Reverse Words in a String III

Identifying Problem Isomorphism

“Reverse Words in a String III” can be mapped to the following problems:

“Reverse String” (#344) on LeetCode
“Reverse String II” (#541) on LeetCode
“Reverse Words in a String” (#151) on LeetCode

Reasoning:

All these involve string manipulation and particularly reversing elements within a string. The complexity arises from the elements that need to be reversed.

“Reverse String” is the simplest problem. The task is to reverse the whole string, a basic operation that forms the foundation for the other problems.
“Reverse String II” increases the complexity slightly by introducing the concept of reversing every 2k characters for a given k. The process of segmenting the string and reversing each segment adds a bit more complexity.
“Reverse Words in a String” is more complex as it requires the reversal of each word’s order within the string but not the characters within the words themselves. This requires the identification of the start and end of each word, adding a layer of complexity.

Each of these builds on the skill of reversing a sequence, with increased complexity from determining which parts of the sequence to reverse. None of these problems are exact isomorphs of “Reverse Words in a String III”, but they all tackle similar concepts of string manipulation and reversal.

Solution

We need to reverse the order of characters in each word within a sentence while preserving the whitespace and initial word order. We can approach this problem by splitting the sentence into words, reversing each word, and then joining them back together with spaces.

Here is the Python solution:

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class Solution:
    def reverseWords(self, s: str) -> str:
        # Split the sentence into words
        words = s.split()

        # Reverse each word and join them back together with spaces
        reversed_words = [word[::-1] for word in words]

        # Join the reversed words with spaces and return the result
        return ' '.join(reversed_words)

In this solution, we use Python’s built-in split function to break the string s into words. The split function splits a string into a list where each word is a list item.

Then, we use a list comprehension to create a new list where each word is reversed. We reverse each word using the Python slicing syntax word[::-1].

Finally, we use the join function to combine the reversed words back into a string with spaces in between each word.

The time complexity of this solution is O(n), where n is the number of characters in the string, as we perform a constant amount of work for each character. The space complexity is also O(n), as we create a list of words and a list of reversed words.

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class Solution {
    public String reverseWords(String s) {
        int lastSpaceIndex = -1;
        char[] chArray = s.toCharArray();
        int len = s.length();
        for (int strIndex = 0; strIndex <= len; strIndex++) {
            if (strIndex == len || chArray[strIndex] == ' ') {
                int startIndex = lastSpaceIndex + 1;
                int endIndex = strIndex - 1;
                while (startIndex < endIndex) {
                    char temp = chArray[startIndex];
                    chArray[startIndex] = chArray[endIndex];
                    chArray[endIndex] = temp;
                    startIndex++;
                    endIndex--;
                }
                lastSpaceIndex = strIndex;
            }
        }
        return new String(chArray);
    }

}

The Java code reverses each word in a string in-place, preserving the initial order of the words and the whitespace between them. Here’s how you can do the same thing in Python:

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class Solution:
    def reverseWords(self, s: str) -> str:
        s = list(s)
        lastSpaceIndex = -1
        len_s = len(s)
        for strIndex in range(len_s + 1):
            if strIndex == len_s or s[strIndex] == ' ':
                startIndex = lastSpaceIndex + 1
                endIndex = strIndex - 1
                while startIndex < endIndex:
                    s[startIndex], s[endIndex] = s[endIndex], s[startIndex]
                    startIndex += 1
                    endIndex -= 1
                lastSpaceIndex = strIndex
        return ''.join(s)

We convert the string s to a list of characters at the beginning, because strings in Python are immutable.
We use Python’s multiple assignment feature to swap the characters in place, without needing a temporary variable.
We use the join method to convert the list of characters back into a string at the end.

The time and space complexity of this Python solution is the same as the original Java code, namely O(n) time and O(n) space, where n is the number of characters in the string.

This solution uses several basic programming concepts:

Character Arrays (Python Lists): Because strings are immutable in Python, the solution first converts the string into a list of characters. This way, characters can be swapped in-place.
Loops: The solution uses a for loop to iterate through each character in the string (now a list).
Conditionals: Within the loop, there is an if condition to check if the current character is a space or the end of the list. This is how the algorithm identifies individual words to reverse.
Two Pointers: The solution uses a two-pointer approach to reverse the characters within a word. One pointer starts at the beginning of the word, and the other starts at the end. The characters at these two positions are swapped, and the pointers move towards each other. The swapping continues until the two pointers meet or cross each other.
Index Tracking: The solution keeps track of the position of the last space character it encountered (lastSpaceIndex). This is used to identify the start and end of each word.
String Manipulation: The solution involves various string manipulations including accessing characters by index, swapping characters, and finally joining the list of characters back into a string.

These building blocks together form the basis of many string manipulation algorithms and are not just limited to reversing words in a sentence. They are essential tools in a programmer’s toolkit.

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def reverseWords_manual(s):
    res = ''
    l, r = 0, 0

    while r < len(s):
        if s[r] != ' ':
            r += 1
        elif s[r] == ' ':
            res += s[l:r + 1][::-1]
            r += 1
            l = r

    res += ' '
    res += s[l:r + 2][::-1]
    return res[1:]

Problem Classification

This problem falls into the domain of “String Manipulation”. String manipulation is a broad category in programming languages that involves performing operations on strings, such as reversing words or characters, changing case, concatenating, splitting, and others.

What:

A string input ’s’ which represents a sentence.
A requirement to reverse the order of characters in each word.
A constraint to preserve the whitespace and initial word order within the sentence.

This can be classified as a “Transformation” problem, as it involves transforming the input string (sentence) into a new string where each word is reversed but the initial order of words is preserved.

In transformation problems, the aim is to transform the input data into a new form following certain rules. In this case, the rule is to reverse each word individually while maintaining the initial order of words and whitespace. This typically involves the use of string manipulation techniques and can often be solved with in-built language methods for string manipulation and iteration constructs to traverse through each word. This makes it a good practice problem for string manipulation and iteration techniques.

Language Agnostic Coding Drills

Identification of distinct coding concepts: The given code implements the following coding concepts:
a. Variable declaration and initialization: The code begins by initializing variables ‘res’, ’l’, and ‘r’ to hold the result, the starting index, and the ending index of each word in the string respectively.
b. Conditional statements: The code uses ‘if’ and ’elif’ conditions to check if the current character is a space or not.
c. While loop: The code uses a ‘while’ loop to iterate over the string.
d. String indexing: The code uses string indexing to access individual characters and slices of the string.
e. String concatenation: The code appends reversed words and spaces to the ‘res’ string.
f. String reversal: The code uses Python’s slice syntax to reverse the substring from index ’l’ to ‘r’ for each word.
Order of increasing difficulty and concept description:
a. Variable declaration and initialization: This is a fundamental concept in any programming language. The challenge level is low.
b. Conditional statements: This is a fundamental concept in programming and is used to perform different computations or actions depending on whether a condition evaluates to true or false. The challenge level is low.
c. String indexing: This involves accessing individual characters in a string using their index. The difficulty level is medium because understanding zero-based indexing can be initially challenging.
d. While loop: This is a fundamental concept in programming. The challenge level is medium, as the logic inside the loop can be complex.
e. String concatenation: This involves joining two or more strings into one. The difficulty level is low.
f. String reversal: This involves reversing the order of characters in a string. In Python, this is done using slice syntax which can be challenging for beginners to understand. Therefore, the difficulty level is high.
Problem-solving approach and strategy:
The problem-solving approach follows these steps:
a. Initialize a result string and two pointers, ’l’ and ‘r’, both pointing to the start of the string.
b. Use a while loop to iterate over the string from left to right. Within the loop, check if the current character at index ‘r’ is a space or not.
c. If it’s not a space, increment ‘r’ to move to the next character.
d. If it’s a space, it means that we’ve found a word. We slice this word from the original string, reverse it, and append it to the result string. After this, move both ’l’ and ‘r’ to the character after the space.
e. After the while loop ends, there will still be the last word left to reverse. We handle this by appending a space and the reversed last word to the result string.
f. Finally, we return the result string from index 1 to the end, because there will be an extra space at the beginning of the result string.

In the final solution, the identified concepts (variables, conditional statements, while loop, string indexing, string concatenation, string reversal) are used together to solve the problem of reversing each word in a sentence while preserving the original order of words and whitespace. Each identified concept contributes a necessary piece to the overall solution.

Targeted Drills in Python

Python coding drills for each identified concept:

a. Variable declaration and initialization:

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# Declare and initialize a variable
num = 10
print(num)  # Output: 10

b. Conditional statements:

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# Declare a variable and use it in a condition
num = 10
if num > 5:
    print("Greater than 5")  # Output: Greater than 5
else:
    print("Not greater than 5")

c. While loop:

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# Use a while loop to print numbers from 1 to 5
i = 1
while i <= 5:
    print(i)  # Output: 1, 2, 3, 4, 5 on separate lines
    i += 1

d. String indexing:

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# Access characters in a string by their index
s = "Hello"
print(s[0])  # Output: H

e. String concatenation:

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# Concatenate two strings
s1 = "Hello"
s2 = " World"
s = s1 + s2
print(s)  # Output: Hello World

f. String reversal:

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# Reverse a string using slicing
s = "Hello"
reversed_s = s[::-1]
print(reversed_s)  # Output: olleH

Problem-specific coding drills:
The problem at hand doesn’t require any specific coding concepts beyond the ones already mentioned.
Integrating all drills together:
The final problem can be solved by combining all the individual coding drills in a particular order. Here’s how you would integrate them:
a. Start by declaring and initializing your variables. In this case, these would be ‘res’, ’l’, and ‘r’.
b. Use a while loop to traverse the string, where ‘r’ serves as the running index.
c. Within the while loop, use a conditional statement to check if the current character is a space. If it’s not, increment ‘r’. If it is, you’ve found a word. Use string indexing to slice the word from the string and string reversal to reverse it. Then, use string concatenation to add the reversed word to ‘res’.
d. Once the while loop finishes, there’s still the last word left to handle. Similar to the steps inside the loop, slice, reverse, and concatenate this final word to ‘res’.
e. Finally, return ‘res’ starting from the second character to remove the leading space.

This way, all the individual drills contribute to the final solution.