What is the time complexity of Decode Ways?

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

What data structures are used to solve Decode Ways?

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

What are the key insights for solving Decode Ways?

Insight 1: Dynamic Programming is suitable because the number of ways to decode a string depends on the number of ways to decode its prefixes. Insight 2: The current number of ways to decode can be determined using only the previous one or two values, suggesting a space-optimized DP solution.

Which companies ask LeetCode 91. Decode Ways in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Commvault and 16 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Decode Ways - LeetCode 91 Solution

Decode Ways - Complete Solution Guide

Decode Ways is LeetCode problem 91, 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

You have intercepted a secret message encoded as a string of numbers. The message is decoded via the following mapping: "1" -> 'A' "2" -> 'B' ... "25" -> 'Y' "26" -> 'Z' However, while decoding the message, you realize that there are many different ways you can decode the message because some codes are contained in other codes ( "2" and "5" vs "25" ). For example, "11106" can be decoded into: "AAJF" with the grouping (1, 1, 10, 6) "KJF" with the grouping (11, 10, 6) The grouping (1, 11, 06) is i

Detailed Explanation

The "Decode Ways" problem presents a scenario where a string of digits needs to be decoded into a sequence of letters, based on a given mapping (1 -> 'A', 2 -> 'B', ..., 26 -> 'Z'). The goal is to determine the number of possible ways to decode the input string. A single digit can represent one letter, or two digits can represent one letter (if the combined number is between 10 and 26, inclusive). The problem statement also specifies some key constraints: the string only contains digits, and leading zeros are invalid when interpreting two-digit combinations (e.g., '06' is invalid). The output should be the total number of valid decoding possibilities.

Solution Approach

The provided solutions use a dynamic programming approach with space optimization. Instead of storing the number of ways to decode each prefix of the string in an array, the solutions maintain only two variables: `one_back` and `two_back`. `one_back` stores the number of ways to decode the substring up to the previous character, and `two_back` stores the number of ways to decode the substring up to two characters before the current one. The algorithm iterates through the string, updating these variables based on whether the current character can be decoded as a single-digit number (1-9) or whether the combination of the previous and current characters can be decoded as a two-digit number (10-26). The final value of `one_back` represents the total number of ways to decode the entire string.

Step-by-Step Algorithm

Step 1: Handle the base cases. If the first character is '0', there are no valid decodings, so return 0. Initialize `one_back` to 1, representing one way to decode an empty string or a single-digit string from 1 to 9. Initialize `two_back` to 1, also representing the base case of an empty string.
Step 2: Iterate through the string from the second character (index 1) to the end.
Step 3: In each iteration, initialize `current` to 0. `current` will store the number of ways to decode the string up to the current character.
Step 4: Check if the current character is not '0'. If it's not '0', it can be decoded as a single-digit number (1-9), so add `one_back` to `current`. This means we can extend all previous decodings by adding the current digit as a separate letter.
Step 5: Check if the previous character is '1' or if the previous character is '2' and the current character is less than or equal to '6'. If this condition is true, the previous and current characters can be combined to form a two-digit number between 10 and 26, inclusive. In this case, add `two_back` to `current`. This means we can extend all decodings from two characters ago by decoding the last two characters together.
Step 6: Update `two_back` to `one_back` and `one_back` to `current`. This prepares for the next iteration by shifting the previous decoding counts.
Step 7: After the loop finishes, return `one_back`, which now holds the total number of ways to decode the entire string.

Key Insights

Insight 1: Dynamic Programming is suitable because the number of ways to decode a string depends on the number of ways to decode its prefixes.
Insight 2: The current number of ways to decode can be determined using only the previous one or two values, suggesting a space-optimized DP solution.
Insight 3: Handling the '0' case is crucial, as a single '0' is not a valid decoding, and a '0' following a '1' or '2' affects the decoding count.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Commvault, Flipkart, Goldman Sachs, Graviton, J.P. Morgan, Lyft, Meta, Microsoft, Moengage, Morgan Stanley, Oracle, Salesforce, Snap, TikTok, Uber, Walmart Labs, Zoho.

Decode Ways | LeetCode 91

Medium

String Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numDecodings(String s) {
        if (s.charAt(0) == '0') {
            return 0;
        }

        // Using O(1) space dynamic programming
        // two_back represents the number of ways to decode s[:i-2]
        // one_back represents the number of ways to decode s[:i-1]
        int two_back = 1;
        int one_back = 1;

        for (int i = 1; i < s.length(); i++) {
            int current = 0;

            // Case 1: Decode s[i] as a single-digit number (1-9)
            if (s.charAt(i) != '0') {
                current = one_back;
            }

            // Case 2: Decode s[i-1:i+1] as a two-digit number (10-26)
            if (s.charAt(i - 1) == '1' || (s.charAt(i - 1) == '2' && s.charAt(i) <= '6')) {
                current += two_back;
            }

            two_back = one_back;
            one_back = current;
        }

        return one_back;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 91: Decode Ways - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

AdobeAmazonAppleBloombergCommvaultFlipkartGoldman SachsGraviton

Stats

Difficulty:Medium

Languages:4

Companies:21

Problem #:91

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

7 steps

1
Step 1: Handle the base cases. If the first character is '0', there are no valid decodings, so return 0. Initialize `one_back` to 1, representing one way to decode an empty string or a single-digit string from 1 to 9. Initialize `two_back` to 1, also representing the base case of an empty string.
2
Step 2: Iterate through the string from the second character (index 1) to the end.
3
Step 3: In each iteration, initialize `current` to 0. `current` will store the number of ways to decode the string up to the current character.
4
Step 4: Check if the current character is not '0'. If it's not '0', it can be decoded as a single-digit number (1-9), so add `one_back` to `current`. This means we can extend all previous decodings by adding the current digit as a separate letter.
5
Step 5: Check if the previous character is '1' or if the previous character is '2' and the current character is less than or equal to '6'. If this condition is true, the previous and current characters can be combined to form a two-digit number between 10 and 26, inclusive. In this case, add `two_back` to `current`. This means we can extend all decodings from two characters ago by decoding the last two characters together.
6
Step 6: Update `two_back` to `one_back` and `one_back` to `current`. This prepares for the next iteration by shifting the previous decoding counts.
7
Step 7: After the loop finishes, return `one_back`, which now holds the total number of ways to decode the entire 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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm processes each character in the string exactly once. The space complexity is constant because the algorithm's memory usage doesn't depend on the size of the input; it only uses a fixed number of variables.

📖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

Dynamic Programming with Array

This approach uses a DP array `dp` of size `n+1`, where `dp[i]` represents the number of ways to decode the first `i` characters of the string. The base case is `dp[0] = 1` (empty string has one way to decode). The recurrence relation is `dp[i] = dp[i-1] + dp[i-2]` if the last one or two characters form a valid decoding.

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

Trade-offs: This approach is easier to understand initially, but it uses O(n) space compared to the O(1) space of the provided solution. It is less efficient in terms of memory usage, especially for very long strings.

Recursion with Memoization

A recursive function can be defined to explore all possible decoding paths. Memoization can be used to store the results of subproblems and avoid redundant computations.

Time: O(n)Space: O(n) due to the memoization table and recursion stack.

Trade-offs: Recursion with memoization is conceptually similar to dynamic programming but can be less efficient due to function call overhead. The space complexity is also O(n) due to the recursion stack and memoization table, making the iterative DP approach more favorable.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the number of ways to decode a string depends on the number of ways to decode its prefixes.
2Insight 2: The current number of ways to decode can be determined using only the previous one or two values, suggesting a space-optimized DP solution.
3Insight 3: Handling the '0' case is crucial, as a single '0' is not a valid decoding, and a '0' following a '1' or '2' affects the decoding count.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Input starts with '0'. The solution correctly handles this by returning 0 immediately, as there are no valid decodings if the first character is '0'.
Edge case 2: Two consecutive digits that form a number greater than 26 (e.g., '27'). The solution handles this correctly because only the single digit is considered, and the two-digit combination is ignored. The condition `s[i-1] == '1' or (s[i-1] == '2' and s[i] <= '6')` ensures that invalid two-digit combinations are not counted.
Edge case 3: A '0' that is not preceded by a '1' or '2'. In this case, `current` will not be updated by `two_back`, ensuring that the number of decoding ways remains unaffected. For example, in '301', '30' is an invalid sequence and it will result in `current = 0` when considering '0', resulting in a sequence break.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the '0' case correctly. If a '0' appears as a single digit (not part of a two-digit combination), it's an invalid decoding and should result in a 0 count.

✗Common mistake 2: Incorrectly calculating the number of ways for two-digit combinations. It's important to ensure that the two-digit number is within the range of 10 to 26, inclusive.

✗Common mistake 3: Forgetting the base case when the string is empty or consists of a single valid digit.

✗Common mistake 4: Using integer arithmetic that could overflow for extremely long strings. While the problem states the answer fits in a 32-bit integer, it's always good to be aware of potential overflow issues for problems involving counting.

✅

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 compression algorithms. Some compression techniques involve encoding data using variable-length codes, similar to the decoding problem. The number of possible decodings can represent the ambiguity in the compressed data.

🏦Real-world scenario 2: Cryptography and secure communication. Encoded messages can sometimes be interpreted in multiple ways, and understanding the possible decodings can be relevant in analyzing the security of the encoding scheme. However, this problem simplifies real-world cryptographic decoding significantly.

🎮Real-world scenario 3: Bioinformatics. In bioinformatics, sequences of DNA or RNA can be represented as strings. Determining the number of possible secondary structures of these sequences can involve similar dynamic programming techniques.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

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: August 31, 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

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

@AlgoMasterJun 18

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

@InterviewProJun 11

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 4

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 Ways"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?