Decode the Slanted Ciphertext - Practice Coding Problems

Decode the Slanted Ciphertext - Problem

String Simulation Medium

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.

Input & Output

Example 1 — Basic Case

$ Input: encodedText = "ch ie pr", rows = 3

› Output: cipher

💡 Note: The matrix is 3×3. Fill row by row: [[c,h, ],[i,e, ],[p,r, ]]. Read diagonally: c→i→p, h→e, (space) gives "cipher"

Example 2 — Single Row

$ Input: encodedText = "iveo eed l te olc", rows = 4

› Output: i love leetcode

💡 Note: Matrix is 4×6. After filling and reading diagonally: i→v→e→o→(space)→(space), giving the original text with trailing spaces removed

Example 3 — One Row Only

$ Input: encodedText = "coding", rows = 1

› Output: coding

💡 Note: With only 1 row, no transformation occurs. The encoded text is the same as original text

Constraints

1 ≤ rows ≤ 1000
1 ≤ encodedText.length ≤ 2000
encodedText consists of lowercase English letters and spaces
originalText does not have trailing spaces

Visualization

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Understanding the Visualization

Encoded Input

Row-wise encoded text: "ch ie pr" with 3 rows

Matrix Reconstruction

Fill 3×3 matrix row by row, then read diagonally

Original Output

Diagonal reading gives: "cipher"

Key Takeaway

🎯 Key Insight: The slanted cipher fills diagonally but reads row-wise - we reverse by filling row-wise and reading diagonally

Asked in

G Google 12 M Microsoft 8 a Amazon 6

The key insight is understanding the slanted transposition cipher pattern: text is filled diagonally but read row-wise. To decode, we reverse this by filling a matrix row-wise and reading diagonally. The optimal approach uses direct position calculation to avoid storing the full matrix. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Direct Position Calculation	O(n)	O(1)	Calculate original character positions directly without building the full matrix
Brute Force Matrix Reconstruction	O(n)	O(n)	Build the full matrix by placing characters row by row, then read diagonally

Direct Position Calculation — Algorithm Steps

Calculate matrix dimensions: cols = ceil(len(encodedText) / rows)
For each diagonal starting position, calculate corresponding encoded positions
Map each original position to its encoded position using mathematical formula
Build result by reading characters in diagonal order and remove trailing spaces

Visualization

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Step-by-Step Walkthrough

Calculate Dimensions

Determine matrix size: cols = ceil(n/rows)

Position Formula

For diagonal (r,c): encoded_pos = r * cols + c

Direct Access

Read characters directly using calculated positions

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* encodedText, int rows) {
    int n = strlen(encodedText);
    if (rows == 1) {
        char* result = malloc((n + 1) * sizeof(char));
        strcpy(result, encodedText);
        return result;
    }
    
    int cols = (n + rows - 1) / rows;
    
    char* result = malloc((n + 1) * sizeof(char));
    int resultIdx = 0;
    
    // Read diagonally without building matrix
    for (int startCol = 0; startCol < cols; startCol++) {
        int r = 0, c = startCol;
        while (r < rows && c < cols) {
            // Calculate position in encoded text
            int pos = r * cols + c;
            if (pos < n) {
                result[resultIdx++] = encodedText[pos];
            } else {
                result[resultIdx++] = ' ';
            }
            r++;
            c++;
        }
    }
    result[resultIdx] = '\0';
    
    // Remove trailing spaces
    while (resultIdx > 0 && result[resultIdx - 1] == ' ') {
        result[--resultIdx] = '\0';
    }
    
    return result;
}

int main() {
    char encodedText[1001];
    fgets(encodedText, sizeof(encodedText), stdin);
    encodedText[strcspn(encodedText, "\n")] = 0;
    
    int rows;
    scanf("%d", &rows);
    
    char* result = solution(encodedText, rows);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through characters with direct position calculation

✓ Linear Growth

Space Complexity

O(1)

Only uses variables for calculation, no matrix storage

✓ Linear Space

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Ln 1, Col 1

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Position Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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