3. Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.

Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.

Q&A

Similar Problems

Given the problem , identify and list down 10 similar problems on LeetCode. These should cover similar concepts or require similar problem-solving approaches as the provided problem. Please also give a brief reason as to why you think each problem is similar to the given problem.