grid47 Exploring patterns and algorithms
Nov 6, 2024
5 min read
Solution to LeetCode 6: Zigzag Conversion Problem
Given a string, write it in a zigzag pattern on a specified number of rows and then read the rows line by line to create the final string. Return this transformed string.
Problem
Approach
Steps
Complexity
Input: The input consists of a string and an integer specifying the number of rows for the zigzag pattern.
Example: Input: s = 'HELLOZIGZAG', numRows = 3
Constraints:
• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000
Output: Return the string after converting it into the zigzag pattern and reading it row by row.
Example: Output: 'HLZGELOIZALG'
Constraints:
• The output must be a single string.
Goal: Transform the string into the zigzag pattern and read row by row to form the final output.
Steps:
• Initialize an array of strings to represent each row in the zigzag pattern.
• Iterate through the input string and append characters to appropriate rows in a zigzag manner.
• Switch direction (downwards or upwards) when the top or bottom row is reached.
• Concatenate the strings from all rows to create the final result.
Goal: The constraints ensure the input string and the number of rows are within valid ranges.
Steps:
• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000
Assumptions:
• The number of rows will always be less than or equal to the length of the string.
• If numRows is 1, the result is the same as the input string.
• Input: Input: s = 'HELLOZIGZAG', numRows = 3
• Explanation: The zigzag pattern for 3 rows creates the final string 'HLZGELOIZALG'.
• Input: Input: s = 'DSACODE', numRows = 4
• Explanation: The zigzag pattern for 4 rows creates the final string 'DCOESADE'.
Approach: The solution uses a simulation approach to mimic the zigzag traversal and builds each row incrementally.
Observations:
• The zigzag traversal alternates direction (downwards or upwards) when boundaries are reached.
• Using an array of strings for each row simplifies the logic for appending characters.
• We can iterate through the string once while maintaining a direction flag and row index.
Steps:
• Check if numRows is 1; if true, return the input string as is.
• Initialize an array of strings, one for each row.
• Iterate through the string while keeping track of the current row and direction.
• Append the current character to the appropriate row and adjust the direction if needed.
• Finally, concatenate all rows to form the final string.
Empty Inputs:
• No empty inputs due to constraints.
Large Inputs:
• The solution should handle strings of length 1000 and up to 1000 rows efficiently.
Special Values:
• If numRows is 1, the output is the same as the input string.
Constraints:
• Ensure memory and time usage are optimized for large input sizes.
string convert(string str, int n) {
int len = str.size();
if(n ==1) return str;
vector<string> s(n, "");
bool down =true; int ridx =1;
for (int i =0; i < len; i++) {
s[ridx -1] += str[i];
if (down) ridx++;
else ridx--;
if ( (ridx == n) || (ridx ==1) )
down =!down;
}
string res ="";
for(int i =0; i < n; i++)
res += s[i];
return res;
}
1 : Function Declaration
string convert(string str, int n) {
Declare the `convert` function, which takes a string `str` and an integer `n` as input and returns a string.
2 : Variable Initialization
int len = str.size();
Calculate the length of the input string `str` and store it in the `len` variable.
3 : Conditional Return
if(n ==1) return str;
If `n` is 1, return the input string `str` as it is.
4 : Array Initialization
vector<string> s(n, "");
Initialize a vector `s` of size `n` to store the characters of each row. Each element is initially an empty string.
5 : Variable Initialization
bool down =true; int ridx =1;
Initialize `down` to `true` to indicate the direction of zigzagging, and `ridx` to 1 to track the current row index.
6 : Loop Iteration
for (int i =0; i < len; i++) {
Iterate over each character in the input string.
7 : String Manipulation
s[ridx -1] += str[i];
Append the current character `str[i]` to the string at index `ridx - 1` in the `s` vector.
8 : Conditional Update
if (down) ridx++;
If `down` is true, increment `ridx` to move to the next row.
9 : Conditional Update
else ridx--;
If `down` is false, decrement `ridx` to move to the previous row.
10 : Conditional Update
if ( (ridx == n) ||
Check if we've reached the bottom row.
11 : Conditional Update
(ridx ==1) )
Check if we've reached the top row.
12 : Conditional Update
down =!down;
Toggle the `down` flag to change the direction of zigzagging.
13 : Variable Initialization
string res ="";
Initialize an empty string `res` to store the final result.
14 : Loop Iteration
for(int i =0; i < n; i++)
Iterate over each row in the `s` vector.
15 : String Manipulation
res += s[i];
Append the current row `s[i]` to the `res` string.
16 : Return Value
return res;
Return the final zigzagged string `res`.
Best Case: O(s.length)
Average Case: O(s.length)
Worst Case: O(s.length)
Description: Each character in the string is processed exactly once.
Best Case: O(s.length)
Worst Case: O(s.length)
Description: Space is used for storing the strings for each row.
