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
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