What is the time complexity of Reverse Degree of a String?

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

What data structures are used to solve Reverse Degree of a String?

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

What are the key insights for solving Reverse Degree of a String?

Insight 1: The core calculation involves determining the reversed alphabet index for each character. This can be efficiently done using the ASCII values of the characters. Insight 2: The problem requires maintaining the 1-based index of each character within the string during iteration.

Which companies ask LeetCode 3498. Reverse Degree of a String in interviews?

This problem is frequently asked at Capgemini.

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Reverse Degree of a String - LeetCode 3498 Solution

Reverse Degree of a String - Complete Solution Guide

Reverse Degree of a String is LeetCode problem 3498, 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 a string s , calculate its reverse degree . The reverse degree is calculated as follows: For each character, multiply its position in the reversed alphabet ( 'a' = 26, 'b' = 25, ..., 'z' = 1) with its position in the string (1-indexed) . Sum these products for all characters in the string. Return the reverse degree of s . Example 1: Input: s = "abc" Output: 148 Explanation: Letter Index in Reversed Alphabet Index in String Product 'a' 26 1 26 'b' 25 2 50 'c' 24 3 72 The reversed degree is

Detailed Explanation

The problem asks us to calculate the 'reverse degree' of a given string 's'. This involves iterating through the string, and for each character, determining its position in the reversed alphabet (where 'a' is 26, 'b' is 25, and so on to 'z' which is 1). We then multiply this reversed alphabet position by the character's position in the string (starting from 1) and sum up these products across all characters in 's' to get the final reverse degree. The input is a string 's' consisting of lowercase English letters, and the output is an integer representing the reverse degree of 's'.

Solution Approach

The provided solutions iterate through the input string 's'. In each iteration, they calculate the 'reversed alphabet index' of the current character using its ASCII value and the ASCII value of 'a' or 'z'. Then, they multiply this index by the position of the character within the string (i+1, since the problem is 1-indexed) and add the result to a running total, `reverse_degree`. Finally, the accumulated sum (reverse degree) is returned.

Step-by-Step Algorithm

Step 1: Initialize a variable `reverse_degree` to 0. This variable will store the calculated reverse degree of the string.
Step 2: Iterate through the input string `s` using a loop. For each character in the string, get the index `i` of the character.
Step 3: Calculate the reverse alphabet index for the current character. This can be done by subtracting the ASCII value of 'a' from the ASCII value of the character and subtracting the result from 25, then adding 1. Alternatively, subtract the current character's ASCII value from 'z' and add 1.
Step 4: Multiply the reverse alphabet index by (i + 1), which represents the 1-based index of the character in the string.
Step 5: Add the result of the multiplication to the `reverse_degree` variable.
Step 6: After iterating through all characters in the string, return the final `reverse_degree` value.

Key Insights

Insight 1: The core calculation involves determining the reversed alphabet index for each character. This can be efficiently done using the ASCII values of the characters.
Insight 2: The problem requires maintaining the 1-based index of each character within the string during iteration.
Insight 3: The problem constraints (lowercase English letters, string length limit) are small enough that integer overflow and very large computations are not concerns.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Simulation.

Companies

Asked at: Capgemini.

Reverse Degree of a String | LeetCode 3498

Easy

String Simulation

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

Solution Code

Available in:

class Solution {
    public int reverseDegree(String s) {
        int reverseDegree = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            int alphabetPosition = 26 - (c - 'a');
            reverseDegree += alphabetPosition * (i + 1);
        }
        return reverseDegree;
    }
}

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Problem Info

Description

Problem 3498: Reverse Degree of a String - Detailed explanation and solution approaches.

Topics

String Simulation

Companies

Capgemini

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3498

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 a variable `reverse_degree` to 0. This variable will store the calculated reverse degree of the string.
2
Step 2: Iterate through the input string `s` using a loop. For each character in the string, get the index `i` of the character.
3
Step 3: Calculate the reverse alphabet index for the current character. This can be done by subtracting the ASCII value of 'a' from the ASCII value of the character and subtracting the result from 25, then adding 1. Alternatively, subtract the current character's ASCII value from 'z' and add 1.
4
Step 4: Multiply the reverse alphabet index by (i + 1), which represents the 1-based index of the character in the string.
5
Step 5: Add the result of the multiplication to the `reverse_degree` variable.
6
Step 6: After iterating through all characters in the string, return the final `reverse_degree` value.

💡

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 we process each character in the string exactly once. The space complexity is constant because we only use a fixed number of variables that do not scale with the input size.

📖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 List Comprehension (Python)

In Python, one could use list comprehension to calculate the products and then sum them up. This can be slightly more concise, but doesn't change the fundamental approach.

Time: O(n)Space: O(n). The list comprehension creates a temporary list of length n.

Trade-offs: While more concise, list comprehensions create a temporary list, resulting in O(n) space complexity whereas the provided Python solution has O(1) space. It is negligibly slower or faster due to internal Python optimizations.

✨

Key Insights

1Insight 1: The core calculation involves determining the reversed alphabet index for each character. This can be efficiently done using the ASCII values of the characters.
2Insight 2: The problem requires maintaining the 1-based index of each character within the string during iteration.
3Insight 3: The problem constraints (lowercase English letters, string length limit) are small enough that integer overflow and very large computations are not concerns.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty string. The provided solutions work correctly for an empty string because the loop will not execute, and the initial value of `reverse_degree` (which is 0) will be returned.
Edge case 2: String with a single character. The solution correctly calculates the reverse degree for a single-character string by calculating its reverse alphabet index and multiplying it by 1.
Edge case 3: String with all the same characters, e.g., 'aaaa'. The algorithm works as expected because it considers each character and its position.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting that the string index is 1-based as specified in the problem, leading to incorrect calculations. Remember to add 1 to the index `i` when multiplying it with the reversed alphabet index.

✗Common mistake 2: Incorrectly calculating the reversed alphabet index. It is crucial to subtract the ASCII value of 'a' from the ASCII value of the character and then account for the reversed order. Many solutions might simply subtract the character's ASCII value from 26 without proper adjustment.

✗Common mistake 3: Integer overflow. Though unlikely given constraints, for larger constraints always consider the upper bound and use appropriate data types (e.g., long or long long in Java/C++).

✅

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: Text analysis and scoring. The concept of assigning weights to characters based on their position in the alphabet or their frequency can be applied in sentiment analysis or keyword extraction to determine the overall significance of words in a document.

🏦Real-world scenario 2: Simple encoding/decoding schemes. While this specific reverse degree calculation is not a robust encryption method, the principle of mapping characters to numerical values and performing calculations can be used as a basic form of data encoding for specific purposes.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 18, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

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 "Reverse Degree of a String"?

For this easy-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?