trjhtr

I have a solution to the following problem that I'd like to know if there's a way I could decrease runtime complexity for.

The directions are: reorder the letters of the first string to match the order of the letters in the second string. If a letter in the first string is not present in the second string, add it to the end of the reordering.

Example input: 'shouts', 'soup'
Example output: 'ssouht'

My code is as follows:

const reOrderLetters = (string1,string2) => 
 let tailString = '';
 let newString = '';
 const mapped = ;

 for (let i=0;i<string1.length;i++) 
 let char = string1[i];
 if (string2.indexOf(char) !== -1) 
 if (!mapped[char]) 
 mapped[char] = char;
 else 
 mapped[char] = mapped[char] += char;
 
 else 
 tailString += char;
 
 
 for (let i=0;i<string2.length;i++) 
 let char = string2[i];
 if (mapped[char]) 
 if (mapped[char].length > 1 && string2.slice(i+1).indexOf(char) !== -1) 
 newString += mapped[char].slice(mapped[char].length-1);
 mapped[char] = mapped[char].slice(0,mapped[char].length-1);
 else 
 newString += mapped[char];
 
 
 
 return newString += tailString;

Your code is pretty good, except for using indexOf(char), as that can be quite expensive: O(string2.length). Here is how I would do it to avoid using that. Assuming hashmap access is O(1) and n being the max length between both strings, my implementation runs in O(n)

function characterCount (str) 
 var chars = ;
 
 for (var i = 0; i < str.length; i ++) 
 chars[str[i]] = (chars[str[i]] 
 
 return chars;




function reOrderLetters (str1, str2) 
 var count1 = characterCount(str1); // O(n)
 var count2 = characterCount(str2); // O(n)
 var result = ;
 var seenCount = 
 
 for (var i = 0; i < str2.length; i++) 0) + 1; 
 
 if (count1.hasOwnProperty(char)) 
 var remaining = count1[char] % count2[char];
 var toRepeat = Math.floor(count1[char] / count2[char]);
 	
 if (seenCount[char] <= remaining) 
 toRepeat = Math.ceil(count1[char] / count2[char]); 
 
 
 result.push(char.repeat(toRepeat)); 
 	
 
 
 for (var i = 0; i < str1.length; i++) 
 var char = str1[i];
 
 if (count2.hasOwnProperty(char) == false) 
 result.push(char.repeat(count1[char]));
 count2[char] = true;
 
 
 
 return result.join("");



console.log(reOrderLetters("shouts", "soups") == "sousht")
console.log(reOrderLetters("aa", "aaaa") == "aa")
console.log(reOrderLetters("aabbbba", "abab") == "aabbabb")