What is the time complexity of Percentage of Letter in String?

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

What data structures are used to solve Percentage of Letter in String?

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

What are the key insights for solving Percentage of Letter in String?

Insight 1: The solution requires iterating through the string to count the occurrences of the target character. Insight 2: Integer division is crucial for rounding down the percentage to the nearest whole number.

Which companies ask LeetCode 2278. Percentage of Letter in String in interviews?

This problem is frequently asked at American Express.

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Percentage of Letter in String - LeetCode 2278 Solution

Q: How to solve Percentage of Letter in String?

The provided solutions use a straightforward iterative approach. They iterate through each character of the input string `s`. If a character matches the input `letter`, a counter is incremented. After iterating through the entire string, the percentage is calculated by multiplying the count by 100 and dividing by the string's length. Integer division is used to ensure the result is rounded down. The Python and C solutions explicitly use integer division (`//` in Python, `/` in C after casting to integer), while Java and C++ rely on implicit type conversion and casting to int to achieve the same result.

Percentage of Letter in String - Complete Solution Guide

Percentage of Letter in String is LeetCode problem 2278, 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 and a character letter , return the percentage of characters in s that equal letter rounded down to the nearest whole percent. Example 1: Input: s = "foobar", letter = "o" Output: 33 Explanation: The percentage of characters in s that equal the letter 'o' is 2 / 6 * 100% = 33% when rounded down, so we return 33. Example 2: Input: s = "jjjj", letter = "k" Output: 0 Explanation: The percentage of characters in s that equal the letter 'k' is 0%, so we return 0. Constraints: 1 <= s.

Detailed Explanation

The problem asks you to calculate the percentage of times a specific character appears in a given string. The input consists of a string `s` (containing lowercase English letters) and a character `letter` (also a lowercase English letter). The output should be the percentage of characters in `s` that are equal to `letter`, rounded down to the nearest whole number. For example, if `s = "foobar"` and `letter = "o"`, the output should be 33 because 'o' appears twice in a string of length 6, and (2/6) * 100 = 33.33..., which rounds down to 33.

Solution Approach

The provided solutions use a straightforward iterative approach. They iterate through each character of the input string `s`. If a character matches the input `letter`, a counter is incremented. After iterating through the entire string, the percentage is calculated by multiplying the count by 100 and dividing by the string's length. Integer division is used to ensure the result is rounded down. The Python and C solutions explicitly use integer division (`//` in Python, `/` in C after casting to integer), while Java and C++ rely on implicit type conversion and casting to int to achieve the same result.

Step-by-Step Algorithm

Step 1: Initialize a counter variable `count` to 0.
Step 2: Iterate through each character `char` in the input string `s`.
Step 3: If `char` is equal to the input `letter`, increment `count`.
Step 4: After the loop, calculate the percentage: `(count * 100) / len(s)` (Python and C) or `(int)(((double)count / s.length()) * 100)` (Java and C++).
Step 5: Return the calculated percentage.

Key Insights

Insight 1: The solution requires iterating through the string to count the occurrences of the target character.
Insight 2: Integer division is crucial for rounding down the percentage to the nearest whole number.
Insight 3: Handling the edge case of an empty string or a string where the target character doesn't appear is important to avoid division by zero errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: American Express.

Percentage of Letter in String | LeetCode 2278

Easy

String

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

Solution Code

Available in:

class Solution {
    public int percentageLetter(String s, char letter) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == letter) {
                count++;
            }
        }
        return (int) (((double) count / s.length()) * 100);
    }
}

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Problem Info

Description

Problem 2278: Percentage of Letter in String - Detailed explanation and solution approaches.

Topics

String

Companies

American Express

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2278

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: Initialize a counter variable `count` to 0.
2
Step 2: Iterate through each character `char` in the input string `s`.
3
Step 3: If `char` is equal to the input `letter`, increment `count`.
4
Step 4: After the loop, calculate the percentage: `(count * 100) / len(s)` (Python and C) or `(int)(((double)count / s.length()) * 100)` (Java and C++).
5
Step 5: Return the calculated percentage.

💡

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 iterate through the string once. The space complexity is constant because we only use a few integer variables to store the count and the length of the string. The space used does 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 `count()` method

In Python, the `count()` method can be used to directly count occurrences of a character in a string. This can make the code slightly more concise.

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

Trade-offs: The `count()` method might be slightly faster in Python due to potential optimizations within the built-in function, but the difference is likely negligible for most cases.

✨

Key Insights

1Insight 1: The solution requires iterating through the string to count the occurrences of the target character.
2Insight 2: Integer division is crucial for rounding down the percentage to the nearest whole number.
3Insight 3: Handling the edge case of an empty string or a string where the target character doesn't appear is important to avoid division by zero errors.

📚

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 solution handles this correctly; if the string is empty, the count will be 0, and the percentage will also be 0, avoiding division by zero.
Edge case 2: The target letter is not present in the string. The solution handles this correctly by returning 0. The count remains 0, and the final calculation results in 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of an empty string, leading to a division-by-zero error.

✗Common mistake 2: Using floating-point division and not explicitly rounding down the result, potentially leading to incorrect output.

✅

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: Analyzing text data to determine the frequency of specific characters, for example in natural language processing or text analytics.

🏦Real-world scenario 2: Monitoring logs or system output to determine the percentage of error messages or events of a certain type.

💡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: September 27, 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 13

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

@TechLeadJul 6

❓

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 "Percentage of Letter in 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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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