Problem Statement:

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

Solution:

We only need to notice that the row number for any character is decided by its modulus wrt $m=2*|s|-2$. This can be easily observed once you notice that the pattern starts repeating after it reaches the top row after one zigzag. Hence $m$ is the length of one zigzag. Next we need to check if each position is coming in the downward phase or the upward phase of zigzag and accordingly update that row.

 
class Solution {
public:
    string convert(string s, int numRows) 
    {
        if (numRows==1) return s;
        int m = 2*numRows - 2;
        vector<string> rows(numRows, "");
        for (int i=0; i<s.length(); i++)
        {
            int r = i%m;
            if (r<numRows) rows[r].push_back(s[i]);
            else rows[m-r].push_back(s[i]);
        }
        string res = "";
        for (string r: rows) res+=r;
        return res;
    }
};

Time complexity: $O(n)$