Problem

A string originalText is encoded using a slanted transposition cipher to a string encodedText with the help of a matrix having a fixed number of rows rows.

originalText is placed first in a top-left to bottom-right manner.

The blue cells are filled first, followed by the red cells, then the yellow cells, and so on, until we reach the end of originalText. The arrow indicates the order in which the cells are filled. All empty cells are filled with ' '. The number of columns is chosen such that the rightmost column will not be empty after filling in originalText.

encodedText is then formed by appending all characters of the matrix in a row-wise fashion.

The characters in the blue cells are appended first to encodedText, then the red cells, and so on, and finally the yellow cells. The arrow indicates the order in which the cells are accessed.

For example, if originalText = "cipher" and rows = 3, then we encode it in the following manner:

The blue arrows depict how originalText is placed in the matrix, and the red arrows denote the order in which encodedText is formed. In the above example, encodedText = "ch ie pr".

Given the encoded string encodedText and number of rows rows, return the original string originalText.

Note: originalText does not have any trailing spaces ' '. The test cases are generated such that there is only one possible originalText.

Examples

Example 1

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Input: encodedText = "ch   ie   pr", rows = 3
Output: "cipher"
Explanation: This is the same example described in the problem description.

Example 2

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![](https://assets.leetcode.com/uploads/2021/10/26/exam1.png)

Input: encodedText = "iveo    eed   l te   olc", rows = 4
Output: "i love leetcode"
Explanation: The figure above denotes the matrix that was used to encode originalText. 
The blue arrows show how we can find originalText from encodedText.

Example 3

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![](https://assets.leetcode.com/uploads/2021/10/26/eg2.png)

Input: encodedText = "coding", rows = 1
Output: "coding"
Explanation: Since there is only 1 row, both originalText and encodedText are the same.

Constraints

  • 0 <= encodedText.length <= 10^6
  • encodedText consists of lowercase English letters and ' ' only.
  • encodedText is a valid encoding of some originalText that does not have trailing spaces.
  • 1 <= rows <= 1000
  • The testcases are generated such that there is only one possible originalText.

Solution

Method 1 – Matrix Simulation and Diagonal Traversal

Intuition

The encoded text is formed by writing the original text diagonally in a matrix and then reading it row by row. To decode, we reconstruct the matrix row-wise and then read the diagonals to recover the original text.

Approach

  1. Compute the number of columns as cols = len(encodedText) // rows.
  2. Fill a matrix of size rows x cols with the characters from encodedText row by row.
  3. Traverse the matrix diagonally: for each diagonal starting from the first column, collect characters from (i, j) where i is the row and j = i + d is the column.
  4. Concatenate the collected characters.
  5. Remove any trailing spaces from the result.

Code

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class Solution {
public:
    string decodeCiphertext(string s, int rows) {
        int n = s.size();
        if (rows == 1) return s;
        int cols = n / rows;
        string ans;
        for (int d = 0; d < cols; ++d) {
            for (int i = 0; i < rows; ++i) {
                int j = d + i;
                if (j < cols) {
                    char c = s[i * cols + j];
                    ans += c;
                }
            }
        }
        while (!ans.empty() && ans.back() == ' ') ans.pop_back();
        return ans;
    }
};
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func decodeCiphertext(s string, rows int) string {
    n := len(s)
    if rows == 1 {
        return s
    }
    cols := n / rows
    ans := make([]byte, 0, n)
    for d := 0; d < cols; d++ {
        for i := 0; i < rows; i++ {
            j := d + i
            if j < cols {
                ans = append(ans, s[i*cols+j])
            }
        }
    }
    // Remove trailing spaces
    for len(ans) > 0 && ans[len(ans)-1] == ' ' {
        ans = ans[:len(ans)-1]
    }
    return string(ans)
}
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class Solution {
    public String decodeCiphertext(String s, int rows) {
        int n = s.length();
        if (rows == 1) return s;
        int cols = n / rows;
        StringBuilder ans = new StringBuilder();
        for (int d = 0; d < cols; d++) {
            for (int i = 0; i < rows; i++) {
                int j = d + i;
                if (j < cols) {
                    ans.append(s.charAt(i * cols + j));
                }
            }
        }
        // Remove trailing spaces
        int len = ans.length();
        while (len > 0 && ans.charAt(len - 1) == ' ') {
            ans.setLength(--len);
        }
        return ans.toString();
    }
}
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class Solution {
    fun decodeCiphertext(s: String, rows: Int): String {
        if (rows == 1) return s
        val n = s.length
        val cols = n / rows
        val ans = StringBuilder()
        for (d in 0 until cols) {
            for (i in 0 until rows) {
                val j = d + i
                if (j < cols) {
                    ans.append(s[i * cols + j])
                }
            }
        }
        return ans.trimEnd().toString()
    }
}
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class Solution:
    def decodeCiphertext(self, s: str, rows: int) -> str:
        if rows == 1:
            return s
        n: int = len(s)
        cols: int = n // rows
        ans: list[str] = []
        for d in range(cols):
            for i in range(rows):
                j = d + i
                if j < cols:
                    ans.append(s[i * cols + j])
        return ''.join(ans).rstrip()
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impl Solution {
    pub fn decode_ciphertext(s: String, rows: i32) -> String {
        if rows == 1 { return s; }
        let n = s.len();
        let cols = n / rows as usize;
        let s = s.as_bytes();
        let mut ans = Vec::with_capacity(n);
        for d in 0..cols {
            for i in 0..rows as usize {
                let j = d + i;
                if j < cols {
                    ans.push(s[i * cols + j]);
                }
            }
        }
        while ans.last() == Some(&b' ') {
            ans.pop();
        }
        String::from_utf8(ans).unwrap()
    }
}
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class Solution {
    decodeCiphertext(s: string, rows: number): string {
        if (rows === 1) return s;
        const n = s.length;
        const cols = Math.floor(n / rows);
        let ans = '';
        for (let d = 0; d < cols; d++) {
            for (let i = 0; i < rows; i++) {
                const j = d + i;
                if (j < cols) {
                    ans += s[i * cols + j];
                }
            }
        }
        return ans.replace(/\s+$/, '');
    }
}

Complexity

  • ⏰ Time complexity: O(n), where n is the length of encodedText, since we visit each character once.
  • 🧺 Space complexity: O(n), for storing the matrix and the answer string.