Percentage of Letter in String

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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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

✓ Built-in Count Method

⏱️ Time: O(n) Space: O(1)

Instead of manually looping, use the built-in count method available in most languages to count occurrences of the target character. This results in cleaner, more readable code while maintaining the same time complexity.

Single Pass Count

⏱️ Time: O(n) Space: O(1)

Loop through each character in the string, count how many match the target letter, then calculate percentage by dividing count by string length and multiplying by 100. Finally, floor the result to get whole percent.

Built-in Count Method — Algorithm Steps

Use built-in count method to count letter occurrences
Calculate percentage: (count / length) * 100
Floor the result to nearest whole percent

Visualization

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Step-by-Step Walkthrough

Built-in Count

Use language's optimized count method

Direct Calculation

Apply percentage formula immediately

Floor Result

Round down to 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];
    char* ptr = s;
    while ((ptr = strchr(ptr, target)) != NULL) {
        count++;
        ptr++;
    }
    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)

Built-in count method still needs to scan entire string

✓ Linear Growth

Space Complexity

O(1)

No extra space needed beyond variables

✓ Linear Space

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Related Problems

Common Approaches

Built-in Count Method — Algorithm Steps

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Code -

Time & Space Complexity

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