LeetCode 2075 - Decode the Slanted Ciphertext

The problem requires decoding a string that was encoded using a slanted transposition cipher. In this cipher, the original text is written diagonally in a grid with a fixed number of rows.

leetcodemediumstringsimulation

LeetCode Problem 2075

Difficulty: 🟡 Medium
Topics: String, Simulation

Solution

Problem Understanding

The problem requires decoding a string that was encoded using a slanted transposition cipher. In this cipher, the original text is written diagonally in a grid with a fixed number of rows. Specifically, the placement starts at the top-left and moves down and to the right in a zig-zag manner. After filling the grid, the encoded string is formed by reading the grid row by row, including spaces for unfilled cells.

The input consists of an encoded string encodedText and an integer rows, which specifies the number of rows used during encoding. The expected output is the original string originalText, without trailing spaces.

Key points include:

  • encodedText can be up to 10^6 characters long, so inefficient approaches may time out.
  • Only lowercase letters and spaces appear in encodedText.
  • There is always a unique valid originalText.
  • Edge cases include rows = 1 where the string is unchanged, or empty strings.

The task requires reconstructing the original diagonal reading order from the row-wise encoded string.

Important constraints:

  • 1 <= rows <= 1000 ensures a manageable number of rows.
  • Trailing spaces are removed in the original text.
  • The encoding guarantees that the last column in the grid is non-empty, avoiding ambiguity.

Edge cases that could trip a naive approach include single-row encoding, very short strings, or strings with multiple consecutive spaces.

Approaches

The brute-force approach involves reconstructing the exact grid used in encoding. This requires first determining the number of columns, filling the grid row by row with the encoded string, and then reading diagonally to reconstruct the original text. While correct, this approach uses O(rows * cols) space, which may exceed limits for very large strings.

The key insight for an optimal solution is that we do not need to store the entire grid. We can simulate the diagonal traversal directly by iterating over the starting points of each row and moving diagonally down and to the right, collecting characters until we reach the end of the string. This reduces space complexity to O(n) for the output string only.

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(rows * cols) Build full grid and read diagonally. Works but uses extra space.
Optimal O(n) O(n) Directly simulate diagonal traversal without storing the full grid.

Algorithm Walkthrough

  1. Determine the number of columns in the grid. Since the string was filled row by row, the number of columns is len(encodedText) // rows.
  2. Initialize an empty list to collect characters of originalText.
  3. Iterate over each row r from 0 to rows - 1.
  4. For each starting row r, traverse diagonally down and right by incrementing both the row and column by 1 at each step. Collect the character from the corresponding position in encodedText.
  5. Continue until reaching the last row or last column.
  6. Join the collected characters into a string and strip any trailing spaces.

Why it works: The algorithm directly mimics the diagonal reading pattern used in the encoding process. Each diagonal corresponds to one position in the original text sequence. By iterating row by row as starting points, all diagonals are covered exactly once, guaranteeing reconstruction of the original text.

Python Solution

class Solution:
    def decodeCiphertext(self, encodedText: str, rows: int) -> str:
        if rows == 1:
            return encodedText.rstrip()
        
        n = len(encodedText)
        cols = n // rows
        result = []
        
        for r in range(rows):
            c = 0
            row, col = r, 0
            while row < rows and col < cols:
                idx = row * cols + col
                result.append(encodedText[idx])
                row += 1
                col += 1
        
        return ''.join(result).rstrip()

Explanation: The code first handles the edge case where there is only one row. It then calculates the number of columns by dividing the total length by the number of rows. Using nested iteration, it simulates the diagonal traversal starting from each row, appending characters to the result list. Finally, it joins the list into a string and removes trailing spaces.

Go Solution

func decodeCiphertext(encodedText string, rows int) string {
    if rows == 1 {
        return strings.TrimRight(encodedText, " ")
    }

    n := len(encodedText)
    cols := n / rows
    result := make([]byte, 0, n)

    for r := 0; r < rows; r++ {
        row, col := r, 0
        for row < rows && col < cols {
            idx := row*cols + col
            result = append(result, encodedText[idx])
            row++
            col++
        }
    }

    return strings.TrimRight(string(result), " ")
}

Go-specific notes: Go requires converting the result slice of bytes back to a string. Trimming trailing spaces uses strings.TrimRight. Slices are used to efficiently append characters without knowing the exact final size, though the capacity is preallocated for performance.

Worked Examples

Example 1: encodedText = "ch ie pr", rows = 3

  • Grid size: 3 rows, 4 columns (len(encodedText)//rows)
  • Grid simulation:
c0 c1 c2 c3
c h
i e
p r
  • Diagonal traversal:
  1. Start row 0: indices (0,0),(1,1),(2,2) → "c", "i", "p"
  2. Start row 1: indices (1,0),(2,1) → "h", "e"
  3. Start row 2: indices (2,0) → "r"
  • Combine → "cipher"

Example 2: encodedText = "iveo eed l te olc", rows = 4

  • Columns = len(encodedText)//rows = 24//4 = 6
  • Grid traversal diagonally reconstructs "i love leetcode".

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each character is visited exactly once in diagonal traversal.
Space O(n) Only the result string of length n is stored, no full grid needed.

The key to the optimal complexity is simulating the traversal without creating a full 2D matrix, reducing space usage while still visiting every character exactly once.

Test Cases

# Basic examples
assert Solution().decodeCiphertext("ch   ie   pr", 3) == "cipher"  # from example 1
assert Solution().decodeCiphertext("iveo    eed   l te   olc", 4) == "i love leetcode"  # example 2
assert Solution().decodeCiphertext("coding", 1) == "coding"  # example 3

# Edge cases
assert Solution().decodeCiphertext("", 1) == ""  # empty string
assert Solution().decodeCiphertext("a", 1) == "a"  # single character
assert Solution().decodeCiphertext("a", 2) == "a"  # single character with more rows
assert Solution().decodeCiphertext("a b  c  ", 3) == "abc"  # trailing spaces in encodedText
Test Why
"ch ie pr", 3 Validates diagonal reconstruction for multiple rows and spaces
"iveo eed l te olc", 4 Tests longer sentence with multiple words and spaces
"coding", 1 Single-row edge case where encoded = original
"" Empty string edge case
"a" with 1 and 2 rows Single-character handling
"a b c ", 3 Encoded string with trailing spaces

Edge Cases

Single row (rows = 1): In this case, the encoding does not change the string. Any solution must check for this condition to avoid unnecessary computation.

Empty string: The input may be empty, in which case the output must also be an empty string. Accessing indices without a check could cause errors.

Trailing spaces in encodedText: Spaces in the encoded string that were added during padding must not appear in the output. The solution uses rstrip() (Python) or TrimRight (Go) to remove trailing spaces safely, ensuring compliance with the problem constraints.