Reverse Words in a String II | CodePath Cliffnotes

Problem Highlights

🔗 Leetcode Link: Reverse Words in a String II
💡 Problem Difficulty: Medium
⏰ Time to complete: 30 mins
🛠️ Topics: Array, Two Pointer
🗒️ Similar Questions: Reverse Words in a String, Rotate Array

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

Established a set (2-3) of test cases to verify their own solution later.

Established a set (1-2) of edge cases to verify their solution handles complexities.

Have fully understood the problem and have no clarifying questions.

Have you verified any Time/Space Constraints for this problem?

Does the input have leading or trailing spaces?
- No
Does input have more than a single space between a word?
- All the words in the input are guaranteed to be separated by a single space.
Can the input be empty?
- The size range of the input is from 1 to 10^5. Therefore the input can never be empty

HAPPY CASE
Input: s = ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Input: ["h","e","l","l","o"," ","w","o","r","l","d"]
Output: ["w","o","r","l","d"," ","h","e","l","l","o"]

EDGE CASE (Multiple Spaces)
Input: s = ["a"]
Output: ["a"]

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Array/Strings, common solution patterns include:

Sort
- Does sorting help us achieve what we need in order to solve the problem? We can not do this since we have to retain the order in this problem.
Two pointer solutions (left and right pointer variables)
- Two pointer may help us. This solution can help us to maintain the order can solve this problem in place.
Storing the elements of the array in a HashMap or a Set
- A hashset will be not helpful here. We have to solve this problem in place, so this method will not help us.
Traversing the array with a sliding window
- Will viewing pieces of the input at a time help us? This can not help solve this problem since we can not do anything even if we have a sliding window.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Reverse the Whole List and Then Reverse Each Word in the List

1. Reverse the whole input List in place
2. Go through the input List
3. Determine where the word end - find each word in the input List
4. Reverse Each Word in the input List in place

⚠️ Common Mistakes

Remember to use two pointer to achieve O(1) space

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def reverseWords(self, s: List[str]) -> None:
        # 1. Reverse the whole input List in place
        self.reverse(s, 0, len(s) - 1)
        
        # 4. Reverse Each Word in the input List in place
        self.reverse_each_word(s)

    def reverse(self, l: list, left: int, right: int) -> None:
        while left < right:
            l[left], l[right] = l[right], l[left]
            left, right = left + 1, right - 1
            
    def reverse_each_word(self, l: list) -> None:
        n = len(l)
        start = end = 0

        # 2. Go through the input List
        while start < n:
            # 3. Determine where the word end - find each word in the input List
            while end < n and l[end] != ' ':
                end += 1
            # reverse the word
            self.reverse(l, start, end - 1)
            # move to the next word
            start = end + 1
            end += 1

class Solution {
    public void reverseWords(char[] s) {
        // 1. Reverse the whole input List in place
        reverse(s, 0, s.length - 1);

        // 4. Reverse Each Word in the input List in place
        reverseEachWord(s);
    }

    public void reverse(char[] s, int left, int right) {
        while (left < right) {
            char tmp = s[left];
            s[left++] = s[right];
            s[right--] = tmp;
        }
    }

    public void reverseEachWord(char[] s) {
        int n = s.length;
        int start = 0, end = 0;

        // 2. Go through the input List
        while (start < n) {
            // 3. Determine where the word end - find each word in the input List
            while (end < n && s[end] != ' ') ++end;
            // reverse the word
            reverse(s, start, end - 1);
            // move to the next word
            start = end + 1;
            ++end;
        }
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

Trace through your code with an input to check for the expected output
Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume N represents the number of items in the array.

Time Complexity: O(n), we need to visit every item in the array to reverse the list and reverse each word in the list one more time.
Space Complexity: O(1), we only need several pointers.

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