Problem #6: Zigzag Conversion

Intuition & Strategy Breakdown

Simulate the zigzag placement using an array of strings representing rows. Track the current row index and direction (downward or upward), appending characters as you iterate through the string.

Python Solution

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1 or numRows >= len(s):
            return s
        rows = [''] * numRows
        curr_row = 0
        going_down = False
        for c in s:
            rows[curr_row] += c
            if curr_row == 0 or curr_row == numRows - 1:
                going_down = not going_down
            curr_row += 1 if going_down else -1
        return ''.join(rows)

Java Solution

class Solution {
    public String convert(String s, int numRows) {
        if (numRows == 1 || numRows >= s.length()) return s;
        StringBuilder[] rows = new StringBuilder[numRows];
        for (int i = 0; i < numRows; i++) rows[i] = new StringBuilder();
        
        int currRow = 0;
        boolean goingDown = false;
        for (char c : s.toCharArray()) {
            rows[currRow].append(c);
            if (currRow == 0 || currRow == numRows - 1) goingDown = !goingDown;
            currRow += goingDown ? 1 : -1;
        }
        
        StringBuilder res = new StringBuilder();
        for (StringBuilder row : rows) res.append(row);
        return res.toString();
    }
}