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
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
1 public String convertSimulation(String s, int nRows) { 2 if (s == null || s.length() == 0 || nRows <= 0) { 3 return ""; 4 } 5 6 if (nRows == 1) { 7 return s; 8 } 9 10 int size = s.length(); 11 List<StringBuilder> buckets = new ArrayList(); 12 13 for (int i = 0; i < nRows; i++) { 14 buckets.add(new StringBuilder()); 15 } 16 17 int i = 0; 18 int index = 0; 19 boolean fromTopToBottom = true; 20 while (i < size) { 21 char c = s.charAt(i); 22 23 if (fromTopToBottom) { 24 if (index == nRows - 1) { 25 fromTopToBottom = false; 26 buckets.get(index).append(c); 27 index--; 28 } else { 29 buckets.get(index).append(c); 30 index++; 31 } 32 } else { 33 if (index == 0) { 34 fromTopToBottom = true; 35 buckets.get(index).append(c); 36 index++; 37 } else { 38 buckets.get(index).append(c); 39 index--; 40 } 41 } 42 i++; 43 } 44 45 String res = ""; 46 for (StringBuilder sb : buckets) { 47 res += sb.toString(); 48 } 49 50 return res; 51 }
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
1 public String convertSimulationOptimizedImplementation(String s, int nRows) { 2 if (s == null || s.length() == 0 || nRows <= 0) { 3 return ""; 4 } 5 6 if (nRows == 1) { 7 return s; 8 } 9 10 int size = s.length(); 11 // directly allocate a string builder array 12 StringBuilder[] buckets = new StringBuilder[nRows]; 13 for (int i = 0; i < nRows; i++) { 14 buckets[i] = new StringBuilder(); 15 } 16 17 18 int i = 0; 19 int index = 0; 20 boolean fromTopToBottom = true; 21 while (i < size) { 22 char c = s.charAt(i); 23 buckets[index].append(c); 24 index += fromTopToBottom ? 1 : -1; 25 26 if (index == nRows - 1 || index == 0) { 27 fromTopToBottom = !fromTopToBottom; 28 } 29 30 i++; 31 } 32 33 String res = ""; 34 for (StringBuilder sb : buckets) { 35 res += sb.toString(); 36 } 37 38 return res; 39 }
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