What is the time complexity of Decode Ways II?

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

What data structures are used to solve Decode Ways II?

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

What are the key insights for solving Decode Ways II?

Insight 1: Dynamic programming is the best approach because the number of ways to decode a string depends on the number of ways to decode its prefixes. This allows us to build up the solution iteratively. Insight 2: We need to handle the '*' character carefully, as it can represent multiple digits. We must correctly calculate the number of possibilities that the '*' contributes in both single-digit and two-digit decoding.

Which companies ask LeetCode 639. Decode Ways II in interviews?

This problem is frequently asked at PhonePe.

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Decode Ways II - LeetCode 639 Solution

Decode Ways II - Complete Solution Guide

Decode Ways II is LeetCode problem 639, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A message containing letters from A-Z can be encoded into numbers using the following mapping: 'A' -> "1" 'B' -> "2" ... 'Z' -> "26" To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into: "AAJF" with the grouping (1 1 10 6) "KJF" with the grouping (11 10 6) Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is

Detailed Explanation

The problem asks us to find the number of ways to decode a string consisting of digits and '*' characters, where each digit maps to a letter A-Z (1->A, 2->B, ..., 26->Z). The '*' character can represent any digit from 1 to 9. We need to return the number of possible decodings modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming with constant space optimization. We maintain two variables, `e0` and `e1`, representing the number of ways to decode the string up to the (i-1)th and ith positions respectively. For each position i, we calculate the number of ways to decode the string up to that position (`e2`) based on the characters at positions (i-1) and i. Then, we update `e0` and `e1` for the next iteration.

Step-by-Step Algorithm

Step 1: Initialize `e0` to 1, which represents the base case of an empty string having one way to decode.
Step 2: Handle the first character of the string. If it is '0', there are no ways to decode the string, so return 0. If it is '*', `e1` is 9 (since '*' can represent 1-9), otherwise `e1` is 1.
Step 3: Iterate through the string starting from the second character (index 1).
Step 4: For each character `c2` at index i, calculate `e2` based on `c2` and the previous character `c1` at index i-1.
Step 5: Calculate the number of ways to decode `c2` as a single character. If `c2` is '*', multiply the number of ways by 9 (as it represents digits 1-9). If `c2` is a digit greater than '0', the number of ways is the same as the previous single character case. Otherwise, it's zero.
Step 6: Calculate the number of ways to decode `c1` and `c2` together as a two-digit number. This requires handling several cases: - If `c1` is '1', and `c2` is '*', then we have 9 additional ways to decode (11, 12, ..., 19). - If `c1` is '1', and `c2` is a digit, we have one additional way to decode (if it valid). - If `c1` is '2', and `c2` is '*', then we have 6 additional ways to decode (21, 22, ..., 26). - If `c1` is '2', and `c2` is a digit between '0' and '6', we have one additional way to decode. - If `c1` is '*', and `c2` is '*', then we have 15 additional ways to decode (11-19, 21-26). This is because '*' can be '1' or '2' for the first digit and 1-9 or 1-6 respectively, for the second digit. - If `c1` is '*', and `c2` is a digit between '0' and '6', we have 2 additional ways to decode (1x, 2x). - If `c1` is '*', and `c2` is a digit greater than '6', we have 1 additional way to decode (1x).
Step 7: Update `e0` to `e1` and `e1` to `e2` for the next iteration.
Step 8: After iterating through the string, return `e1` as the final number of ways to decode the string, modulo 10^9 + 7.

Key Insights

Insight 1: Dynamic programming is the best approach because the number of ways to decode a string depends on the number of ways to decode its prefixes. This allows us to build up the solution iteratively.
Insight 2: We need to handle the '*' character carefully, as it can represent multiple digits. We must correctly calculate the number of possibilities that the '*' contributes in both single-digit and two-digit decoding.
Insight 3: The problem involves considering pairs of characters and single characters, requiring accurate case analysis based on the possible values of the '*' character and its position.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: PhonePe.

Decode Ways II | LeetCode 639

Hard

String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int numDecodings(String s) {
        int MOD = (int)1e9 + 7;
        
        long e0 = 1;
        
        if (s.charAt(0) == '0') {
            e0 = 0;
            long e1 = 0;
            return (int) e1;
        } else if (s.charAt(0) == '*') {
            long e1 = 9;
            e0 = 1;
            for (int i = 1; i < s.length(); i++) {
                char c1 = s.charAt(i-1);
                char c2 = s.charAt(i);

                long e2 = 0;
                
                if (c2 == '*') {
                    e2 = (9 * e1) % MOD;
                } else if (c2 > '0') {
                    e2 = e1;
                }
                
                if (c1 == '1') {
                    if (c2 == '*') {
                        e2 = (e2 + 9 * e0) % MOD;
                    } else {
                        e2 = (e2 + e0) % MOD;
                    }
                } else if (c1 == '2') {
                    if (c2 == '*') {
                        e2 = (e2 + 6 * e0) % MOD;
                    } else if ('0' <= c2 && c2 <= '6') {
                        e2 = (e2 + e0) % MOD;
                    }
                } else if (c1 == '*') {
                    if (c2 == '*') {
                        e2 = (e2 + 15 * e0) % MOD;
                    } else if ('0' <= c2 && c2 <= '6') {
                        e2 = (e2 + 2 * e0) % MOD;
                    } else {
                        e2 = (e2 + e0) % MOD;
                    }
                }
                
                e0 = e1;
                e1 = e2;
            }
            
            return (int) e1;
        } else {
            long e1 = 1;
            e0 = 1;
             for (int i = 1; i < s.length(); i++) {
                char c1 = s.charAt(i-1);
                char c2 = s.charAt(i);

                long e2 = 0;
                
                if (c2 == '*') {
                    e2 = (9 * e1) % MOD;
                } else if (c2 > '0') {
                    e2 = e1;
                }
                
                if (c1 == '1') {
                    if (c2 == '*') {
                        e2 = (e2 + 9 * e0) % MOD;
                    } else {
                        e2 = (e2 + e0) % MOD;
                    }
                } else if (c1 == '2') {
                    if (c2 == '*') {
                        e2 = (e2 + 6 * e0) % MOD;
                    } else if ('0' <= c2 && c2 <= '6') {
                        e2 = (e2 + e0) % MOD;
                    }
                } else if (c1 == '*') {
                    if (c2 == '*') {
                        e2 = (e2 + 15 * e0) % MOD;
                    } else if ('0' <= c2 && c2 <= '6') {
                        e2 = (e2 + 2 * e0) % MOD;
                    } else {
                        e2 = (e2 + e0) % MOD;
                    }
                }
                
                e0 = e1;
                e1 = e2;
            }
            
            return (int) e1;
        }

    }
}

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Problem Info

Description

Problem 639: Decode Ways II - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

PhonePe

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:639

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

8 steps

1
Step 1: Initialize `e0` to 1, which represents the base case of an empty string having one way to decode.
2
Step 2: Handle the first character of the string. If it is '0', there are no ways to decode the string, so return 0. If it is '*', `e1` is 9 (since '*' can represent 1-9), otherwise `e1` is 1.
3
Step 3: Iterate through the string starting from the second character (index 1).
4
Step 4: For each character `c2` at index i, calculate `e2` based on `c2` and the previous character `c1` at index i-1.
5
Step 5: Calculate the number of ways to decode `c2` as a single character. If `c2` is '*', multiply the number of ways by 9 (as it represents digits 1-9). If `c2` is a digit greater than '0', the number of ways is the same as the previous single character case. Otherwise, it's zero.
6
Step 6: Calculate the number of ways to decode `c1` and `c2` together as a two-digit number. This requires handling several cases: - If `c1` is '1', and `c2` is '*', then we have 9 additional ways to decode (11, 12, ..., 19). - If `c1` is '1', and `c2` is a digit, we have one additional way to decode (if it valid). - If `c1` is '2', and `c2` is '*', then we have 6 additional ways to decode (21, 22, ..., 26). - If `c1` is '2', and `c2` is a digit between '0' and '6', we have one additional way to decode. - If `c1` is '*', and `c2` is '*', then we have 15 additional ways to decode (11-19, 21-26). This is because '*' can be '1' or '2' for the first digit and 1-9 or 1-6 respectively, for the second digit. - If `c1` is '*', and `c2` is a digit between '0' and '6', we have 2 additional ways to decode (1x, 2x). - If `c1` is '*', and `c2` is a digit greater than '6', we have 1 additional way to decode (1x).
7
Step 7: Update `e0` to `e1` and `e1` to `e2` for the next iteration.
8
Step 8: After iterating through the string, return `e1` as the final number of ways to decode the string, modulo 10^9 + 7.

💡

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 solution processes each character of the string in a single pass. The space complexity is constant because only a fixed number of variables (`e0`, `e1`, `e2`) are used, irrespective of the input string's length.

📖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

Top-Down Dynamic Programming (Memoization)

A recursive approach with memoization to store the number of ways to decode substrings. This approach can be more intuitive for some people.

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

Trade-offs: This approach has a slightly higher space complexity due to the memoization table and recursion stack, but can be easier to understand for some.

✨

Key Insights

1Insight 1: Dynamic programming is the best approach because the number of ways to decode a string depends on the number of ways to decode its prefixes. This allows us to build up the solution iteratively.
2Insight 2: We need to handle the '*' character carefully, as it can represent multiple digits. We must correctly calculate the number of possibilities that the '*' contributes in both single-digit and two-digit decoding.
3Insight 3: The problem involves considering pairs of characters and single characters, requiring accurate case analysis based on the possible values of the '*' character and its position.

📚

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: Empty string. This is handled implicitly by the initialization of `e0` to 1, as the loop does not execute.
Edge case 2: String starting with '0'. The code explicitly checks for this case and returns 0, as '0' cannot be decoded as a single character.
Edge case 3: String containing consecutive '*' characters. The solution accurately calculates the combinations possible when '*' appears adjacent to other '*' or digit characters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the '*' character. Failing to account for all the possible digits that '*' can represent leads to an incorrect number of decodings.

✗Common mistake 2: Not applying the modulo operator (%) at each step. The problem requires returning the answer modulo 10^9 + 7 to prevent integer overflow. Forgetting this step leads to incorrect results for large inputs.

✗Common mistake 3: Incorrectly considering the edge cases where '0' appears in the string, particularly after a '1' or '2'. '0' cannot be decoded independently, so '10' and '20' are valid, but '0x' is not.

✅

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. Decoding encoded messages can be applied to compressing data where certain characters or strings are represented by shorter codes.

🏦Real-world scenario 2: Cryptography. Decoding secret messages involves similar techniques of mapping encoded symbols back to their original meaning. This problem shares aspects with basic encryption and decryption methods.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: September 8, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 II"?

For this hard-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?