What is the time complexity of Number of Changing Keys?

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

What data structures are used to solve Number of Changing Keys?

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

What are the key insights for solving Number of Changing Keys?

Ignoring case: The problem explicitly states that case does not matter when determining key changes. This simplifies the solution by converting all characters to lowercase or uppercase before comparison. Iterative approach: The solution can be efficiently implemented using a single iteration through the input string, comparing each character to the previous one.

Which companies ask LeetCode 3019. Number of Changing Keys in interviews?

This problem is frequently asked at Autodesk.

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Number of Changing Keys - LeetCode 3019 Solution

Number of Changing Keys - Complete Solution Guide

Number of Changing Keys is LeetCode problem 3019, 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

You are given a 0-indexed string s typed by a user. Changing a key is defined as using a key different from the last used key. For example, s = "ab" has a change of a key while s = "bBBb" does not have any. Return the number of times the user had to change the key. Note: Modifiers like shift or caps lock won't be counted in changing the key that is if a user typed the letter 'a' and then the letter 'A' then it will not be considered as a changing of key. Example 1: Input: s = "aAbBcC" Output: 2

Detailed Explanation

The problem asks to count the number of times a user changes keys while typing a string. A key change is defined as pressing a different letter than the last one pressed, ignoring case. For example, pressing 'a' then 'A' is not a key change, but pressing 'a' then 'b' is. The input is a string `s` containing uppercase and lowercase English letters, and the output is the number of key changes.

Solution Approach

The provided solutions use an iterative approach. They first handle the base case of empty or single-character strings. Then, they iterate through the string, converting each character to lowercase (or uppercase). Each character is compared to the previous character. If they are different, a key change is counted. The final count of key changes is returned.

Step-by-Step Algorithm

Handle the base case: If the string length is 0 or 1, return 0 (no key changes).
Initialize variables: Store the count of key changes (initially 0) and the previous character (the first character of the string converted to lowercase).
Iterate through the string: Starting from the second character, iterate through the remaining characters.
Compare current and previous characters: Convert the current character to lowercase and compare it to the previous character.
Increment count: If the characters are different, increment the key change count and update the previous character.
Return count: After iterating through the entire string, return the final key change count.

Key Insights

Ignoring case: The problem explicitly states that case does not matter when determining key changes. This simplifies the solution by converting all characters to lowercase or uppercase before comparison.
Iterative approach: The solution can be efficiently implemented using a single iteration through the input string, comparing each character to the previous one.
Base case handling: Empty or single-character strings should be handled as special cases to avoid index out-of-bounds errors or incorrect counts.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Autodesk.

Number of Changing Keys | LeetCode 3019

Easy

String

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

Solution Code

Available in:

class Solution {
    public int countKeyChanges(String s) {
        int count = 0;
        if (s.length() == 0) {
            return 0;
        }
        char prev = Character.toLowerCase(s.charAt(0));
        for (int i = 1; i < s.length(); i++) {
            char curr = Character.toLowerCase(s.charAt(i));
            if (curr != prev) {
                count++;
                prev = curr;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 3019: Number of Changing Keys - Detailed explanation and solution approaches.

Topics

String

Companies

Autodesk

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3019

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
Handle the base case: If the string length is 0 or 1, return 0 (no key changes).
2
Initialize variables: Store the count of key changes (initially 0) and the previous character (the first character of the string converted to lowercase).
3
Iterate through the string: Starting from the second character, iterate through the remaining characters.
4
Compare current and previous characters: Convert the current character to lowercase and compare it to the previous character.
5
Increment count: If the characters are different, increment the key change count and update the previous character.
6
Return count: After iterating through the entire string, return the final key change count.

💡

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 algorithm iterates through the string once, performing constant-time operations (character comparison and lowercase conversion) in each iteration. No additional data structures proportional to the input size are used, resulting in constant space complexity.

📖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

A recursive solution could also be implemented, but it would be less efficient than the iterative approach due to the overhead of function calls. The base cases would be the same, and the recursive step would compare the current character to the result of the recursive call on the rest of the string.

Time: O(n)Space: O(n) due to recursive call stack.

Trade-offs: The iterative approach is preferred for its efficiency in both time and space complexity. The recursive approach is less efficient and less readable for this specific problem.

✨

Key Insights

1Ignoring case: The problem explicitly states that case does not matter when determining key changes. This simplifies the solution by converting all characters to lowercase or uppercase before comparison.
2Iterative approach: The solution can be efficiently implemented using a single iteration through the input string, comparing each character to the previous one.
3Base case handling: Empty or single-character strings should be handled as special cases to avoid index out-of-bounds errors or incorrect counts.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Empty string: The code correctly handles an empty string by returning 0.
Single-character string: The code correctly handles a single-character string by returning 0.
All same characters (case-insensitive): The code correctly identifies no key changes if all characters are the same (ignoring case).

⚠️

Common Mistakes to Avoid

✗Forgetting base cases: Failing to handle empty or single-character strings can lead to errors.

✗Incorrect case handling: Not converting characters to the same case before comparison will lead to incorrect results.

✗Off-by-one errors: Incorrect loop boundaries or indexing could lead to missing key changes or extra counts.

✅

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?

💼Analyzing typing patterns: This algorithm could be used to analyze typing patterns of users, identifying potential areas for improvement in typing efficiency.

🏦Data compression: While not a direct application, the concept of identifying repetitive sequences (or lack thereof) is relevant in data compression algorithms.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: October 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

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 Changing Keys"?

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

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