Last updated on January 20th, 2025 at 04:16 am

Here, we see the Zigzag Conversion LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

String

Level of Question

Medium

Zigzag Conversion LeetCode Solution

1. 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"

2. Coding Pattern Used in Solution

The coding pattern used in the provided code is “Zigzag Conversion”. This pattern involves arranging characters in a zigzag pattern across multiple rows and then reading them row by row to form the final string. It is a specialized problem-solving approach that doesn’t directly map to the common patterns listed.

3. Code Implementation in Different Languages

3.1 Zigzag Conversion 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 < numRows; i++) {
            for (int j = 0; j + i < n; j += cycleLen) {
                ret += s[j + i];
                if (i != 0 && i != numRows - 1 && j + cycleLen - i < n)
                    ret += s[j + cycleLen - i];
            }
        }
        return ret;
    }
};

3.2 Zigzag Conversion 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 < numRows; i++) {
            for (int j = 0; j + i < n; j += cycleLen) {
                ret.append(s.charAt(j + i));
                if (i != 0 && i != numRows - 1 && j + cycleLen - i < n)
                    ret.append(s.charAt(j + cycleLen - i));
            }
        }
        return ret.toString();
    }
}

3.3 Zigzag Conversion 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(''), '');
};

3.4 Zigzag Conversion Python

class Solution:
    def convert(self, s: str, numRows: int) -> 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. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n)	O(n)
Java	O(n)	O(n)
JavaScript	O(n)	O(n)
Python	O(n)	O(n)

• The code implements a Zigzag Conversion pattern with a time complexity of O(n) and a space complexity of O(n) across all languages.
• It efficiently processes the input string to construct the zigzag pattern row by row.