# Javascript Interview Questions(Part-4) – User friendly Tech help

JavaScript Coding interview questions with solutions:-
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nQuestion1:-n

You would like to set a password for a bank account. However, there are three restrictions on the format of the password:
it has to contain only alphanumerical characters (a−z, A−Z, 0−9);there should be an even number of letters;there should be an odd number of digits.
You are given a string S consisting of N characters. String S can be divided into words by splitting it at, and removing, the spaces. The goal is to choose the longest word that is a valid password. You can assume that if there are K spaces in string S then there are exactly K + 1 words.
For example, given "test 5 a0A pass007 ?xy1", there are five words and three of them are valid passwords: "5", "a0A" and "pass007". Thus the longest password is "pass007" and its length is 7. Note that neither "test" nor "?xy1" is a valid password, because "?" is not an alphanumerical character and "test" contains an even number of digits (zero).
Write a function:
int solution(char *S);
that, given a non-empty string S consisting of N characters, returns the length of the longest word from the string that is a valid password. If there is no such word, your function should return −1.
For example, given S = "test 5 a0A pass007 ?xy1", your function should return 7, as explained above.Assume that:N is an integer within the range [1..200];string S consists only of printable ASCII characters and spaces.In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

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nSolution1:-

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function solution(S) {
// write your code in JavaScript (Node.js 6.4.0)
var pwds = S.split(" ");
var pwdLen = -1;
pwds.forEach(function(pwd){
if(pwd.match(/^[a-zA-Z0-9]*\$/)){
if(pwd.length%2!=1)
return;
var chrCnt = pwd.match(/[a-zA-Z]/g || []).length;
if(chrCnt%2!=0)
return;
if (pwd.length > pwdLen)
pwdLen = pwd.length; } });
return pwdLen;}

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Question2:-
nGiven four integers, display the maximum time possible in 24 hour format HH:MM. For example, if you are give A = 1, B = 9, C = 9, D = 2 then output should be 19:29. Max time can be 23:59 and min time can be 00:00.

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If it is not possible to construct 24 hour time then return error. For example, given A = 1, B = 9, C = 7, D = 9 an error should be returned since minimum time represented by these integers is 17:99 which is “NOT POSSIBLE”?

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Solution2:-

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function solution(A, B, C, D) {
var arr = [];
var digit = [];
var timeString = "";
//create array from given numbers arr.push(A); arr.push(B); arr.push(C); arr.push(D); digit[0] = findMax(arr,2); digit[1] = digit[0]==2?findMax(arr,3):findMax(arr,9); digit[2] = findMax(arr,5); digit[3] = findMax(arr, 9);
//final number
if(digit[0] ==-1 ||digit[1] ==-1||digit[2] ==-1||digit[3] ==-1){
return "NOT POSSIBLE"; }
timeString = digit[0]+""+digit[1]+":"+digit[2]+""+digit[3];
return timeString;}
//finding the less than equal number and return it
function findMax(arr, find){
if(arr.length!=4) {
return -1; }
var numToFind = -1;
var indexToRemove = -1;
//iterate arrary
for(var i = 0; i < arr.length;i++){
if(arr[i] numToFind) { numToFind = arr[i]; indexToRemove =i; } } }
if (indexToRemove == -1)
return -1; arr[indexToRemove] = -1;
return numToFind; }

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Question3:-Given DOM tree I need to find the maximum depth of the nested ul/ol tags.
nExample:-

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• Item:
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1. Point:
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• elem1
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2. n

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• elem2
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• simple list1
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The depth would be 3

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Solution3:-

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function solution() {
var len, max_depth=0;
\$('li:not(:has(ol)):not(:has(ul))').each(function(){
len = \$(this).parents('ul,ol').length;
if(len > max_depth) max_depth =len; });
//console.log(max_depth);
return max_depth;}

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