Leetcode 2075: Decode the Slanted Ciphertext
Given an encoded string and the number of rows used to encode the original text, recover the original string. The encoded text is produced by filling a matrix in a slanted transposition cipher and reading it row by row.
Problem
Approach
Steps
Complexity
Input: The input consists of an encoded string `encodedText` and an integer `rows` representing the number of rows used to encode the original text.
Example: encodedText = 'hello wor ld ', rows = 3
Constraints:
• 0 <= encodedText.length <= 10^6
• encodedText consists of lowercase English letters and spaces
• encodedText is a valid encoding of some originalText that does not have trailing spaces
• 1 <= rows <= 1000
Output: Return the decoded original text as a string.
Example: Output: 'helloworld'
Constraints:
Goal: To decode the encoded text back to its original form using the given number of rows.
Steps:
• Calculate the number of columns in the matrix from the length of the encoded text and the number of rows.
• Reconstruct the matrix by placing the characters of the encoded text according to the slanted transposition cipher pattern.
• Read the matrix diagonally and reconstruct the original text.
Goal: The constraints specify the valid range for the input size and number of rows.
Steps:
• 0 <= encodedText.length <= 10^6
• 1 <= rows <= 1000
• encodedText consists of lowercase English letters and spaces
• encodedText is a valid encoding with no trailing spaces
Assumptions:
• The encoded text does not contain trailing spaces.
• Input: Example 1
• Explanation: In this example, the encoded string is 'hello wor ld ' with 3 rows. By filling the matrix diagonally, we can recover the original string 'helloworld'.
• Input: Example 2
• Explanation: For the input 'abc defghi' and 3 rows, the matrix looks like: 'abc', 'def', 'ghi'. Reading diagonally, we recover 'abcdefghi'.
Approach: The solution involves reconstructing the matrix from the encoded string and reading it diagonally to recover the original string.
Observations:
• We need to reverse the encoding process by filling a matrix with the encoded text and reading it diagonally.
• The problem can be solved by calculating the matrix dimensions and then reversing the transposition cipher to get the original string.
Steps:
• Determine the number of columns based on the encoded string length and rows.
• Fill the matrix following the slanted transposition cipher pattern.
• Read the matrix diagonally to recover the original text.
Empty Inputs:
• If the encoded text is empty, the original text is also empty.
Large Inputs:
• Ensure that the algorithm works efficiently for the maximum input size (encodedText.length up to 10^6).
Special Values:
• When there is only one row, the encoded text is the same as the original text.
Constraints:
• The algorithm must handle the given constraints and ensure that the reconstructed text is valid.
string decodeCiphertext(string et, int rows) {
cout << et.size() << " " << rows;
string res = "";
int col = et.size() / rows;
int idx = 0;
string space = "";
// cout << col << " " << rows << " ";
while(idx < col - rows + 2) {
for(int i = 0; i < rows && idx + i + i * col < et.size(); i++) {
if(et[idx + i + i * col] == ' ')
space += ' ';
else {
res += (space + et[idx + i + i * col]);
space = "";
}
}
idx++;
}
return res;
}
1 : Function Declaration
string decodeCiphertext(string et, int rows) {
This is the function signature for `decodeCiphertext`, which accepts a string `et` (the encoded text) and an integer `rows` (the number of rows in the cipher matrix). It returns the decoded string.
2 : Variable Initialization
string res = "";
This line initializes an empty string `res`, which will hold the decoded result.
3 : Column Calculation
int col = et.size() / rows;
This line calculates the number of columns in the cipher matrix by dividing the size of the encoded string `et` by the number of rows.
4 : Index Initialization
int idx = 0;
This line initializes an index `idx` to track the position of the current character being processed in the cipher.
5 : Space Initialization
string space = "";
This initializes a `space` string to hold spaces that are encountered during decoding, ensuring that they are correctly placed in the result.
6 : While Loop
while(idx < col - rows + 2) {
This `while` loop iterates over the columns and adjusts the `idx` value to ensure we process the matrix correctly for decoding.
7 : For Loop
for(int i = 0; i < rows && idx + i + i * col < et.size(); i++) {
This nested `for` loop iterates through the rows and checks if the current character index falls within the bounds of the encoded string.
8 : Space Check
if(et[idx + i + i * col] == ' ')
This checks if the current character in the encoded string is a space.
9 : Space Accumulation
space += ' ';
If the character is a space, it is added to the `space` string to maintain the spaces in the decoded message.
10 : Character Processing
else {
If the character is not a space, it is added to the decoded result string `res`.
11 : Result Update
res += (space + et[idx + i + i * col]);
This line appends the current character (along with any accumulated spaces) to the result string `res`.
12 : Space Reset
space = "";
This resets the `space` string after processing a non-space character.
13 : Index Increment
idx++;
This increments the `idx` value to move to the next character in the encoded string.
14 : Return Statement
return res;
This line returns the decoded string stored in `res`.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in terms of the size of the input string, as we process each character exactly once.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is linear in terms of the size of the input string, as we store the matrix and the result.
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