# ZigZag Conversion – LeetCode Solutions

LeetCode has a hard coding Problem in Its’ Algorithm Section: “ZigZag Conversion of Given String”. Today We are going to solve this problem. LeetCode Link of the Problem is HERE.

#### Question

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);
```
##### Examples
```Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
```
```Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I
```
```Input: s = "A", numRows = 1
Output: "A"
```
##### Conditions
• `1 <= s.length <= 1000`
• `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
• `1 <= numRows <= 1000`

#### Solution to Zigzag Conversion

The Skeleton Code given in LeetCode is

``````class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
``````

Firstly, Check for the Least Possible Solution. In which we are given an input String with Number of Rows = 1, In this case, it should return the same input string.

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

Secondly, Create an array of size equal to String length.

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

l = len(s)

arr=["" for x in range(l)]
``````

Set the row to 0 and the orientation to “down.” The direction determines whether we should go in rows up or down. Go over the input string(loop).

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

l = len(s)

arr=["" for x in range(l)]
down = True
row = 0
for i in range(l):

``````

Add the current character to the current row’s string. Change the orientation to ‘up’ if the row number is n-1. Change the orientation to ‘down’ if the row number is 0. Row++ if the direction is down. Otherwise, row–. Return the String.

Complete Code:

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

l = len(s)

arr=["" for x in range(l)]
row = 0

for i in range(l):

arr[row] += s[i]

if row == numRows - 1:
down = False

elif row == 0:
down = True

if down:
row += 1
else:
row -= 1

# Print concatenation
# of all rows
final = ""
for i in range(l):
final += arr[i]
return final``````