In this post, we are going to solve the 6. Zigzag Conversion problem of Leetcode. This problem 6. Zigzag Conversion is a Leetcode medium level problem. Let’s see code, 6. Zigzag Conversion.

Problem

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

Now, let’s see the code of 6. Zigzag Conversion – Leetcode Solution.

Zigzag Conversion – Leetcode Solution

6. Zigzag Conversion – Solution in Java

class Solution {
    public String convert(String s, int nRows) {
        char[] c = s.toCharArray();
        int len = c.length;
        StringBuffer[] sb = new StringBuffer[nRows];
        for (int i = 0; i < sb.length; i++) sb[i] = new StringBuffer();

        int i = 0;
        while (i < len) {
            for (int idx = 0; idx < nRows && i < len; idx++) // vertically down
                sb[idx].append(c[i++]);
            for (int idx = nRows-2; idx >= 1 && i < len; idx--) // obliquely up
                sb[idx].append(c[i++]);
        }
        for (int idx = 1; idx < sb.length; idx++)
            sb[0].append(sb[idx]);
        return sb[0].toString();
    }   
}

6. Zigzag Conversion – Solution in C++

class Solution {
public:
    string convert(string s, int numRows) {
        vector<string> vs(numRows, "");
        int n = s.length(), i = 0;
        while (i < n) {
            for (int j = 0; j < numRows && i < n; j++)
                vs[j].push_back(s[i++]);
            for (int j = numRows - 2; j >= 1 && i < n; j--)
                vs[j].push_back(s[i++]);
        }
        string zigzag;
        for (string v : vs) zigzag += v;
        return zigzag;
    } 
};

6. Zigzag Conversion – Solution in Python

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if numRows == 1 or numRows >= len(s):
            return s

        L = [''] * numRows
        index, step = 0, 1

        for x in s:
            L[index] += x
            if index == 0:
                step = 1
            elif index == numRows -1:
                step = -1
            index += step

        return ''.join(L)

Note: This problem 6. Zigzag Conversion is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.