How do you solve Zigzag Conversion on LeetCode?

Zigzag Conversion (LeetCode #6) can be solved using Brute Force or Optimal Solution (Hash Map). The optimal approach is Optimal Solution (Hash Map) with O(n) time complexity and O(n) space complexity. Key insight: The zigzag pattern can be visualized as a wave, where characters are placed in a downward slope and then an upward slope. Recognizing this pattern helps in determining the row placement of each character.

What algorithm or data structure is used to solve Zigzag Conversion?

Zigzag Conversion uses the following concepts: String. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Zigzag Conversion?

When you first see this problem, visualize the zigzag pattern on paper to understand how characters are placed. Communicate your approach clearly, explaining how you will manage the direction of traversal and how you will store the characters. Mention edge cases such as when numRows is 1 or greater than the length of the string, as these can simplify the problem.

What are common mistakes when solving Zigzag Conversion?

Forgetting to handle the case when numRows is 1, which should return the original string immediately. Not correctly managing the direction of traversal, leading to incorrect row placements.

Zigzag Conversion — LeetCode #6 (Medium)

Tags: String

Related patterns: Hash Map, Array, Two Pointers

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves simulating the zigzag pattern by iterating through the string and placing characters in the correct rows. This method is straightforward but inefficient due to the multiple passes required to construct the final string.

The time complexity is O(n²) because for each character, we may need to traverse the rows multiple times to construct the final string. The space complexity is O(1) since we are using a fixed amount of space regardless of input size.

Step 1: Create an array of strings, one for each row.
Step 2: Iterate through the characters of the input string, placing each character in the appropriate row based on the current direction (down or up).
Step 3: After filling all rows, concatenate the strings from each row to form the final output.

Let's dry-run the first example with s = 'PAYPALISHIRING' and numRows = 3: 1. Initialize rows = ['', '', ''], current_row = 0, going_down = False. 2. Process 'P': rows = ['P', '', ''], current_row = 0, going_down = True. 3. Process 'A': rows = ['P', 'A', ''], current_row = 1, going_down = True. 4. Process 'Y': rows = ['P', 'A', 'Y'], current_row = 2, going_down = False. 5. Process 'P': rows = ['P', 'A', 'Y'], current_row = 1, going_down = False. 6. Process 'A': rows = ['P', 'A', 'Y'], current_row = 0, going_down = True. 7. Process 'L': rows = ['P', 'A', 'Y'], current_row = 1, going_down = True. 8. Process 'I': rows = ['P', 'A', 'Y'], current_row = 2, going_down = False. 9. Process 'S': rows = ['P', 'A', 'Y'], current_row = 1, going_down = False. 10. Process 'H': rows = ['P', 'A', 'Y'], current_row = 0, going_down = True. 11. Process 'I': rows = ['P', 'A', 'Y'], current_row = 1, going_down = True. 12. Process 'R': rows = ['P', 'A', 'Y'], current_row = 2, going_down = False. 13. Process 'I': rows = ['P', 'A', 'Y'], current_row = 1, going_down = False. 14. Process 'N': rows = ['P', 'A', 'Y'], current_row = 0, going_down = True. 15. Process 'G': rows = ['P', 'A', 'Y'], current_row = 1, going_down = True. Final rows: ['PAHN', 'APLSIIG', 'YIR'] => Result: 'PAHNAPLSIIGYIR'.

Optimal Solution (Hash Map) approach

Time complexity: O(n). Space complexity: O(n).

The optimal approach uses a hash map to store characters in their respective rows, allowing for efficient access and concatenation. This method significantly reduces the number of passes needed to construct the final string.

The time complexity is O(n) because we only need to traverse the string once. The space complexity is O(n) due to the storage of characters in the rows.

Step 1: Create a list of strings (or a hash map) to hold characters for each row.
Step 2: Iterate through the string, placing each character in the appropriate row based on the current direction.
Step 3: After filling all rows, concatenate the strings to form the final output.

Key Insights

The zigzag pattern can be visualized as a wave, where characters are placed in a downward slope and then an upward slope. Recognizing this pattern helps in determining the row placement of each character.
Understanding the direction of traversal (down and up) is crucial. This can be managed with a simple boolean flag that toggles when reaching the top or bottom row.

Common Mistakes

Forgetting to handle the case when numRows is 1, which should return the original string immediately.
Not correctly managing the direction of traversal, leading to incorrect row placements.

Interview Tips

When you first see this problem, visualize the zigzag pattern on paper to understand how characters are placed.
Communicate your approach clearly, explaining how you will manage the direction of traversal and how you will store the characters.
Mention edge cases such as when numRows is 1 or greater than the length of the string, as these can simplify the problem.

Zigzag Conversion

Medium

String↗Hash Map↗Array↗Two Pointers↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Hash Map)★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

⚙️

Algorithm

3 steps

1Step 1: Create a list of strings (or a hash map) to hold characters for each row.
2Step 2: Iterate through the string, placing each character in the appropriate row based on the current direction.
3Step 3: After filling all rows, concatenate the strings to form the final output.

solution.py20 lines

1# Full working Python code with comments
2
3def convert(s: str, numRows: int) -> str:
4    if numRows == 1 or numRows >= len(s):
5        return s
6
7    rows = [''] * numRows
8    current_row = 0
9    going_down = False
10
11    for char in s:
12        rows[current_row] += char
13        if current_row == 0:
14            going_down = True
15        elif current_row == numRows - 1:
16            going_down = False
17        current_row += 1 if going_down else -1
18
19    return ''.join(rows)
20

ℹ

Complexity note: The time complexity is O(n) because we only need to traverse the string once. The space complexity is O(n) due to the storage of characters in the rows.

1The zigzag pattern can be visualized as a wave, where characters are placed in a downward slope and then an upward slope. Recognizing this pattern helps in determining the row placement of each character.
2Understanding the direction of traversal (down and up) is crucial. This can be managed with a simple boolean flag that toggles when reaching the top or bottom row.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.