What is the time complexity of Second Largest Digit in a String?

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

What data structures are used to solve Second Largest Digit in a String?

This problem uses Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Second Largest Digit in a String?

Insight 1: Using a set to store unique digits avoids duplicate counting and simplifies finding the second largest. Insight 2: Sorting the set of digits allows for easy retrieval of the second largest element.

Which companies ask LeetCode 1796. Second Largest Digit in a String in interviews?

This problem is frequently asked at Softwire.

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Second Largest Digit in a String - LeetCode 1796 Solution

Q: How to solve Second Largest Digit in a String?

The provided solutions use different approaches but achieve the same result. The Python solution utilizes a set to store unique digits, then sorts them in descending order to easily access the second largest. The Java, C++, and C solutions iteratively track the largest and second largest digits encountered, updating them as needed. Both approaches efficiently handle the problem.

Second Largest Digit in a String - Complete Solution Guide

Second Largest Digit in a String is LeetCode problem 1796, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an alphanumeric string s , return the second largest numerical digit that appears in s , or -1 if it does not exist . An alphanumeric string is a string consisting of lowercase English letters and digits. Example 1: Input: s = "dfa12321afd" Output: 2 Explanation: The digits that appear in s are [1, 2, 3]. The second largest digit is 2. Example 2: Input: s = "abc1111" Output: -1 Explanation: The digits that appear in s are [1]. There is no second largest digit. Constraints: 1 <= s.length <=

Detailed Explanation

The problem asks you to find the second largest digit in a given alphanumeric string. An alphanumeric string contains only lowercase English letters and digits. The function should return the second largest digit found in the string. If there is only one unique digit or no digits at all, it should return -1.

Solution Approach

The provided solutions use different approaches but achieve the same result. The Python solution utilizes a set to store unique digits, then sorts them in descending order to easily access the second largest. The Java, C++, and C solutions iteratively track the largest and second largest digits encountered, updating them as needed. Both approaches efficiently handle the problem.

Step-by-Step Algorithm

Step 1: Iterate through the input string `s`.
Step 2: For each character, check if it's a digit using `isdigit()` (or equivalent).
Step 3: If it's a digit, convert it to an integer and either add it to a set (Python solution) or update the `largest` and `secondLargest` variables (Java, C++, C solutions).
Step 4: (Python Solution) After iterating, check if the set has at least two elements. If not, return -1. Otherwise, sort the set and return the second element.
Step 4: (Java, C++, C Solutions) After iterating, return the value of `secondLargest`.

Key Insights

Insight 1: Using a set to store unique digits avoids duplicate counting and simplifies finding the second largest.
Insight 2: Sorting the set of digits allows for easy retrieval of the second largest element.
Insight 3: Handling edge cases where there are fewer than two unique digits is crucial (returning -1).

Complexity Analysis

Time Complexity: O(n + klogk)

Space Complexity: O(k)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Softwire.

Second Largest Digit in a String | LeetCode 1796

Easy

Hash Table String

Time: O(n + klogk)Space: O(k)

Solution Code

Available in:

class Solution {
    public int secondHighest(String s) {
        int largest = -1;
        int secondLargest = -1;

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                int digit = c - '0';
                if (digit > largest) {
                    secondLargest = largest;
                    largest = digit;
                } else if (digit < largest && digit > secondLargest) {
                    secondLargest = digit;
                }
            }
        }

        return secondLargest;
    }
}

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Problem Info

Description

Problem 1796: Second Largest Digit in a String - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

Softwire

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1796

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

5 steps

1
Step 1: Iterate through the input string `s`.
2
Step 2: For each character, check if it's a digit using `isdigit()` (or equivalent).
3
Step 3: If it's a digit, convert it to an integer and either add it to a set (Python solution) or update the `largest` and `secondLargest` variables (Java, C++, C solutions).
4
Step 4: (Python Solution) After iterating, check if the set has at least two elements. If not, return -1. Otherwise, sort the set and return the second element.
5
Step 4: (Java, C++, C Solutions) After iterating, return the value of `secondLargest`.

💡

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 + klogk)

💾

Space Complexity

O(k)

📝

Why?

Python's time complexity comes from iterating through the string (O(n)) and sorting the set of unique digits (O(k log k)), where k is the number of unique digits. Java, C++, and C solutions only iterate through the string once (O(n)) and use constant extra space. The space complexity in Python is determined by the size of the set of unique digits, which can be at most 10. Java, C++, and C have constant space complexity because they only use a few integer 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

Using an array to count digit occurrences

Create an array of size 10 to store the count of each digit (0-9). Iterate through the string, incrementing the count for each digit encountered. Then, find the second largest index with a count greater than 0.

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

Trade-offs: This approach has the same time complexity as the optimal Java/C++/C solutions but might be slightly less readable. It's a good alternative if you prioritize constant space complexity and are comfortable with array-based counting.

✨

Key Insights

1Insight 1: Using a set to store unique digits avoids duplicate counting and simplifies finding the second largest.
2Insight 2: Sorting the set of digits allows for easy retrieval of the second largest element.
3Insight 3: Handling edge cases where there are fewer than two unique digits is crucial (returning -1).

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: The input string contains no digits. The solutions correctly handle this by returning -1.
Edge case 2: The input string contains only one unique digit. The solutions correctly handle this by returning -1.
Edge case 3: The input string contains digits larger than 9. The solutions correctly handle this because they only consider digits 0-9.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case where there is only one unique digit or no digits at all.

✗Common mistake 2: Inefficiently searching for the second largest digit without sorting or using a more optimized tracking method (e.g., nested loops).

✅

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 validation – checking if a user input contains a specific number of unique digits within a certain range.

🏦Real-world scenario 2: Log analysis – identifying the second most frequent digit in a log file containing timestamps or numerical data to potentially spot anomalies.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: November 24, 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.

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Second Largest Digit in a String"?

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

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