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[LeetCode] 006. Zigzag Conversion

Problem (Medium)

006. 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

Approach 1: My Solution

Idea

Straight forward idea as stated in the problem:

  • create a 2-D array with numRows rows and width withd, where width is roughly len(s) // (numRows-1) + 2.
  • Put the string character by character into the 2-D array in the zigzag order.
  • Concatenate the characters back to a new string based on the order stated in the problem.

Solution

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1
            return s
        width = len(s) // (numRows-1) + 2
        m = [[None for i in range(width)] for j in range(numRows)]

        i, j, k, up = 0, 0, 0, 0
        while k < len(s):
            m[i][j] = s[k]
            if i == numRows-1:
                up = 1
                j += 1
            if i == 0 and k != 0:
                up = 0
                j += 1

            if up:
                i -= 1
            else:
                i += 1
            k += 1
            
        res = ''
        for row in m:
            res += ''.join([i for i in row if i])
        return res

Complexity

  • Time: $O(n)$
  • Space: $O(nr)$

Approach 2: Improve the first solution

Idea

Instead of initialize a 2-D array, just initialize a 1-D array in which each item represent the string for a row in the zigzag pattern.

Solution

class Solution2:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s
        rows = [''] * numRows
        cur_row, down = 0, 0

        for c in s:
            rows[cur_row] += c
            if cur_row == 0 or cur_row == numRows-1:
                down ^= 1
            cur_row += 1 if down else -1
        
        return ''.join([row for row in rows])

Complexity

  • Time: $O(n)$
  • Space: $O(n)$

Approach 3: Calculate index from original string to the new directly

Idea

Visit all characters in row 0 first, then row 1, then row 2, and so on …

Solution

class Solution2:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1: return s

        res = ''
        cycle_len = 2*numRows -2
        for i in range(numRows):
            for j in range(0, len(s) - i, cycle_len):
                res += s[j+i]
                if i != 0 and i != numRows -1 and j + cycle_len -i < len(s):
                    res += s[j + cycle_len -i]
        return res

Complexity

  • Time: $O(n)$
  • Space: $O(n)$



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