Find missing alphabets to make a string panagram

An algorithm to find the missing alphabets to make a string Panagram.

We will find all the missing alphabets that will make the string panagram and return the missing characters in alphabetical order.

Example

Input:
"Hi from learnersbucket"
"Learn just don't study"

Output:
"dghjpqvwxyz"
"bcfghiklmpqvwxz"

A simple Solution O(n) to find the missing alphabets to make string panagram.

Implementation

• We will create an array of 26 numbers and keep the track of present alphabets by marking them true.
• To convert the alphabets to numbers will calculate the ASCII difference of the number from the small ‘a’.
• Once we have marked the present alphabets we can easily print the missing alphabets in order.
• Note we are going to ignore the case here.
• Everything will be written in ES6.

let missingAlphabets = (str) => {
  //create a new array and initilize it with 0
  let arr = new Array(26);
  arr.fill(0, 0, 25);
  
  //convert the string to lowercase
  str.toLowerCase();
 
  //Mark the present string as true
  for(let i = 0; i < str.length; i++){
     if (str[i] >= 'a' && str[i] <= 'z') {
          arr[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] = true; 
     }
  }
  
 //Create the string of the missing alphabets
  let missing = '';
  for(let i = 0; i < arr.length; i++){
     if(!arr[i]){
       missing += String.fromCharCode(97 + i);
     }
  }
  
 return missing;
}

Input:
console.log(missingAlphabets("Learn just don't study"));
console.log(missingAlphabets("Hi from learnersbucket"));

Output:
"bcfghiklmpqvwxz"
"dghjpqvwxyz"

Time Complexity: O(n).
Space Complexity: O(1).

Time and Space complexity

• We are iterating through each character in the given string and marking its corresponding number as true in the array that will take O(n), Then we are looping through the array and converting the false value to form a string which will take O(26), so Time complexity O(n + 26) = O(n).
• As we are using constant space (An array of 26 length no matter how big is the given string), so Space complexity is O(1).