6. Zigzag Conversion LeetCode Solution

In this guide we will provide 6. Zigzag Conversion LeetCode Solution with best time and space complexity. The solution to Zigzag Conversion problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Table of Contents

6. Zigzag Conversion LeetCode Solution image

Problem Statement of Zigzag Conversion

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

Complexity Analysis

  • Time Complexity: O(|\texttt{s}|)
  • Space Complexity: O(|\texttt{s}|)

6. Zigzag Conversion LeetCode Solution in C++

class Solution {
 public:
  string convert(string s, int numRows) {
    string ans;
    vector<vector<char>> rows(numRows);
    int k = 0;
    int direction = (numRows == 1) - 1;

    for (const char c : s) {
      rows[k].push_back(c);
      if (k == 0 || k == numRows - 1)
        direction *= -1;
      k += direction;
    }

    for (const vector<char>& row : rows)
      for (const char c : row)
        ans += c;

    return ans;
  }
};
/* code provided by PROGIEZ */

6. Zigzag Conversion LeetCode Solution in Java

class Solution {
  public String convert(String s, int numRows) {
    StringBuilder sb = new StringBuilder();
    List<Character>[] rows = new List[numRows];
    int k = 0;
    int direction = numRows == 1 ? 0 : -1;

    for (int i = 0; i < numRows; ++i)
      rows[i] = new ArrayList<>();

    for (final char c : s.toCharArray()) {
      rows[k].add(c);
      if (k == 0 || k == numRows - 1)
        direction *= -1;
      k += direction;
    }

    for (List<Character> row : rows)
      for (final char c : row)
        sb.append(c);

    return sb.toString();
  }
}
// code provided by PROGIEZ

6. Zigzag Conversion LeetCode Solution in Python

class Solution:
  def convert(self, s: str, numRows: int) -> str:
    rows = [''] * numRows
    k = 0
    direction = (numRows == 1) - 1

    for c in s:
      rows[k] += c
      if k == 0 or k == numRows - 1:
        direction *= -1
      k += direction

    return ''.join(rows)
#code by PROGIEZ

Additional Resources

See also  22. Generate Parentheses LeetCode Solution

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