Baozi Leetcode solution 6: ZigZag Conversion

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"

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

 Problem link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

Thought Process

It's a very good implementation problem, by simply simulating each character in the string to each row, it would take extra O(N) space to hold up the array where N is the string size. However, there is definitely a difference between a good and poor implementation.

 

There is also a math formula we can use, please see attached pdf for solutions.

Solutions

 

Simulation poor implementation

public String convertSimulation(String s, int nRows) {
    if (s == null || s.length() == 0 || nRows <= 0) {
        return "";
    }

    if (nRows == 1) {
        return s;
    }

    int size = s.length();
    List<StringBuilder> buckets = new ArrayList();

    for (int i = 0; i < nRows; i++) {
        buckets.add(new StringBuilder());
    }

    int i = 0; 
    int index = 0;
    boolean fromTopToBottom = true;
    while (i < size) {
        char c = s.charAt(i);

        if (fromTopToBottom) {
            if (index == nRows - 1) {
                fromTopToBottom = false;
                buckets.get(index).append(c);
                index--;
            } else {
                buckets.get(index).append(c);
                index++;
            }
        } else {
            if (index == 0) {
                fromTopToBottom = true;
                buckets.get(index).append(c);
                index++;
            } else {
                buckets.get(index).append(c);
                index--;
            }
        }
        i++;
    }

    String res = "";
    for (StringBuilder sb : buckets) {
        res += sb.toString();
    }

    return res;
}

 

Time Complexity: O(N) where N is the string size

Space Complexity: O(N) since we used an extra array to store each character in the string

 

Simulation good implementation

public String convertSimulationOptimizedImplementation(String s, int nRows) {
    if (s == null || s.length() == 0 || nRows <= 0) {
        return "";
    }

    if (nRows == 1) {
        return s;
    }

    int size = s.length();
    // directly allocate a string builder array
    StringBuilder[] buckets = new StringBuilder[nRows];
    for (int i = 0; i < nRows; i++) {
        buckets[i] = new StringBuilder();
    }


    int i = 0;
    int index = 0;
    boolean fromTopToBottom = true;
    while (i < size) {
        char c = s.charAt(i);
        buckets[index].append(c);
        index += fromTopToBottom ? 1 : -1;

        if (index == nRows - 1 || index == 0) {
            fromTopToBottom = !fromTopToBottom;
        }

        i++;
    }

    String res = "";
    for (StringBuilder sb : buckets) {
        res += sb.toString();
    }

    return res;
}

 

Time Complexity: O(N) where N is the string size

Space Complexity: O(N) since we used an extra array to store each character in the string

References

• Leetcode official solution