What is the time complexity of Decode the Slanted Ciphertext?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Decode the Slanted Ciphertext?

This problem uses String, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Decode the Slanted Ciphertext?

Insight 1: The original text is placed diagonally in the matrix, and the encoded text is formed by reading it row-wise. The key is to reverse this process. Insight 2: The number of columns in the matrix is determined by `cols = len(encodedText) / rows`. This is crucial for determining the diagonal jumps in the encoded text.

Which companies ask LeetCode 2075. Decode the Slanted Ciphertext in interviews?

This problem is frequently asked at Grammarly.

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Decode the Slanted Ciphertext - LeetCode 2075 Solution

Decode the Slanted Ciphertext - Complete Solution Guide

Decode the Slanted Ciphertext is LeetCode problem 2075, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to decode a string that was encoded using a slanted transposition cipher. The original text is written diagonally into a matrix of `rows` rows. Empty cells in the matrix are padded with spaces. The encoded text is formed by reading the matrix row by row. Given the `encodedText` and the number of `rows`, we need to reconstruct the original text without any trailing spaces.

Solution Approach

The solution reverses the encoding process. It first calculates the number of columns in the matrix. Then, it iterates through the columns of the imagined matrix. For each column index `i`, it starts at that index in the `encodedText` and iteratively appends the characters located diagonally (by incrementing by `cols + 1`) to the result string. Finally, trailing spaces are removed from the result.

Step-by-Step Algorithm

Step 1: Check if the encodedText is empty. If so, return an empty string.
Step 2: Calculate the number of columns: `cols = len(encodedText) / rows`.
Step 3: Handle the case where the number of rows is 1. If rows == 1, the encodedText is the same as the original text so just return it.
Step 4: Initialize an empty string `res` to store the decoded text.
Step 5: Iterate from `i = 0` to `cols - 1` (columns).
Step 6: Inside the outer loop, initialize `curr = i`.
Step 7: Start a while loop that continues as long as `curr < len(encodedText)`.
Step 8: Append the character at `encodedText[curr]` to the `res` string.
Step 9: Update `curr` by adding `cols + 1` to move diagonally.
Step 10: After the loops complete, remove any trailing spaces from the `res` string.
Step 11: Return the `res` string.

Key Insights

Insight 1: The original text is placed diagonally in the matrix, and the encoded text is formed by reading it row-wise. The key is to reverse this process.
Insight 2: The number of columns in the matrix is determined by `cols = len(encodedText) / rows`. This is crucial for determining the diagonal jumps in the encoded text.
Insight 3: The diagonal elements are accessed with an increment of `cols + 1` in the encoded string.
Insight 4: Trailing spaces must be removed from the decoded string.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Simulation.

Companies

Asked at: Grammarly.

Decode the Slanted Ciphertext | LeetCode 2075

Medium

String Simulation

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public String decodeCiphertext(String encodedText, int rows) {
        int n = encodedText.length();
        if (n == 0) {
            return "";
        }

        int cols = n / rows;
        if (cols == 0 && n > 0) { // a single partial column
            return encodedText.stripTrailing();
        }
        
        // If rows is 1, the text is not transformed.
        if (rows == 1) {
            return encodedText;
        }
        
        StringBuilder res = new StringBuilder();
        for (int i = 0; i < cols; i++) {
            int curr = i;
            while (curr < n) {
                res.append(encodedText.charAt(curr));
                curr += cols + 1;
            }
        }
        
        return res.toString().stripTrailing();
    }
}

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Problem Info

Description

Problem 2075: Decode the Slanted Ciphertext - Detailed explanation and solution approaches.

Topics

String Simulation

Companies

Grammarly

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2075

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Check if the encodedText is empty. If so, return an empty string.
2
Step 2: Calculate the number of columns: `cols = len(encodedText) / rows`.
3
Step 3: Handle the case where the number of rows is 1. If rows == 1, the encodedText is the same as the original text so just return it.
4
Step 4: Initialize an empty string `res` to store the decoded text.
5
Step 5: Iterate from `i = 0` to `cols - 1` (columns).
6
Step 6: Inside the outer loop, initialize `curr = i`.
7
Step 7: Start a while loop that continues as long as `curr < len(encodedText)`.
8
Step 8: Append the character at `encodedText[curr]` to the `res` string.
9
Step 9: Update `curr` by adding `cols + 1` to move diagonally.
10
Step 10: After the loops complete, remove any trailing spaces from the `res` string.
11
Step 11: Return the `res` string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates `cols` times, and the inner loop iterates approximately `rows` times. Since `cols * rows = n`, the overall complexity is O(n). The space complexity is O(n) because we create a new string `res` that can potentially store all characters from the encoded string.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Matrix Reconstruction

Instead of directly calculating the indices, one can reconstruct the matrix as a 2D array, fill it with the characters from the encoded text, and then read the original text diagonally. This approach is more intuitive, especially for beginners, but may have a higher space complexity.

Time: O(n)Space: O(n)

Trade-offs: Easier to understand, but potentially higher space complexity if the matrix is explicitly created and stored.

✨

Key Insights

1Insight 1: The original text is placed diagonally in the matrix, and the encoded text is formed by reading it row-wise. The key is to reverse this process.
2Insight 2: The number of columns in the matrix is determined by `cols = len(encodedText) / rows`. This is crucial for determining the diagonal jumps in the encoded text.
3Insight 3: The diagonal elements are accessed with an increment of `cols + 1` in the encoded string.
4Insight 4: Trailing spaces must be removed from the decoded string.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty input string. The code handles this by returning an empty string.
Edge case 2: Number of rows is 1. In this case, the encoded text is the same as the original text. The code handles this by directly returning the encoded text.
Edge case 3: Input where the number of rows divides the length of encodedText with remainder. This produces additional space characters at the end that need to be removed.
Edge case 4: Input with only spaces. The trailing space removal must function correctly to return the empty string.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not removing trailing spaces. The problem statement specifies that the original text does not have trailing spaces. Forgetting to trim these spaces will lead to incorrect results.

✗Common mistake 2: Incorrectly calculating the number of columns. It's crucial to use `cols = len(encodedText) / rows`.

✗Common mistake 3: Off-by-one errors in index calculations within the loops. The increment `cols + 1` must be used to correctly traverse the diagonals.

✗Common mistake 4: Not handling the edge case where rows = 1. The algorithm for decoding does not work in this situation.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data encryption and security. Transposition ciphers, although simple, demonstrate the basic principles of encoding and decoding information.

🏦Real-world scenario 2: Image processing. Similar concepts of matrix manipulation and data rearrangement are used in image processing tasks such as image encryption and watermarking.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Decode the Slanted Ciphertext"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?