Percentage of Letter in String - Practice Coding Problems

Percentage of Letter in String - Problem

String Easy

You're analyzing text data and need to calculate character frequencies! Given a string s and a specific character letter, determine what percentage of the string is made up of that character.

Your task is to return the percentage of characters in s that equal letter, rounded down to the nearest whole percent.

Example: In the string "programming" with letter 'g', there are 2 'g's out of 11 total characters. That's 18.18%, which rounds down to 18%.

Input & Output

example_1.py — Basic Case

$ Input: s = "foobar", letter = "o"

› Output: 33

💡 Note: The letter 'o' appears 2 times in "foobar" which has 6 characters. (2/6) * 100 = 33.33%, rounded down to 33%.

example_2.py — No Matches

$ Input: s = "jjjj", letter = "k"

› Output: 0

💡 Note: The letter 'k' does not appear in "jjjj", so the percentage is 0%.

example_3.py — All Matches

$ Input: s = "aaaa", letter = "a"

› Output: 100

💡 Note: All 4 characters are 'a', so the percentage is (4/4) * 100 = 100%.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters
letter is a lowercase English letter

Visualization

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Understanding the Visualization

Scan Survey

Go through each response (character) in the dataset (string)

Count Target

Keep track of how many chose our specific option (target letter)

Calculate Rate

Divide count by total responses and multiply by 100 for percentage

Floor Result

Round down to nearest whole percent as required

Key Takeaway

🎯 Key Insight: This problem only requires counting one specific character and applying a simple percentage formula. The integer division automatically handles the floor operation, making the solution both elegant and efficient.

Asked in

a Amazon 15 G Google 12 ⊞ Microsoft 8 f Meta 5

The optimal solution uses a single pass through the string to count occurrences of the target letter. We then calculate the percentage using integer division to automatically floor the result. Time complexity is O(n) and space complexity is O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass Solution (Optimal)	O(n)	O(1)	Count target character frequency in one pass through the string
Brute Force (Nested Character Matching)	O(n)	O(1)	For each character, check against target letter multiple times

Single Pass Solution (Optimal) — Algorithm Steps

Initialize counter for target letter
Iterate through string once
Increment counter when target letter is found
Calculate percentage using integer division for floor effect
Return the result

Visualization

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Step-by-Step Walkthrough

Start Scan

Begin with count = 0, scan left to right

Count Matches

Increment counter when target letter found

Complete Pass

Finish scanning entire string

Calculate Result

Apply percentage formula with floor division

Code -

solution.c — C

int percentageLetter(char* s, char letter) {
    int count = 0;
    int length = strlen(s);
    
    // Single pass counting
    for (int i = 0; i < length; i++) {
        if (s[i] == letter) {
            count++;
        }
    }
    
    // Integer division automatically floors
    return (count * 100) / length;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable

✓ Linear Space

28.4K Views

Medium Frequency

~5 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass Solution (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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