> For the complete documentation index, see [llms.txt](https://programming.arora-aditya.com/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://programming.arora-aditya.com/leetcode/string/6.-zigzag-conversion.md).

# 6. ZigZag Conversion

### Approach 1: Iterate character by character

```python
class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if numRows<=1: return s
        ans=[""]*numRows
        i=0
        for c in s:
            ans[i]+=c
            if i==numRows-1:
                d=-1
            if i==0:
                d=1
            i+=d
        return ''.join(ans)
```

Iterate over the characters one by one, while maintaining a counter for current row number. Keep incrementing the counter one by one and when you reach `numRows` start subtracting 1 each time, and append the character to the calculated index and return the join of all these rows as a string

> **Time Complexity: O(n)**
>
> **Space Complexity: O(n)**