Reverse String II

10 Prerequisite LeetCode Problems

“541. Reverse String II” asks to reverse every k characters for a string. This problem involves string manipulation and understanding of substrings.

A similar problem is “344. Reverse String”. This problem requires reversing a string, which is a simpler version of the task in “Reverse String II”. While “Reverse String” involves reversing the whole string, “Reverse String II” takes it a step further by reversing every k characters of the string.

Another isomorphic problem is “557. Reverse Words in a String III” on LeetCode. In this problem, each word in the input string is reversed. This is a bit more complex than “Reverse String” because you have to identify each word and reverse it, but it’s similar to “Reverse String II” as both problems involve reversing sections of the input string.

“344. Reverse String” is the simplest, followed by “541. Reverse String II”, and then “557. Reverse Words in a String III” is the most complex.

Though the problems are not identical, they share the core concept of string reversal, making them approximately isomorphic. Understanding how to solve these problems can deepen your knowledge of string manipulation in different contexts.

This problem involves string manipulation. The Python string has a built-in function called slice that can help us perform the necessary operations.

The idea is to break the string into chunks of size 2k, reverse the first k characters and concatenate them to the result. We also need to handle the edge case where the remaining string has less than k characters.

Here is the Python code for the same:

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class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        res = ""
        for i in range(0, len(s), 2*k):
            res += s[i:i+k][::-1] + s[i+k:i+2*k]
        return res

In the code above, s[i:i+k][::-1] selects the first k characters of the chunk and reverses them, and s[i+k:i+2*k] selects the second k characters of the chunk and leaves them in their original order. The results are then concatenated to form the final result.

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class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        arr = list(s)
        n = len(arr)
        i = 0
        while i < n:
            j = min(i + k - 1, n - 1)
            self.swap(arr, i, j)
            i += 2 * k
        return ''.join(arr)

    def swap(self, arr: list, l: int, r: int) -> None:
        while l < r:
            arr[l], arr[r] = arr[r], arr[l]
            l += 1
            r -= 1

Problem Classification

This problem falls under the domain of “String Manipulation”. It involves the task of reordering characters within a string according to specified rules.

What components:

  1. A string s - This is the input string that needs to be manipulated.
  2. An integer k - This value dictates how many characters in the string should be reversed at a time.
  3. Every 2k characters - This indicates the frequency with which the reversing operation is performed on the string. Every 2k characters, we reverse the first k.
  4. Conditions for fewer than k characters, fewer than 2k but at least k characters - These conditions outline how to handle cases where the remaining characters in the string are less than k or 2k but at least k.

This problem can be classified as a “Pattern Recognition” problem, as it involves identifying the repeating pattern of reversing every k characters in each block of 2k characters. This classification is suitable because solving the problem involves recognizing this pattern and applying it to the given string.

The key to understanding this problem is recognizing the pattern of operations that need to be applied to the string (i.e., reverse the first k characters for every 2k characters). Additionally, the problem has defined handling methods when the remaining string has fewer than k characters or between k and 2k characters. These need to be taken into account when developing the solution.

Language Agnostic Coding Drills

  1. Dissect the code and identify each distinct concept it contains:

    • Variable declaration and initialization: The variables arr, n, and i are declared and initialized.
    • Array data structures: The string s is converted into an array of characters (arr) for easier manipulation.
    • Function definition: Two functions reverseStr and swap are defined to solve the problem.
    • Control structures: A while loop is used to iterate through the array in chunks of 2k characters.
    • Conditional statements: The min function is used to ensure the end index of the array slice doesn’t exceed the length of the array.
    • Array manipulation: The swap function is used to reverse the order of characters in a given slice of the array.
    • Index manipulation: The code updates the index i after each iteration to move to the next 2k characters in the array.

  2. List out coding concepts or drills in order of increasing difficulty:

    • Variable declaration and initialization: Declaring and initializing variables. This is the simplest concept and is a building block for more complex operations.
    • Array data structures: Understanding how to work with arrays.
    • Function definition: Defining and using functions.
    • Control structures: Using while loops to control program flow.
    • Conditional statements: Using min to prevent index out of bounds errors.
    • Array manipulation: Reversing a subsection within an array.
    • Index manipulation: Incrementing an index to move to different parts of the array. This requires understanding of array’s slicing and indexing.

  3. Describe the problem-solving approach that leads from the problem statement to the final solution:

    The problem statement requires us to reverse every k characters within each 2k character slice of the input string. The solution starts by converting the string into an array for easier manipulation. Then it defines a swap function to reverse a given slice of the array.

    In the main reverseStr function, it initiates a while loop that continues as long as the index i is within the bounds of the array. In each iteration, it calculates the end index j for the current slice, taking care to ensure j doesn’t exceed the bounds of the array. Then it calls swap to reverse the characters from i to j in the array. After reversing, it increments i by 2k to move to the next 2k character slice.

    The loop continues until all 2k character slices have been processed. Finally, it joins the array back into a string and returns the result.

    Each coding drill contributes to this overall solution. The variable declaration and initialization and array data structures drills allow us to manipulate the string as an array. The function definition drill lets us encapsulate the reversing operation in the swap function. The control structures and index manipulation drills enable us to iterate over the array in 2k character slices. The conditional statements and array manipulation drills help us to safely reverse each k character slice within the array.

Targeted Drills in Python

  1. Writing separate pieces of Python code encapsulating each identified concept

    • Variable declaration and initialization

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      a = 5
b = "Hello"

    • Array data structures

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      arr = ['a', 'b', 'c', 'd', 'e']

    • Function definition

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      def greet(name):
    print(f"Hello, {name}!")

    • Control structures (while loop)

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      i = 0
while i < 5:
    print(i)
    i += 1

    • Conditional statements

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      if a < b:
    print("a is less than b")
else:
    print("a is not less than b")

    • Array manipulation (reversing a sublist)

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      arr = [1, 2, 3, 4, 5]
arr[1:3] = arr[1:3][::-1]

    • Index manipulation

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      arr = [1, 2, 3, 4, 5]
for i in range(0, len(arr), 2):
    print(arr[i])

  2. Identifying and writing coding drills for problem-specific concepts

    The specific problem asks for the reversal of every k characters for every 2k characters counting from the start of the string. Thus, two problem-specific drills will be essential here:

    • Understanding string to list and list to string conversions

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      s = "hello"
s_list = list(s)  # Convert string to list
s_str = ''.join(s_list)  # Convert list back to string

    • Swapping elements in an array using indices

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      def swap(arr, l, r):
    while l < r:
        arr[l], arr[r] = arr[r], arr[l]
        l += 1
        r -= 1

  3. Describing how these pieces can be integrated together

    To solve the initial problem, these drills can be assembled in the following order:

    • Begin by converting the string into a list (drill on understanding string to list conversions).
    • Initialize your starting index to 0 (drill on variable declaration and initialization).
    • Use a while loop (drill on control structures) to go through the list. Inside this loop:
      • Calculate the end index of the subsection to be reversed using the min function (drill on conditional statements).
      • Reverse the subsection of the list from the start index to the calculated end index (drill on array manipulation and the problem-specific swap function).
      • Increment the starting index by 2k (drill on index manipulation).
    • Convert the list back to a string (drill on understanding list to string conversions) and return it.