leetcode-study

6. Zigzag Conversion

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        """
        Converts the given string 's' into a zigzag pattern with 'numRows' rows, then
        reads the pattern row by row and returns the result as a single string.
        
        Parameters:
            s (str): The input string to be converted.
            numRows (int): The number of rows in the zigzag pattern.
        
        Returns:
            str: The resulting string after applying the zigzag conversion.
        """
        # When numRows is 1 or greater than/equal to the length of s,
        # no zigzag conversion is needed.
        if numRows == 1 or numRows >= len(s):
            return s
        
        # Create a list to hold strings for each row.
        rows = [''] * numRows  
        curr_row = 0          # Pointer for the current row.
        going_down = False    # Direction flag: True means moving downwards.
        
        # Process each character in the input string.
        for char in s:
            # Append the character to the current row.
            rows[curr_row] += char
            
            # If we hit the top or bottom of the zigzag pattern, we flip the direction.
            if curr_row == 0 or curr_row == numRows - 1:
                going_down = not going_down
            
            # Update the current row pointer based on the direction.
            curr_row += 1 if going_down else -1
        
        # Combine all rows to form the final string.
        return ''.join(rows)

Summary of Techniques and Approaches:

• Simulation with Direction Toggle: The solution simulates the zigzag pattern by maintaining a current row pointer and toggling the direction using a boolean flag. This approach is widely applicable to problems that require simulating movement through rows or columns.

• Edge Case Handling: The early return for cases where numRows is 1 or greater than or equal to the string length prevents unnecessary processing. Always check for such edge cases to simplify the solution.

• Efficient Data Collection: By using a list of strings for each row and joining them at the end, the solution maintains simplicity and efficiency. This technique is useful when dealing with segmented data that needs to be combined later.

• Incremental Construction: Building the result incrementally as you iterate over the input avoids extra passes over the data, leading to an optimal time complexity of O(n). Look for opportunities to accumulate results during a single pass when possible.

These approaches can be applied to a broad range of algorithmic problems, especially those involving transformation or reordering of data based on simulation of movement or state changes.