**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

Given a string `s`

and a character `letter`

, return* the percentage of characters in *

`s`

`letter`

**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.

```
class Solution {
public:
int percentageLetter(string s, char letter) {
int count=0;
for(int i=0;i<s.length();i++)
{ if(s[i]==letter)
{
count++;
}
}
return (count*100)/s.length();
}
};
```

```
class Solution:
def percentageLetter(self, s: str, letter: str) -> int:
return (s.count(letter)*100)//len(s)
```

```
class Solution {
public int percentageLetter(String s, char letter) {
int count=0; //count acts as a counter variabe to count the give character "letter" in the string
//Traverse the string s and each time the character int the string is equal to "letter"
//increment the count variable by 1.
for(char ch:s.toCharArray())
{
if(ch==letter)
count++;
}
//If the count is zero, then the percentage is also zero
if(count==0)
return 0;
// to find the total characters in the strings or we can say to find the length of the string
int total=s.length();
// Calculate the total percentage using the standard percentage formula i.e (given*100)/total
int per=(count*100)/total;
return per;
}
}
```

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