LeetCode 6: Zigzag Conversion

Problem Description

LeetCode Problem 6

The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: “PAHNAPLSIIGYIR”

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

Input: s = “A”, numRows = 1
Output: “A”

Constraints:

• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ’,’ and ’.‘.
• 1 <= numRows <= 1000

Difficulty: Medium

Tags: string

Rating: 34.78%

Complexity Analysis

The solution has the following complexity characteristics:

• Time Complexity: $O(n²)$
• Space Complexity: $O(n)$

Note: This is an automated analysis and may not capture all edge cases or specific algorithmic optimizations.

Solution

Here’s my Python solution to this problem:

#Problem 6: ZigZag Conversion

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1 or numRows >= len(s): return s
        
        n = len(s)
        cycle = 2 * numRows - 2
        o = [] 

        for i in range(0, n, cycle):
            o.append(s[i])

        for r in range(1, numRows-1):
            for i in range(r, n, cycle):
                o.append(s[i])
                second_idx = i + cycle - 2*r
                if second_idx < n:
                    o.append(s[second_idx])

        for i in range(numRows-1, n, cycle):
            o.append(s[i])

        return ''.join(o)


class Solution:
    def convert(s: str, numRows: int) -> str:
        if numRows == 1 or numRows >= len(s):
            return s
        
        rows = [[] for _ in range(numRows)]
        current_row = 0
        step = 1  # 1 means moving down, -1 means moving up
        
        for char in s:
            rows[current_row].append(char)
            
            if current_row == numRows - 1:
                step = -1
            elif current_row == 0:
                step = 1
                
            current_row += step
        
        return ''.join([''.join(row) for row in rows])