Percentage of Letter in String - Practice Coding Problems

Percentage of Letter in String - Problem

String Easy

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.

Input & Output

Example 1 — Basic Case

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

› Output: 33

💡 Note: There are 2 'o's in "foobar" out of 6 characters. Percentage = (2/6) * 100 = 33.33, floored to 33.

Example 2 — All Same Character

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

› Output: 0

💡 Note: There are 0 'k's in "jjjj". Percentage = (0/4) * 100 = 0.

Example 3 — Perfect Match

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

› Output: 100

💡 Note: All 4 characters are 'j'. Percentage = (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

Input

String and target character

Count

Count occurrences of target character

Calculate

Apply percentage formula and floor result

Key Takeaway

🎯 Key Insight: Simple counting with percentage math - floor division gives us the rounded down result directly

Asked in

G Google 15 M Meta 12

The key insight is to count character occurrences and apply simple percentage math with floor operation. Best approach uses built-in count methods for cleaner code. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass Count	O(n)	O(1)	Iterate through string once, count matching characters, calculate percentage
Built-in Count Method	O(n)	O(1)	Use language built-in count method for cleaner code

Single Pass Count — Algorithm Steps

Iterate through each character in string
Count occurrences of target letter
Calculate percentage: (count / length) * 100
Floor the result to nearest whole percent

Visualization

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Step-by-Step Walkthrough

Scan String

Go through each character counting matches

Calculate

Divide count by length and multiply by 100

Floor Result

Round down to nearest whole percent

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s, char* letter) {
    int count = 0;
    int len = strlen(s);
    char target = letter[0];
    for (int i = 0; i < len; i++) {
        if (s[i] == target) {
            count++;
        }
    }
    int percentage = (count * 100) / len;
    return percentage;
}

int main() {
    char s[1000], letter[10];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(letter, sizeof(letter), stdin);
    letter[strcspn(letter, "\n")] = 0;
    int result = solution(s, letter);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using variables for count and calculation

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass Count — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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