What is the time complexity of Number of Unique Good Subsequences?

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

What data structures are used to solve Number of Unique Good Subsequences?

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

What are the key insights for solving Number of Unique Good Subsequences?

Insight 1: The core idea is to use dynamic programming to track the number of subsequences ending with '0' and '1'. This helps avoid generating and storing all possible subsequences, which would be inefficient. Insight 2: We only need to keep track of the number of subsequences ending with '0' and '1' at each step because adding a '1' to subsequences ending with '0' or '1' always results in a valid subsequence, and similarly, adding a '0' to existing subsequences ending with '0' or '1' results in valid ones. The 'unique' constraint is implicitly handled.

Which companies ask LeetCode 1987. Number of Unique Good Subsequences in interviews?

This problem is frequently asked at Oracle.

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Number of Unique Good Subsequences - LeetCode 1987 Solution

Number of Unique Good Subsequences - Complete Solution Guide

Number of Unique Good Subsequences is LeetCode problem 1987, 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

You are given a binary string binary . A subsequence of binary is considered good if it is not empty and has no leading zeros (with the exception of "0" ). Find the number of unique good subsequences of binary . For example, if binary = "001" , then all the good subsequences are ["0", "0", "1"] , so the unique good subsequences are "0" and "1" . Note that subsequences "00" , "01" , and "001" are not good because they have leading zeros. Return the number of unique good subsequences of binary . S

Detailed Explanation

The problem asks us to find the number of unique good subsequences of a given binary string. A subsequence is derived from a string by deleting some or no characters without changing the order of the remaining characters. A 'good' subsequence must not be empty and should not have leading zeros, with the exception of the subsequence '0' itself. The goal is to count the number of *unique* such good subsequences, modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming with two variables: `ends_with_0` and `ends_with_1`. These variables keep track of the number of unique good subsequences ending with '0' and '1', respectively. The algorithm iterates through the binary string, updating these variables based on the current digit. If the digit is '1', the number of subsequences ending with '1' becomes the sum of the previous number of subsequences ending with '0', '1', and the '1' itself (representing a new subsequence consisting only of '1'). If the digit is '0', the number of subsequences ending with '0' becomes the sum of the previous number of subsequences ending with '0' and '1'. Finally, a boolean variable `has_zero` is used to indicate if the string contains a zero. If so, we add 1 to the final count to account for the single '0' subsequence. The modulo operation is applied after each calculation to prevent integer overflow.

Step-by-Step Algorithm

Step 1: Initialize `ends_with_0` and `ends_with_1` to 0. These represent the number of unique good subsequences ending with '0' and '1' respectively.
Step 2: Initialize `has_zero` to 0 (or false). This flag will be set to 1 (or true) if the input string contains at least one '0'.
Step 3: Iterate through the input binary string, character by character.
Step 4: If the current digit is '1', update `ends_with_1` to `(ends_with_0 + ends_with_1 + 1) % MOD`. This accounts for adding '1' to all existing subsequences (ending in '0' or '1') and adding the single '1' subsequence.
Step 5: If the current digit is '0', update `ends_with_0` to `(ends_with_0 + ends_with_1) % MOD`. This accounts for adding '0' to all existing subsequences (ending in '0' or '1'). Also, set `has_zero` to 1.
Step 6: After iterating through the entire string, the final answer is `(ends_with_0 + ends_with_1 + has_zero) % MOD`. This is the sum of all subsequences ending with '0' and '1', plus 1 if the string contained a '0' (to account for the lone '0' subsequence).

Key Insights

Insight 1: The core idea is to use dynamic programming to track the number of subsequences ending with '0' and '1'. This helps avoid generating and storing all possible subsequences, which would be inefficient.
Insight 2: We only need to keep track of the number of subsequences ending with '0' and '1' at each step because adding a '1' to subsequences ending with '0' or '1' always results in a valid subsequence, and similarly, adding a '0' to existing subsequences ending with '0' or '1' results in valid ones. The 'unique' constraint is implicitly handled.
Insight 3: The edge case of the single '0' subsequence needs special consideration. We use a `has_zero` flag to track whether the original string contains a zero and add it to the final count if it does.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Oracle.

Number of Unique Good Subsequences | LeetCode 1987

Hard

String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int numberOfUniqueGoodSubsequences(String binary) {
        int MOD = (int)1e9 + 7;
        
        int ends_with_0 = 0;
        int ends_with_1 = 0;
        int has_zero = 0;
        
        for (int i = 0; i < binary.length(); i++) {
            char digit = binary.charAt(i);
            if (digit == '1') {
                ends_with_1 = (ends_with_0 + ends_with_1 + 1) % MOD;
            } else {
                ends_with_0 = (ends_with_0 + ends_with_1) % MOD;
                has_zero = 1;
            }
        }
        
        return (ends_with_0 + ends_with_1 + has_zero) % MOD;
    }
}

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Problem Info

Description

Problem 1987: Number of Unique Good Subsequences - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

Oracle

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1987

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

6 steps

1
Step 1: Initialize `ends_with_0` and `ends_with_1` to 0. These represent the number of unique good subsequences ending with '0' and '1' respectively.
2
Step 2: Initialize `has_zero` to 0 (or false). This flag will be set to 1 (or true) if the input string contains at least one '0'.
3
Step 3: Iterate through the input binary string, character by character.
4
Step 4: If the current digit is '1', update `ends_with_1` to `(ends_with_0 + ends_with_1 + 1) % MOD`. This accounts for adding '1' to all existing subsequences (ending in '0' or '1') and adding the single '1' subsequence.
5
Step 5: If the current digit is '0', update `ends_with_0` to `(ends_with_0 + ends_with_1) % MOD`. This accounts for adding '0' to all existing subsequences (ending in '0' or '1'). Also, set `has_zero` to 1.
6
Step 6: After iterating through the entire string, the final answer is `(ends_with_0 + ends_with_1 + has_zero) % MOD`. This is the sum of all subsequences ending with '0' and '1', plus 1 if the string contained a '0' (to account for the lone '0' subsequence).

💡

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 code iterates through each character of the input string exactly once. The space complexity is constant because the amount of memory used does not depend on the size of the input 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

Recursive Approach with Memoization

A recursive approach could be used to generate all subsequences and then filter them based on the 'good' criteria. Memoization could be used to avoid redundant computations of subsequences, but this would likely be much slower and more memory intensive.

Time: O(2^n) in the worst case without memoization, O(n^2) with memoization.Space: O(n) for recursion stack, potentially more with memoization.

Trade-offs: This approach is generally less efficient than the iterative DP solution due to overhead from recursion and the complexity of managing and storing subsequences.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming to track the number of subsequences ending with '0' and '1'. This helps avoid generating and storing all possible subsequences, which would be inefficient.
2Insight 2: We only need to keep track of the number of subsequences ending with '0' and '1' at each step because adding a '1' to subsequences ending with '0' or '1' always results in a valid subsequence, and similarly, adding a '0' to existing subsequences ending with '0' or '1' results in valid ones. The 'unique' constraint is implicitly handled.
3Insight 3: The edge case of the single '0' subsequence needs special consideration. We use a `has_zero` flag to track whether the original string contains a zero and add it to the final count if it does.

📚

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 string is "0". The solution correctly returns 1, because only the subsequence "0" is considered good.
Edge case 2: Input string is "1". The solution correctly returns 1, because only the subsequence "1" is considered good.
Edge case 3: Input string is "111". The solution correctly returns 2, representing unique subsequences "1" and "11" and "111".
Edge case 4: Input string contains very long sequences of 0s and 1s. The modulo operation ensures that the result does not overflow.
Edge case 5: Input string is "0000". The solution correctly returns 1 because there is only one unique good subsequence which is "0".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to take the modulo at each step, which can lead to integer overflow and incorrect results.

✗Common mistake 2: Not handling the case where the string contains only '0's correctly. The 'has_zero' flag is crucial for this.

✗Common mistake 3: Trying to generate all subsequences explicitly. This leads to a time complexity issue for larger input strings. The problem requires you to find UNIQUE subsequences, meaning storing each of them isn't necessary, just the COUNTS of them.

✗Common mistake 4: Incorrectly identifying what constitutes a good subsequence (i.e., forgetting that '0' is allowed, or allowing leading zeros).

✅

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: This problem showcases a simple dynamic programming approach applicable in sequence analysis where the goal is to count unique patterns or combinations of elements in a sequence while adhering to specific constraints (e.g., network packet analysis, DNA sequence analysis).

🏦Real-world scenario 2: Counting distinct valid paths in a directed graph, where some nodes have restrictions or costs associated with traversing them. The 'good subsequence' analogy corresponds to a valid path adhering to these restrictions.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: June 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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 "Number of Unique Good Subsequences"?

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.

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