Zigzag Conversion LeetCode Solution

Last updated on July 15th, 2024 at 12:54 am

Here, We see Zigzag Conversion LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

List of all LeetCode Solution

Topics

String

Companies

Level of Question

Medium

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"```
```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"```

1. Zigzag Conversion Leetcode Solution C++

```class Solution {
public:
string convert(string s, int numRows) {

if (numRows == 1) return s;

string ret;
int n = s.size();
int cycleLen = 2 * numRows - 2;

for (int i = 0; i &lt; numRows; i++) {
for (int j = 0; j + i &lt; n; j += cycleLen) {
ret += s[j + i];
if (i != 0 &amp;&amp; i != numRows - 1 &amp;&amp; j + cycleLen - i &lt; n)
ret += s[j + cycleLen - i];
}
}
return ret;
}
};```

1.1 Explanation

1. Edge Case: If numRows is 1, return the original string s.
2. Initialize: Create an empty string ret to store the result. Calculate cycleLen, which is the length of one cycle in the zigzag pattern.
3. Nested Loops:
• Outer loop: Iterate through each row from 0 to numRows – 1.
• Inner loop: Iterate through the string s in steps of cycleLen.
• Add characters to ret from the main column (s[j + i]) and the diagonal column (s[j + cycleLen – i]) if applicable.

• O(n).

• O(n).

2. Zigzag Conversion Leetcode Solution Java

```class Solution {
public String convert(String s, int numRows) {

if (numRows == 1) return s;

StringBuilder ret = new StringBuilder();
int n = s.length();
int cycleLen = 2 * numRows - 2;

for (int i = 0; i &lt; numRows; i++) {
for (int j = 0; j + i &lt; n; j += cycleLen) {
ret.append(s.charAt(j + i));
if (i != 0 &amp;&amp; i != numRows - 1 &amp;&amp; j + cycleLen - i &lt; n)
ret.append(s.charAt(j + cycleLen - i));
}
}
return ret.toString();
}
}```

2.1 Explanation

1. Edge Case: If numRows is 1, return the original string s.
2. Initialize: Create a StringBuilder ret to store the result. Calculate cycleLen, which is the length of one cycle in the zigzag pattern.
3. Nested Loops:
• Outer loop: Iterate through each row from 0 to numRows – 1.
• Inner loop: Iterate through the string s in steps of cycleLen.
• Add characters to ret from the main column (s.charAt(j + i)) and the diagonal column (s.charAt(j + cycleLen – i)) if applicable.

• O(n).

• O(n).

3. Zigzag Conversion Leetcode Solution JavaScript

```var convert = function(s, numRows) {
if (numRows === 1 || s.length < numRows) return s;

let rows = [];
let converted = '';
let reverse = false;
let count = 0;

for (let i = 0; i < numRows; i++) rows[i] = [];
for (let i = 0; i < s.length; i++) {
rows[count].push(s[i]);
reverse ? count-- : count++;
if (count === numRows - 1 || count === 0) reverse = !reverse;
}
return rows.reduce((converted, cur) => converted + cur.join(''), '');
};
```

3.1 Explanation

1. Edge Case: If numRows is 1 or the length of s is less than numRows, return the original string s.
2. Initialize: Create an array rows to store characters for each row. Create a variable converted to store the final result. Initialize reverse to handle the direction change in the zigzag pattern and count to track the current row.
3. Fill Rows:
• Iterate through each character in the string s.
• Push characters into the corresponding row in rows.
• Change direction (reverse) when reaching the top or bottom row.
4. Join Rows: Combine all rows into the final result string using reduce.

• O(n).

• O(n).

4. Zigzag Conversion Leetcode Solution Python

```class Solution:
def convert(self, s: str, numRows: int) -&gt; str:
if numRows == 1:
return s

row_arr = [""] * numRows
row_idx = 1
going_up = True

for ch in s:
row_arr[row_idx-1] += ch
if row_idx == numRows:
going_up = False
elif row_idx == 1:
going_up = True

if going_up:
row_idx += 1
else:
row_idx -= 1

return "".join(row_arr)```

4.1 Explanation

1. Edge Case: If numRows is 1, return the original string s.
2. Initialize: Create a list row_arr to store strings for each row. Initialize row_idx to track the current row and going_up to handle the direction change in the zigzag pattern.
3. Fill Rows:
• Iterate through each character in the string s.
• Add characters to the corresponding row in row_arr.
• Change direction (going_up) when reaching the top or bottom row.
4. Join Rows: Combine all rows into the final result string using “”.join.

• O(n).

• O(n).
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