January 26, 2014
More Questions CSS Interview Questions, HTML Interview Questions
if you are little more comfortable or claim to be comfortable with javascript, these questions would not be enough for you. more coming
Question: How would you verify a prime number?
Answer: a prime number is only divisible by itself and 1. So, i will run a while loop and increase by 1. (look at the code example. If you dont get it. this is not your cake. do learn javaScript basics and come back.)
function isPrime(n){
var divisor = 2;
while (n > divisor){
if(n % divisor == 0){
return false;
}
else
divisor++;
}
return true;
}
> isPrime(137);
= true
> isPrime(237);
= false
Interviewer: Can you make this better?
You: yes. the divisor are increased 1 at a time. after 3 i can increase by 2. if a number is divisible by any even number, it will be divisible by 2.
Extra: if you dont have to increase the divisor up to the number. you can stop much earlier. let me explain it in the following steps (just seat back and read as many times as needed)
Don't worry! if you didnt get it. just leave it. If you are not applying for a Research post, you would be alright.
function isPrime(n)
{
var divisor = 3,
limit = Math.sqrt(n);
//check simple cases
if (n == 2 || n == 3)
return true;
if (n % 2 == 0)
return false;
while (divisor <= limit)
{
if (n % divisor == 0)
return false;
else
divisor += 2;
}
return true;
}
> isPrime(137);
= true
> isPrime(237);
= false
Question: How could you find all prime factors of a number?
Answer: Run a while loop. start dividing by two and if not divisible increase divider until u r done.
function primeFactors(n){
var factors = [],
divisor = 2;
while(n>2){
if(n % divisor == 0){
factors.push(divisor);
n= n/ divisor;
}
else{
divisor++;
}
}
return factors;
}
> primeFactors(69);
= [3, 23]
Interviewer:What is the run time complexity? can you make this better
You: this is O(n). you can increase divisor by 2 form divisor = 3. Because, if a number is divisible by any even number it would divisible by 2. Hence, you dont need to divide by even numbers. Besides, you will not have a factor bigger than half of its value. if you want make it complex use tough concept in no. 1 (try-2. if u get it)
Question: How do get nth Fibonacci number?
Answer: I create an array and start from iterate through.
Fibonacci series is one of the most popular interview question for beginners. so, you have to learn this one.
function fibonacci(n){
var fibo = [0, 1];
if (n <= 2) return 1;
for (var i = 2; i <=n; i++ ){
fibo[i] = fibo[i-1]+fibo[i-2];
}
return fibo[n];
}
> fibonacci(12);
= 144
Interviewer: What is the run time complexity?
you: O(n)
Interviewer: can you make it recursive?
function fibonacci(n){
if(n<=1)
return n;
else
return fibonacci(n-1) + fibonacci (n-2);
}
> fibonacci(12);
= 144
Interviewer: What is the run time complexity?
You: O(2n). detail about complexity
Question: How would you find the greatest common divisor of two numbers?
function greatestCommonDivisor(a, b){
var divisor = 2,
greatestDivisor = 1;
//if u pass a -ve number this will not work. fix it dude!!
if (a < 2 || b < 2)
return 1;
while(a >= divisor && b >= divisor){
if(a %divisor == 0 && b% divisor ==0){
greatestDivisor = divisor;
}
divisor++;
}
return greatestDivisor;
}
> greatestCommonDivisor(14, 21);
=7
> greatestCommonDivisor(69, 169);
= 1
Sorry. can't explain it. As i myself dont understand it 80% of the times. my algorithm analysis instructor told about this and stole if from class note (i am a good student, btw!)
function greatestCommonDivisor(a, b){
if(b == 0)
return a;
else
return greatestCommonDivisor(b, a%b);
}
Note: use your brain to understand it.
Question: How would you remove duplicate members from an array?
Answer: will start a while looping and keep an object/ associated array. If i find an element for the first time i will set its value as true (that will tell me element added once.). if i find a element in the exists object, i will not insert it into the return array.
function removeDuplicate(arr){
var exists ={},
outArr = [],
elm;
for(var i =0; i<arr.length; i++){
elm = arr[i];
if(!exists[elm]){
outArr.push(elm);
exists[elm] = true;
}
}
return outArr;
}
> removeDuplicate([1,3,3,3,1,5,6,7,8,1]);
= [1, 3, 5, 6, 7, 8]
Question: How would you merge two sorted array?
Answer: I will keep a pointer for each array and (read the code. and be careful about this one.)
function mergeSortedArray(a, b){
var merged = [],
aElm = a[0],
bElm = b[0],
i = 1,
j = 1;
if(a.length ==0)
return b;
if(b.length ==0)
return a;
/*
if aElm or bElm exists we will insert to merged array
(will go inside while loop)
to insert: aElm exists and bElm doesn't exists
or both exists and aElm < bElm
this is the critical part of the example
*/
while(aElm || bElm){
if((aElm && !bElm) || aElm < bElm){
merged.push(aElm);
aElm = a[i++];
}
else {
merged.push(bElm);
bElm = b[j++];
}
}
return merged;
}
> mergeSortedArray([2,5,6,9], [1,2,3,29]);
= [1, 2, 2, 3, 5, 6, 9, 29]
Question: How would you swap two numbers without using a temporary variable?
Answer: Waste time about thinking it. though u know the answer, act like you are thinking and take your time to answer this one.
function swapNumb(a, b){
console.log('before swap: ','a: ', a, 'b: ', b);
b = b -a;
a = a+ b;
b = a-b;
console.log('after swap: ','a: ', a, 'b: ', b);
}
> swapNumb(2, 3);
= before swap: a: 2 b: 3
= after swap: a: 3 b: 2
bit manipulation: Sorry, i can't explain this to you. Kinjal Dave suggested logical conjunction to understand it. Waste 30 min there at your own risk.
function swapNumb(a, b){
console.log("a: " + a + " and b: " + b);
a = a ^ b;
b = a ^ b;
a = a ^ b;
console.log("a: " + a + " and b: " + b);
}
> swapNumb(2, 3);
= a: 2 and b: 3
= a: 3 and b: 2
Question: How would you reverse a string in JavaScript?
Answer: I can loop through the string and concatenate letters to a new string
function reverse(str){
var rtnStr = '';
for(var i = str.length-1; i>=0;i--){
rtnStr +=str[i];
}
return rtnStr;
}
> reverse('you are a nice dude');
= "edud ecin a era uoy"
Interviewer: You know concatenation performed well in modern browsers but becomes slow in older browsers like IE8. Is there any different way, you can reverse a string?
Answer: sure. i can use an array and also add some checking. if string is null or other than string this will fail. let me do some type check as well. Using this array is like using string buffer in some server side languages.
function reverse(str){
var rtnStr = [];
if(!str || typeof str != 'string' || str.length < 2 ) return str;
for(var i = str.length-1; i>=0;i--){
rtnStr.push(str[i]);
}
return rtnStr.join('');
}
Interviewer: What is the run time complexity?
You: O(n)
Interviewer: Can you make this better?
You: I can loop through half of the index and it will save little bit. (this is kind of useless, might not impress interviewer)
function reverse(str) {
str = str.split('');
var len = str.length,
halfIndex = Math.floor(len / 2) - 1,
revStr;
for (var i = 0; i <= halfIndex; i++) {
revStr = str[len - i - 1];
str[len - i - 1] = str[i];
str[i] = revStr;
}
return str.join('');
}
Interviewer: That works, but can u do it in a recursive way?
You: sure.
function reverse (str) {
if (str === "") {
return "";
} else {
return reverse(str.substr(1)) + str.charAt(0);
}
}
Interviewer: Can you use any build in method to make it little cleaner?
You: yes.
function reverse(str){
if(!str || str.length <2) return str;
return str.split('').reverse().join('');
}
Question: Can you make reverse function as string extension?
Answer: I need to add this function to the String.prototype and instead of using str as parameter, i need to use this
String.prototype.reverse = function (){
if(!this || this.length <2) return this;
return this.split('').reverse().join('');
}
> 'abc'.reverse();
= 'cba'
Question: How would you reverse words in a sentence?
Answer: You have to check for white space and walk through the string. Ask is there could be multiple whitespace.
//have a tailing white space
//fix this later
//now i m sleepy
function reverseWords(str){
var rev = [],
wordLen = 0;
for(var i = str.length-1; i>=0; i--){
if(str[i]==' ' || i==0){
rev.push(str.substr(i,wordLen+1));
wordLen = 0;
}
else
wordLen++;
}
return rev.join(' ');
}
A quick solution with build in methods:
function reverseWords(str){
return str.split(' ').reverse();
}
Question: If you have a string like "I am the good boy". How can you generate "I ma eht doog yob"? Please note that the words are in place but reverse.
Answer: To do this, i have to do both string reverse and word reverse.
function reverseInPlace(str){
return str.split(' ').reverse().join(' ').split('').reverse().join('');
}
> reverseInPlace('I am the good boy');
= "I ma eht doog yob"
Interviewer: ok. fine. can you do it without using build in reverse function?
you: (you mumbled): what the heck!!
//sum two methods.
//you can simply split words by ' '
//and for each words, call reverse function
//put reverse in a separate function
//if u cant do this,
//have a glass of water, and sleep
Question: How could you find the first non repeating char in a string?
Answer: You must ask follow up questions.
Clarification: Is it case sensitive?
Interviewer: interviewer might say no.
Clarification: is it very long string or couple hundred?
Interviewer: Why does that matter
you: for example, if it is a very long string say a million characters and i want to check whether 26 English characters are repeating. I might check whether all characters are duplicated in every 200 letters (for example) other than looping through the whole string. this will save computation time.
Interviewer: For simplicity, you string is "the quick brown fox jumps then quickly blow air"
Clarification: (silly question) ascii or unicode.
function firstNonRepeatChar(str){
var len = str.length,
char,
charCount = {};
for(var i =0; i<len; i++){
char = str[i];
if(charCount[char]){
charCount[char]++;
}
else
charCount[char] = 1;
}
for (var j in charCount){
if (charCount[j]==1)
return j;
}
}
>firstNonRepeatChar('the quick brown fox jumps then quickly blow air');
= "f"
this has one problem. It is not guaranteed that for in loop will be executed in order.
Question: How will you remove duplicate characters from a sting?
You: This is very similar to first non repeating char. You will asks similar question. Is it case sensitive.
If interviewer says, this is case sensitive then life become easier you. If he says no. you can either use string.toLowercase() to make whole string lower. he might not like it. because return string will not posses the same case. So
function removeDuplicateChar(str){
var len = str.length,
char,
charCount = {},
newStr = [];
for(var i =0; i<len; i++){
char = str[i];
if(charCount[char]){
charCount[char]++;
}
else
charCount[char] = 1;
}
for (var j in charCount){
if (charCount[j]==1)
newStr.push(j);
}
return newStr.join('');
}
> removeDuplicateChar('Learn more javascript dude');
= "Lnmojvsciptu"
Note:this has one problem. It is not guaranteed that for in loop will be executed in order.
For case insensitive: when u r setting property of charCount or increase counter, just make the char.toLowerCase(). or you can do something fancy with charCode (if you can deal with it.)
Question: How will you verify a word as palindrome?
Answer: if you reverse a word and it becomes same as the previous word, it is called palindrome.
function isPalindrome(str){
var i, len = str.length;
for(i =0; i<len/2; i++){
if (str[i]!== str[len -1 -i])
return false;
}
return true;
}
> isPalindrome('madam')
= true
> isPalindrome('toyota')
= false
or you can use build in method
function checkPalindrom(str) {
return str == str.split('').reverse().join('');
}
Similar: Find whether a string contains a contiguous palindromic substring in O(n) time. Can you solve the problem in O(1) time?
Question:If you have a function that generate random number between 1 to 5 how could u generate random number 1 to 7 by using that function?
Answer: Very simple. think of some basic arithmetic and you will get it.
function rand5(){
return 1 + Math.random() * 4;
}
function rand7(){
return 5 + rand5() / 5 * 2;
}
Question: from a unsorted array of numbers 1 to 100 excluding one number, how will you find that number.
Explanation: You have an array of numbers 1 to 100 in an array. Only one number is missing in the array. The array is unsorted. Find the missing number.
Answer: You have to act like you are thinking a lot. and then talk about the sum of a linear series of n numbers = n*(n+1)/2
function missingNumber(arr){
var n = arr.length+1,
sum = 0,
expectedSum = n* (n+1)/2;
for(var i = 0, len = arr.length; i < len; i++){
sum += arr[i];
}
return expectedSum - sum;
}
> missingNumber([5, 2, 6, 1, 3]);
= 4
Note: this one will give u missing one number in any array of length
Question: From a unsorted array, check whether there are any two numbers that will sum up to a given number?
Answer: Simplest thing in the world. double loop
function sumFinder(arr, sum){
var len = arr.length;
for(var i =0; i<len-1; i++){
for(var j = i+1;j<len; j++){
if (arr[i] + arr[j] == sum)
return true;
}
}
return false;
}
> sumFinder([6,4,3,2,1,7], 9);
= true
> sumFinder([6,4,3,2,1,7], 2);
= false
Interviewer: What is the time complexity of this function
You: O(n2)
Interviewer: Can you make this better
You: Let me see. I can have an object where i will store the difference of sum and element. And then when i get to a new element and if i find the difference is the object, then i have a pair that sums up to the desired sum.
function sumFinder(arr, sum){
var differ = {},
len = arr.length,
substract;
for(var i =0; i<len; i++){
substract = sum - arr[i];
if(differ[substract])
return true;
else
differ[arr[i]] = true;
}
return false;
}
> sumFinder([6,4,3,2,1,7], 9);
= true
> sumFinder([6,4,3,2,1,7], 2);
= false
similar problem: find the maximum difference of two numbers in an array max difference
Even tougherFinding three elements in an array whose sum is closest to an given number
Question: How would you find the largest sum of any two elements?
Answer: this is actually very simple and straight forward. Just find the two largest number and return sum of them
function topSum(arr){
var biggest = arr[0],
second = arr[1],
len = arr.length,
i = 2;
if (len<2) return null;
if (biggest<second){
biggest = arr[1];
second = arr[0];
}
for(; i<len; i++){
if(arr[i] > biggest){
second = biggest;
biggest = arr[i];
}
else if (arr[i]>second){
second = arr[i];
}
}
return biggest + second;
}
Question: Count Total number of zeros from 1 upto n?
Answer: If n = 50. number of 0 would be 11 (0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100). Please note that 100 has two 0. This one looks simple but little tricky
Explanation: So the tick here is. if you have a number 1 to 50 the value is 5. just 50 divided by 10. However, if the value is 100. the value is 11. you will get by 100/10 = 10 and 10/10. Thats how you will get in the more zeros in one number like (100, 200, 1000)
function countZero(n){
var count = 0;
while(n>0){
count += Math.floor(n/10);
n = n/10;
}
return count;
}
> countZero(2014);
= 223
Question: How can you match substring of a sting?
Answer: Will use to pointer (one for string and another for the substring) while iterating the string. And will have another variable to hold the starting index of the initial match.
function subStringFinder(str, subStr){
var idx = 0,
i = 0,
j = 0,
len = str.length,
subLen = subStr.length;
for(; i<len; i++){
if(str[i] == subStr[j])
j++;
else
j = 0;
//check starting point or a match
if(j == 0)
idx = i;
else if (j == subLen)
return idx;
}
return -1;
}
> subStringFinder('abbcdabbbbbck', 'ab')
= 0
> subStringFinder('abbcdabbbbbck', 'bck')
= 9
//doesn't work for this one.
> subStringFinder('abbcdabbbbbck', 'bbbck')
= -1
Question: How can you fix the last problem?
Answer: figure out yourself.
Question: How would you create all permutation of a string?
Answer: This could be a tough one based on you level of knowledge about algorithm.
function permutations(str){
var arr = str.split(''),
len = arr.length,
perms = [],
rest,
picked,
restPerms,
next;
if (len == 0)
return [str];
for (var i=0; i<len; i++)
{
rest = Object.create(arr);
picked = rest.splice(i, 1);
restPerms = permutations(rest.join(''));
for (var j=0, jLen = restPerms.length; j< jLen; j++)
{
next = picked.concat(restPerms[j]);
perms.push(next.join(''));
}
}
return perms;
}
Explanation:
Need more: CSS Interview Questions, HTML Interview Questions
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