Percentage of Letter in String LeetCode Solution

Problem – Percentage of Letter in String LeetCode Solution

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.

Percentage of Letter in String LeetCode Solution in C++

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();
        
    }
};

Percentage of Letter in String LeetCode Solution in Python

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

Percentage of Letter in String LeetCode Solution in Java

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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