# Solving Leetcode Interviews in Seconds with AI: Percentage of Letter in String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2278" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.length <= 100 s consists of lowercase English letters. letter is a lowercase English letter.  

	# Explanation
	*   Iterate through the input string `s` and count the occurrences of the target character `letter`.
*   Calculate the percentage of `letter` in `s` by dividing the count by the length of `s` and multiplying by 100.
*   Return the integer part of the calculated percentage (rounding down).

*   Runtime Complexity: O(n), where n is the length of the string `s`. Storage Complexity: O(1).

	
	# Code
	```python
	def percentageLetter(s: str, letter: str) -> int:
    """
    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.length <= 100
    s consists of lowercase English letters.
    letter is a lowercase English letter.
    """
    count = 0
    for char in s:
        if char == letter:
            count += 1
    return (count * 100) // len(s)
	```