Generated by Codex with GPT-5

Quick facts

What the problem is really asking

The problem gives a string and a row count, then asks for the string that would appear if the characters were written in a zigzag pattern and read row by row.

For example, with numRows = 3, the characters move:

  • down through the rows
  • then diagonally back up
  • then down again

The actual output is not the 2D drawing. The output is the concatenation of each row after simulating that movement.

So the core task is:

take each character, decide which row it belongs to as the zigzag path moves, then join the rows together.

The first correct idea

One natural approach is to build the full grid, place characters into cells, and then scan the rows.

That does work, but it is more complicated than necessary. Most of the grid is empty space, and maintaining column positions adds bookkeeping that does not help with the final answer.

The important observation is that the final string depends only on which row each character lands in. The exact visual spacing does not matter.

The key insight

The zigzag movement follows a simple repeating rule:

  1. Start at row 0.
  2. Move downward one row at a time.
  3. When the bottom row is reached, reverse direction.
  4. Move upward one row at a time.
  5. When the top row is reached, reverse direction again.

That means the algorithm never needs a matrix. It only needs:

  • one string builder per row
  • a current row pointer
  • a direction flag saying whether the path is moving down or up

Each new character is appended to the current row, then the row pointer is updated.

How to derive the optimal solution in an interview

A clean way to derive it is:

  1. Notice that the output is read row by row.
  2. Realize that each character belongs to exactly one row in that traversal.
  3. Replace the visual matrix with a list of row builders.
  4. Simulate the row index as it bounces between the top and bottom rows.

The two special cases also fall out naturally:

  • If numRows == 1, there is no zigzag at all.
  • If numRows >= len(s), every character stays in its own straight pass, so the answer is just the original string.

Why the algorithm is correct

The invariant is:

after processing the first i characters, each row builder contains exactly the characters that belong to that row in the zigzag pattern for the prefix s[:i], in the correct left-to-right order.

This holds because each step appends the next character to the single row currently visited by the zigzag path. Then the direction update moves the path exactly the same way the drawing would move: reversing at the top and bottom boundaries.

By the time every character has been processed, the row builders contain the exact contents of the zigzag pattern’s rows. Joining those rows therefore produces the required answer.

Best approach

The row-simulation solution is the standard optimal answer for this problem:

  • Time: O(n), because each character is processed once
  • Space: O(n), because the output rows together store all characters

For interview purposes, this is the right solution to give first. It is easy to explain, easy to implement correctly, and directly matches the structure of the problem.

Python solution

The implementation below keeps the row-assignment logic separate from the LeetCode wrapper. It handles the edge cases up front and uses a direction variable that flips only when the path touches the top or bottom row.

from typing import List


def convert_to_zigzag_rows(text: str, row_count: int) -> str:
    """
    Return the row-by-row reading of the zigzag pattern formed by text.
    """
    if row_count <= 1 or row_count >= len(text):
        return text

    rows: List[List[str]] = [[] for _ in range(row_count)]
    current_row = 0
    direction = 1  # 1 means moving down, -1 means moving up.

    for character in text:
        rows[current_row].append(character)

        if current_row == 0:
            direction = 1
        elif current_row == row_count - 1:
            direction = -1

        current_row += direction

    return "".join("".join(row_characters) for row_characters in rows)


class Solution:
    def convert(self, s: str, numRows: int) -> str:
        """LeetCode entry point."""
        return convert_to_zigzag_rows(s, numRows)

Interview follow-ups

Can you solve it without explicitly storing every row as a separate list?

Yes. Another valid approach is to compute the answer row by row using the zigzag cycle length, which is 2 * numRows - 2. The first and last rows contribute characters at regular cycle intervals. Middle rows contribute two characters per cycle: one vertical character and one diagonal character. This works because the zigzag pattern repeats with that fixed cycle size. The time complexity stays O(n), and the extra working memory besides the output buffer can be reduced to O(1). The tradeoff is that the index math is more subtle, so it is easier to make off-by-one mistakes than with the row-simulation approach.

What if the interviewer asks you to print the actual zigzag layout?

Then the problem changes. Now the spacing matters, so just grouping by row is no longer enough. A direct simulation with a 2D grid or with per-row coordinate tracking becomes appropriate. The row-simulation solution for the original problem discards column information on purpose because the final answer does not need it. Once the visual layout is required, that information must be preserved. The implementation becomes more cumbersome and usually uses more space, but it is the right tradeoff for the changed output requirement.

What edge cases are most likely to cause bugs here?

The biggest one is numRows == 1. If that case is not handled early, the direction-flipping logic can break because the top and bottom row are the same row. Another common edge case is when numRows is at least the string length. In that situation, the zigzag never actually turns, so the original string should be returned unchanged. Calling out these cases shows the interviewer that the control flow has been thought through carefully, not just coded from a memorized pattern.

If memory use matters, which implementation would you prefer?

If the goal is simply to return the converted string, both main approaches are already efficient enough for normal interview constraints because the output itself is O(n). In practice, the row-builder simulation is usually the better choice because it is simpler and less error-prone. If the interviewer pushes specifically on auxiliary memory beyond the output buffer, the cycle-based indexing approach is the one to discuss, because it can avoid maintaining multiple row containers. The tradeoff is readability versus tighter control over extra working state.