Last updated on October 10th, 2024 at 12:46 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.

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Zigzag Conversion LeetCode Solution

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 < 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;
    }
};

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.

1.2 Time Complexity

• O(n).

1.3 Space Complexity

• 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 < 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();
    }
}

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.

2.2 Time Complexity

• O(n).

2.3 Space Complexity

• 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.

3.2 Time Complexity

• O(n).

3.3 Space Complexity

• O(n).

4. Zigzag Conversion Leetcode Solution 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)

4.1 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.

4.2 Time Complexity

• O(n).

4.3 Space Complexity

• O(n).