Code Wars: Find the smallest (NOT PASSED YET)

https://www.codewars.com/kata/573992c724fc289553000e95/train/javascript


function smallest(n) {
    let arr = (n).toString().split("");
    let index = 0;
    let smallest = arr[0];
    let smallest2 = arr[1];
    let replaceItem;
    let final;
 
    //find the smallest number that is not the first number in array
    for (i = 1; i < arr.length; i++) {
      if(arr[i] < smallest) {
        index = i;
        smallest = arr[i];
      }
    }
    //if above does not go thru, find next largest number after 1st number
    if (index === 0) {
      for (i = 2; i < arr.length; i++) {
        if(arr[i] < smallest2) {
          index = i;
          smallest2 = arr[i];
        }
      }
    }
    //remove the element retreived from the loops above
    replaceItem = arr.splice(index, 1);
 
    //place the removed element to the first number it is less than
    for (i = 0; i < arr.length; i++) {
      if (replaceItem < arr[i]) {
        arr.splice(i, 0, replaceItem);
        final = i;
        break;
      }
    }
 
    return [parseInt(arr.join(""), 10), index, final]    //parsetInt to remove leading zeros from string
}

Comments

Post a Comment

Popular posts from this blog

Code Wars: longest_palindrome (6 kyu)

Link to code challenge Find the length of the longest substring in the given string s that is the same in reverse. As an example, if the input was “I like racecars that go fast”, the substring (racecar) length would be 7. If the length of the input string is 0, the return value must be 0. //Loop thru string and find longest substring that is also a palindrome //Efficiency- start with longest length first (each "outer" loop still looping to shorter //strings but done prior to "inner" loops which have longer strings- can be more efficient?) longestPalindrome=function(s){ if (s === "") return 0; else if (s.length === 1) return 1; let mainPointer = s.length; let subPointer = 0; let largest = 0; while (subPointer = 0) { //refactored- instead of pushing to new array let newString = s.slice(subPointer, mainPointer); //can this be refactored by comparing the two endpoints of the string but longer code? if ( newStr...
Read more

Code Wars: Data Reverse (6 kyu)

Link to code challenge Code efficiency: Loops through string once (from the end) and with one new variable. String splicing- consider its efficiency? Try not to manipulate data when looping for efficiency. function dataReverse(data) { let arr = []; while (data.length) { arr.push(...data.splice(data.length - 8, data.length)); } return arr; }
Read more

Code Wars: Replace With Alphabet Position

https://www.codewars.com/kata/546f922b54af40e1e90001da/train/javascript Welcome. In this kata you are required to, given a string, replace every letter with its position in the alphabet. If anything in the text isn't a letter, ignore it and don't return it. "a" = 1 ,  "b" = 2 , etc. Example alphabet_position( "The sunset sets at twelve o' clock." ) Should return  "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11"  (as a string) function alphabetPosition(text) {   //split string into array of letters only   let sentence = text.toLowerCase().split("");   for (let i = sentence.length - 1; i >= 0; i--) {     if(sentence[i].match(/[a-z]/) === null) {       sentence.splice(i, 1);     }   }   //create an array of alphabets (97 represents lower case "a", 65 would be "A")   let alphaArr = [];   for (let i = 0; i < 26; i++) {     alpha...
Read more